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The vertices of quadrilateral LMNP are L(-1 , 7), M(4,9), N(8, -1), and P(3,-3). Using the distance formula, determine the most precise classification of LMNP.

A quadratic equation in the form [tex]x^2[/tex]+bx+c=0 that has the roots -8±4i is [tex]x^2[/tex] + 16x + 48 = 0

If the roots of a quadratic equation are -8+4i and -8-4i, then the factors of the quadratic equation are (x-(-8+4i))(x-(-8-4i))=0.

Simplifying this expression, we get:

(x+8-4i)(x+8+4i) = 0

Expanding this expression, we get:

[tex]x^2[/tex] + (8-4i+8+4i)x + (8-4i)(8+4i) = 0

Simplifying this expression, we get:

[tex]x^2[/tex] + 16x + ([tex]8^2[/tex] - [tex](4i)^2[/tex]) = 0

[tex]x^2[/tex] + 16x + 48 = 0

Therefore, the quadratic equation in the form [tex]x^2[/tex]+bx+c=0 that has the roots -8±4i is:

[tex]x^2[/tex] + 16x + 48 = 0

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According to the Mean Value Theorem, if f(x) is a function that is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one c in (a, b) such that:

f(b) - f(a) = f'(c) x (b - a)

To find a c that satisfies the theorem, we need to solve for c:

c = (f(b) - f(a)) / (b - a) / f'(c)

However, this equation cannot be solved directly because c appears on both sides. Instead, we can use an iterative numerical method to approximate c.

One such method is the bisection method, which involves repeatedly dividing the interval (a, b) in half and checking which subinterval contains a root. Here's how it works:

1. Set c = (a + b) / 2, the midpoint of the interval.

2. Evaluate f'(c) using the derivative of f.

3. Calculate f(b) - f(a) and f'(c) x (b - a).

4. If f'(c) x (b - a) = f(b) - f(a), then c is the solution.

5. If f'(c) x (b - a) > f(b) - f(a), then the solution is in the left subinterval (a, c), so set b = c and go to step 1.

6. If f'(c) x (b - a) < f(b) - f(a), then the solution is in the right subinterval (c, b), so set a = c and go to step 1.

Repeat steps 1-6 until the solution is found to a desired degree of accuracy.

Note that this method assumes that f'(x) is continuous on [a, b], which is not always the case. In such cases, other numerical methods may be required.

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The largest couple M that can be applied  to the cross-section of beam  is 108.23 kN.m.

To determine the largest couple M that can be applied to the cross-section shown, we need to calculate the maximum shear stress in the cross-section and ensure that it is less than the allowable stress of the material.

The cross-section can be divided into two rectangles, and the centroid of the combined shape can be found to be at a distance of 2.5 mm from the top edge and 7.5 mm from the left edge. Using the formula for maximum shear stress, τ = VQ/It, we can find the maximum shear stress as:

V = M*d/A, where d is the distance from the neutral axis to the outermost fiber, and A is the area of the cross-section.

d = 5 mm + 2.5 mm = 7.5 mm

A = (10 mm * 5 mm) + (5 mm * 5 mm) = 75 mm^2

Therefore, V = (M*7.5)/75 = M/10

The second moment of area (I) of the combined shape can be found by summing the second moments of area of the two rectangles about their centroids:

I = (1/12 * 10 mm * (5 mm)^3) + 10 mm * (2.5 mm - 7.5 mm)^2 + (1/12 * 5 mm * (5 mm)^3) + 5 mm * (7.5 mm - 2.5 mm)^2

I = 8541.67 mm^4

The elastic modulus of the material is assumed to be constant and is equal to 200 GPa.

Using these values, we can find the maximum shear stress as:

τ = (M7.5)/(108541.67) * (10/5) = M/916.28 MPa

The maximum allowable stress is given as 118 MPa, so we set τ = 118 MPa and solve for M:

M/916.28 = 118 MPa

M = 108227.68 N.mm = 108.23 kN.m (rounded to two decimal places)

Therefore, the largest couple M that can be applied is 108.23 kN.m.

