Someone pls answer all?

Answer:

The cost of 16