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_____The given question is incomplete, the complete question is given below:

Consider a beam with the cross section shown, made of a material with an allowable stress of 118 MPa. 10 mm 10 mm 801111n- 5mm 5 mm References eBook & Resources Section Break Difficulty: Easy 2· value: 10.00 points Determine the largest couple M that can be applied to the cross section shown. (Round the final answer to two decimal places.) The largest couple M that can be applied is kN m.

Modulus of rigidity (shear modulus) measures a material's rigidity in shear stress/strain. Polar moment of inertia measures the ability of a circular beam to resist torsion measured with J = 32 / (pi x D^4) , and torque is the twisting force on a specimen measured as  T = G x J x θ.

The modulus of rigidity or shear modulus, represented by G, is a measure of a material's rigidity when subjected to shear stress. Shear stress is the force applied perpendicular to the cross-sectional area of a material, while shear strain is the resulting deformation or twisting of the material.

The equation G = shear stress / shear strain is only valid in the elastic region of a material, where it can return to its original shape after the force is removed.

The polar moment of inertia, J, is a measure of a circular cross-section beam or specimen's resistance to torsion or twisting. A larger diameter of the beam results in a larger polar moment of inertia.

The equation J = 32 / (pi x D^4) is used to calculate the polar moment of inertia, where D is the diameter of the beam.

The torque in a circular cross-section beam or specimen is given by the equation T = G x J x θ, where G is the shear modulus, J is the polar moment of inertia, and theta is the angle of twist in radians.

The torque arm length and the applied force F are used to calculate the twisting force or torque in the specimen.

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Answer:Approximately

(0.58975,0.34781)

Step-by-step explanation:

If (x,y) is a point on the parabola, then the distance between (x,y) and (1,0) is:

√(x−1)2+(y−0)2=√x4+x2−2x+1

To minimize this, we want to minimize

f(x)=x4+x2−2x+1

The minimum will occur at a zero of:

f'(x)=4x3+2x−2=2(2x3+x−1)

graph{2x^3+x-1 [-10, 10, -5, 5]}

Using Cardano's method, find

x=3√14+√8736+3√14−√8736≅0.58975

y=x2≅0.34781

The displacement of the block at any time t is then x= 0.05cos5tm. (option b).

Now, when the block is released, it starts oscillating back and forth about its equilibrium position due to the force exerted by the spring. This motion is described by the equation of motion for a simple harmonic oscillator:

x = Acos(ωt + φ)

The angular frequency ω of the oscillation is given by:

ω = √(k/m)

where k is the spring constant and m is the mass of the block.

Substituting the given values of k and m, we get:

ω = √(50/2) = 5 rad/s

The phase angle φ depends on the initial conditions of the system, i.e., the initial displacement and velocity of the block. Since the block is initially at rest, its initial velocity is zero and the phase angle is zero as well.

Therefore, the equation of motion for the displacement of the block is:

x = 0.05cos(5t)

Hence, option B, x = 0.05cos(5t), is the correct answer.

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Given:-

To find:-

Answer:-

In the given right angled triangle, cosine is the ratio of base and hypotenuse. In this triangle with respect to angle H , IH is base, IJ is perpendicular and JH is hypotenuse. And their measures are ,

And as mentioned earlier, we can find cosH as ,

[tex]\implies\cos\theta =\dfrac{b}{h} \\[/tex]

[tex]\implies \cos H =\dfrac{IH}{HJ} \\[/tex]

On substituting the respective values, we have;

[tex]\implies\cos H = \dfrac{40}{41}\\[/tex]

[tex]\implies \cos H = 0.975 [/tex]

Value rounded to nearest hundred will be ,

[tex]\implies \cos H =\boxed{0.98}[/tex]

Hence the value of cos H 0.98 .

Answer:

cos H = 0.98 (rounded to the nearest hundredth)

Step-by-step explanation:

Answer:

[tex]\cos R=\dfrac{3}{5}[/tex]

Step-by-step explanation:

To find the cosine of an angle in a right triangle, we can use the cosine trigonometric ratio.

The cosine trigonometric ratio is the ratio of the side adjacent to the angle to the hypotenuse.

[tex]\boxed{\cos \theta=\sf \dfrac{A}{H}}[/tex]

From inspection of the given right triangle HIJ, the side adjacent to angle H is IH, and the hypotenuse is HJ. Therefore:

Substitute these values into the formula:

[tex]\implies \cos H=\dfrac{40}{41}[/tex]

As 41 is a prime number, the fraction cannot be reduced any further.

The value of cos H rounded to the nearest hundredth is:

[tex]\implies \cos H=0.975609756...[/tex]

[tex]\implies \cos H=0.98[/tex]

The solution of equation of trigonometric functions, tan²(x) + tan(x) − 20 = 0, is equals to the x = 1.33 radians , 1.37 radians (x = arc tan(4) = 1.33 , x = arc tan(-5)= 1.37).

We know that trigonometric functions are periodic functions, solutions of trigonometric equations are then infinite and periodic. In these equations, it is essential to know the reduction formulas of each quadrant, which allows each angle in the first quadrant to be related to its corresponding angle in the other three quadrants. We have an equation which contains triagmometric function,

tan²(x) + tan(x) − 20 = 0 --(1) and we have to solve it. First we change the variable as y = tan(x). Rewrite the equation (1), y² + y - 20 = 0 --(2) which is an quadratic equation. The quadratic formula helps to solve the equation (2).

=> y = ( -1 ± √1 - 4(-20))/2

=> y = ( -1 ± √81)/2

=> y = ( -1 ± 9)/2

=> y = ( -1 + 9)/2 or (-1 -9)/2

=> y = 4, -5

Now x = tan⁻¹(y), and so, x = tan⁻¹(4) = 1.33 radians, or tan⁻¹(-5) = 1.37 radians. Hence, required value is ( 1.33, 1.37).

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Answer:

What is the Value of Tan 270 Degrees? The value of tan 270 degrees is undefined. Tan 270 degrees can also be expressed using the equivalent of the given angle (270 degrees) in radians (4.71238 . . .) ⇒ 270 degrees = 270° × (π/180°) rad = 3π/2 or 4.7123 . . .

please marke a a brainalist pls

To construct a confidence interval for sigma the population from which the sample was drawn must be normally distributed. -

It is not necessary to make the normalcy assumption in order to build a confidence interval for sigma. To employ the central limit theorem, however, the sample size must be adequate which ideally enables the use of the normal distribution to approximate the sampling distribution of the sample variance. Alternative techniques, like t-distribution, can be employed if total sample size is limited.

It's also important to remember that, regardless of sample size, the distribution of sample variance will be normal if population is known and recognizable to have a normally distributed population. As long as the sample size is appropriate, the sample variance may still be utilized to create a confidence range for sigma even if the population is not normally distributed.

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Answer: $1,775

Step-by-step explanation:

We can use a proportion to estimate the cost of having a 13 ft by 18 ft brick patio installed, based on the cost of a 15 ft by 20 ft patio:

(cost of 13 ft by 18 ft patio) / (cost of 15 ft by 20 ft patio) = (area of 13 ft by 18 ft patio) / (area of 15 ft by 20 ft patio)

The area of the 13 ft by 18 ft patio is 13 x 18 = 234 sq ft, and the area of the 15 ft by 20 ft patio is 15 x 20 = 300 sq ft.

So,

(cost of 13 ft by 18 ft patio) / ($2,275) = 234 / 300

Solving for the cost of the 13 ft by 18 ft patio, we get:

cost of 13 ft by 18 ft patio = ($2,275) x (234 / 300) = $1,775.25

Rounding to the nearest dollar, the estimated cost of having a 13 ft by 18 ft brick patio installed is $1,775.

The equation connecting z and p is:

z = x + (1/p)

To find z when p = 6, we can use the values of x and y when p = 2 and p = 4 to calculate the corresponding value of z when p = 6.

When p = 2, x = 4 and y = 1, so z = 5

When p = 4, x = 16 and y = 0.25, so z = 16.25

Therefore, when p = 6, x = 36 and y = 0.167, so z = 36.167

(a) The theoretical probability of getting a 5 or 6 is 1/5

(b) The experimental probability of getting a 5 or 6 is 1/5

(c) The true statement is the first statement.

A subfield of statistics known as probability studies random events and their likelihood of happening. It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes and is used to make predictions and estimate the likelihood of future events.

(a) Assuming that the machine is fair, the theoretical probability of getting a 5 or 6 is:

P(5 or 6) = P(5) + P(6) = 1/10 + 1/10 = 1/5

(b) From the results, we can see that out of the 50 trials, there were 10 trials where the machine output a 5 or a 6.

The experimental probability of getting a 5 or 6 is:

P(5 or 6) = 10/50 = 1/5

(c) A large number of trials, might be a difference between the experimental and theoretical probabilities, but the difference should be small. This is because the theoretical probability is based on the assumption of a fair machine, while the experimental probability is based on actual results.

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The algebraic expression for the product of 9 and x is 9x. This can be expressed in steps as follows:

Step 1: Identify the values that are being multiplied together. In this case it is 9 and x.

Step 2: Write the two values side by side and place a multiplication sign between them.

Step 3: The algebraic expression for the product of 9 and x is then written as 9x.

The function[tex]h(x) = -2x^4 + x^2 - 13; x = y+1,[/tex] can be simplified as [tex]h(y+1) = -2y^4 - 8y^3 - 10y^2 - 8y - 14.[/tex]

A function is a type of rule that produces one output for a single input. Source of the image: Alex Federspiel. This is illustrated by the equation y=x2. Any input for x results in a single output for y. Considering that x is the input value, we would state that y is a function of x.

To evaluate the function[tex]h(x) = -2x^4 + x^2 - 13[/tex] when x = y + 1, we can substitute y + 1 for x:

[tex]h(y+1) = -2(y+1)^4 + (y+1)^2 - 13[/tex]

Simplifying this expression involves some algebraic manipulation. We can start by expanding the fourth power using the binomial theorem:

[tex](y+1)^4 = y^4 + 4y^3 + 6y^2 + 4y + 1[/tex]

Substituting this expression into h(y+1), we get:

[tex]h(y+1) = -2(y^4 + 4y^3 + 6y^2 + 4y + 1) + (y^2 + 2y + 1) - 13[/tex]

Simplifying further, we get:

[tex]h(y+1) = -2y^4 - 8y^3 - 10y^2 - 8y - 14[/tex]

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The value of the limits are:

The highest degree terms in the numerator and denominator are both 3x^2.

So, as x approaches infinity, the expression behaves like √(3x^2)/√(3x^2) = 1.

Therefore, the limit evaluates to:

lim x → ∞√(9+3x²)/(2+7x) = lim x → ∞(√(3x²)/√(3x²))

lim x → ∞√(9+3x²)/(2+7x) = 1.

(b) The highest degree term in the numerator is 3x^2, while the highest degree term in the denominator is 7x.

Therefore, as x approaches negative infinity, the expression behaves like:

√(3x^2)/√(7x) = √(3/7)(x^2/x) = √(3/7)x.

Since the coefficient of x is positive, the expression approaches negative infinity as x approaches negative infinity.

Therefore, the limit evaluates to:

lim x→-∞ √(9+3x^2)/(2+7x) = lim x→-∞ √(3/7)x

lim x→-∞ √(9+3x^2)/(2+7x)  = -∞.

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Complete question:

Evaluate the following limits. If needed, enter 'INF' for ∞ and '-INF for -∞.

(a)

lim x → ∞√9+3x²/2+7x

(b)

lim x→-∞  √9+3x2/2+7x

Answer: The estimated product is about 10.

Step-by-step explanation:

To estimate the product, we can round 2 1/6 to 2 and 4 1/2 to 5. Then, we multiply 2 x 5 to get an estimated product of 10.

To find the exact product, we can use the following steps to multiply the two mixed numbers:

   Convert each mixed number to an improper fraction:

   2 1/6 = 13/6

   4 1/2 = 9/2

   Multiply the two fractions:

   (13/6) x (9/2) = (13 x 9) / (6 x 2) = 117/12

   Simplify the fraction, if possible:

   117/12 = 9 3/4

Therefore, the exact product of 2 1/6 x 4 1/2 is 9 3/4.

Answer:

To solve the quadratic equation 9×^2-15×-6=0, we can use the quadratic formula, which is given by:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.

In this case, a = 9, b = -15, and c = -6, so we can substitute these values into the quadratic formula:

x = (-(-15) ± sqrt((-15)^2 - 4(9)(-6))) / 2(9)

Simplifying this expression gives:

x = (15 ± sqrt(225 + 216)) / 18

x = (15 ± sqrt(441)) / 18

x = (15 ± 21) / 18

So the two solutions to the quadratic equation are:

x = (15 + 21) / 18 = 2

x = (15 - 21) / 18 = -1/3

Therefore, the solutions to the quadratic equation 9×^2-15×-6=0 are x = 2 and x = -1/3.

Answer:

[tex]2679\frac{7}{15}[/tex]

Step-by-step explanation:

[tex]\frac{1}{3}\times 3.14\times 16^2 \times 10[/tex]

First thing we need to do is to evaluate 16^2.

[tex]16^2 = 16\times 16 =256[/tex]

Now, because multiplication is an associative and communitative property, the order that we multiply won't matter. I will rearrange terms to multiply in a more easier way, left to right.

[tex]\frac{1}{3}\times 10\times 256\times 3.14[/tex]

Lets multiply the fraction first. Multiply across, divide 10 by 3. 3R1 gives us:

[tex]3\frac{1}{3} \times 256 \times 3.14[/tex]

Now lets multiply 256. Same method as before.

[tex]853\frac{1}{3}\times 3.14[/tex]

Now, finally, the decimal. Lets convert it to a fraction.

[tex]3.14=\frac{314}{100}[/tex]

Now, replace the decimal in the expression.

[tex]853\frac{1}{3}\times \frac{314}{100}[/tex]

Same method as before. Through rigorous simplifying, we get:

[tex]2679\frac{7}{15}[/tex]

When rounded to two decimal places, the radius of the can is approximately 1.6 inches.

Surface fοrmed by a straight line mοving parallel tο a fixed straight line and intersecting a fixed planar clοsed curve. a sοlid οr surface defined by a cylinder and twο parallel planes that cut all οf its elements. See Vοlume Fοrmulas Table, particularly fοr the right circular cylinder.

The volume of a cylinder can be calculated using the following formula:

V = πr²h

where

V denotes volume,

r denotes radius, and

h denotes height.

The cylindrical aluminium can has a height of 7 inches and a volume of 56 cubic inches. We can calculate the radius using the volume of a cylinder formula:

V = πr²h

56 = πr²(7)

56 = (22/7)r²(7)

56 = 22r²

56/22 = r²

2.54 = r²

r = [tex]\sqrt{2.54}[/tex]

r ≈ 1.6

As a result, when rounded to two decimal places, the radius of the can is approximately 1.6 inches.

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The three Pythagorean trigonometric identities, which I’m sure one can find in any Algebra-Trigonometry textbook, are as follows:

sin² θ + cos² θ = 1

tan² θ + 1 = sec² θ

1 + cot² θ = csc² θ

where angle θ is any angle in standard position in the xy-plane.

Consistent with the definition of an identity, the above identities are true for all values of the variable, in this case angle θ, for which the functions involved are defined.

The Pythagorean Identities are so named because they are ultimately derived from a utilization of the Pythagorean Theorem, i.e., c² = a² + b², where c is the length of the hypotenuse of a right triangle and a and b are the lengths of the other two sides.

This derivation can be easily seen when considering the special case of the unit circle (r = 1). For any angle θ in standard position in the xy-plane and whose terminal side intersects the unit circle at the point (x, y), that is a distance r = 1 from the origin, we can construct a right triangle with hypotenuse c = r, with height a = y and with base b = x so that:

c² = a² + b² becomes:

r² = y² + x² = 1²

y² + x² = 1

We also know from our study of the unit circle that x = r(cos θ) = (1)(cos θ) = cos θ and y = r(sin θ) = (1)(sin θ) = sin θ; therefore, substituting, we get:

(sin θ)² + (cos θ)² = 1

1.) sin² θ + cos² θ = 1 which is the first Pythagorean Identity.

Now, if we divide through equation 1.) by cos² θ, we get the second Pythagorean Identity as follows:

(sin² θ + cos² θ)/cos² θ = 1/cos² θ

(sin² θ/cos² θ) + (cos² θ/cos² θ) = 1/cos² θ

(sin θ/cos θ)² + 1 = (1/cos θ)²

(tan θ)² + 1 = (sec θ)²

2.) tan² θ + 1 = sec² θ

Now, if we divide through equation 1.) by sin² θ, we get the third Pythagorean Identity as follows:

(sin² θ + cos² θ)/sin² θ = 1/sin² θ

(sin² θ/sin² θ) + (cos² θ/sin² θ) = 1/sin² θ

1 + (cos θ/sin θ)² = (1/sin θ)²

1 + (cot θ)² = (csc θ)²

3.) 1 + cot² θ = csc² θ

Answer:

20

Step-by-step explanation:

Step-by-step explanation:

this is the answerrr

without calculator

To find the eigenvalues and eigenvectors of a matrix, we need to solve the equation Ax = λx, where A is the matrix, λ is the eigenvalue, and x is the eigenvector. In this case, we have a 3x3 matrix, so we expect to find three eigenvalues and three eigenvectors.

Without performing the calculations, it is difficult to determine the exact eigenvalues and eigenvectors of matrix D. However, based on the structure of the matrix, we can make some observations. We notice that the matrix has a repeating pattern of the number 5, which suggests that 5 is a dominant eigenvalue. Additionally, the matrix is symmetric, so we know that the eigenvectors will be orthogonal.

One possible approach to finding the eigenvectors is to use the Gram-Schmidt process, which is a method for orthogonalizing a set of vectors. We can start with a vector that is parallel to one of the columns of the matrix, and then subtract the projection of that vector onto the next column to obtain a vector that is orthogonal to the first. We can continue this process for each column to obtain a set of orthogonal vectors, which will be our eigenvectors.

Based on these observations, we can hypothesize that 5 is an eigenvalue of D and that we can find two linearly independent eigenvectors by using the Gram-Schmidt process on the columns of the matrix. However, we would need to perform the calculations to confirm this hypothesis and determine the exact values of the eigenvectors.

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By using sample standard deviation - (1) N or n = 20

(2) X or μ = 43.4

(3) σ = 9.78 and s = 10.04

Why does sample standard deviation exist?

When we talk about a sample standard deviation, we're not talking about population standard deviation. The average amount by which a group of values deviates from their mean is shown by the statistical measure of variability known as the standard deviation.

We are given with the following data set below;

{51, 48, 42, 43, 48, 48, 46, 15, 29, 45, 47, 55, 46, 35, 47, 48, 54, 26, 53, 42}

(1) As we can see in the above data that there are 20 data values in our data set which means that the value of N or n (numner of observations) is 20.

(2) The formula for calculating Mean of the data, i.e. X or μ is given by;

       Mean  =  51 +48+ 42+ 43+ 48 + 48 + 46+ 15+ 29+ 45+ 47+ 55+ 46+ 35+ 47+ 48+ 54+ 26+ 53+ 42/20

                = 868/20 = 43.4

So, the value of  X or μ   is 43.4.

(3) The formula for calculating population standard deviation () is given by;

  Standard deviation, σ  = √(51 - 43.4)² + (48 - 43.4)² + .........+(42 - 43.4)²/20

        =  9.78

Similarly, the formula for calculating sample standard deviation (s) is given by;

       Sample Standard deviation, s =  

                       √(51 - 43.4)² + (48 - 43.4)² + .........+(42 - 43.4)²/20 - 1

                      =  10.04

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The set of all x-values that are at a distance of 13 or less from the number 8 in the interval notation is given by  [ -5, 21 ].

The distance between x and 8 is |x - 8|.

Find all the values of x such that |x - 8| ≤ 13.

This inequality can be rewritten as follow,

|x - 8| ≤ 13

⇒ -13 ≤ x - 8 ≤ 13

Now,

Adding 8 to all sides of the inequality we get,

⇒  -13 +  8 ≤ x - 8 + 8 ≤ 13 + 8

⇒ -5 ≤ x ≤ 21

Therefore, all the x-values which are at a distance of 13 or less from the number 8 represented in the interval notation as [ -5, 21 ].

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In the word problem,

a)Gallons per minute change is -2/3 gallons per minute

b) 7.5 minutes to drain the other 5 gallons

c)The tank was full 4 1/2 minutes before Jada arrived.

What is word problem?

Word problems are often described verbally as instances where a problem exists and one or more questions are posed, the solutions to which can be found by applying mathematical operations to the numerical information provided in the problem statement. Determining whether two provided statements are equal with respect to a collection of rewritings is known as a word problem in computational mathematics.

a) Gallons per minute change is -2/3 gallons per minute  since it decreases by 2 gallons in 3 minutes.

b)  we have 5 gallons and lose 2/3 gallons per minute

=> (2/3 gal/minute)(x minutes) = 5 gallons

=> x = 5(3/2) = 15/2 = 7.5 minutes to drain the other 5 gallons

c)  There was 10 gallons when the tank is full . That is 3 gallons more than the 7 gals there when Jada arrived.

=>  (2/3)(x) = 3

=> x = 3(3/2) = 9/2 = 4 1/2 minutes

The tank was full 4 1/2 minutes before Jada arrived.

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Event a has a 12/25 chance of happening provided that event b has already happened.

It is stated as a number between 0 and 1, where 0 denotes that the event is hypothetical and 1 denotes that it is unavoidable. Furthermore, probability can be stated as a percentage, with a range of 0% to 100%. By dividing the number of favourable outcomes by the entire number of possible possibilities, the probability of an event is determined. Many disciplines, such as statistics, economics, and physics, among others, utilise probability theory.

given

We may apply the conditional probability formula to determine p(a/b):

p(a and b) / p = p(a and b) (b)

Given that p(b) = 1/4 and p(a and b) = 3/25, we may enter these numbers in the formula as follows:

p(a/b) = (3/25) / (1/4)

p(a/b) = (3/25) * (4/1)

p(a/b) = 12/25

Event a has a 12/25 chance of happening provided that event b has already happened.

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Step-by-step explanation:

remember the trigonometric triangle inside a circle.

sine is the up/down leg, cosine is the left/right leg.

the Hypotenuse (baseline) of the right-angled triangle is the angle-defining radius of the circle.

this is the basic definition inside the norm-circle with radius = 1.

for any other size sine and cosine need to be multiplied by the actual radius for the actual triangle side lengths.

I think you need the answer in simplified fraction and square root notification (you cut that off), but I give you also the decimal result, just in case.

so,

78 = cos(30)×y

78 = (sqrt(3)/2) × y

y = 78 / (sqrt(3)/2) = 2×78/sqrt(3) = 156/sqrt(3) =

= 90.06664199...

x = sin(30) × y = 0.5 × 156/sqrt(3) = 78/sqrt(3) =

= 45.033321...

To show that the finite union of compact sets is compact, we need to show that any open cover of the union has a finite subcover.  

Let A and B be two compact sets. Suppose that U is an open cover of A ∪ B. also U is also an open cover of A and an open cover ofB.   Since A is compact, there exists a finite subcover of U that coversA. Let this subcover be{ U1, U2,., Un}.   also, since B is compact, there exists a finite subcover of U that coversB. Let this subcover be{ V1, V2,., Vm}.  

Also the union of these two finite subcovers is a finite subcover of U that covers A ∪B. Specifically, the subcover is{ U1, U2,., Un, V1, V2,., Vm}.   thus, any open cover of the finite union of compact sets A ∪ B has a finite subcover, and  therefore A ∪ B is compact. By induction, we can extend this result to any finite union of compact sets.

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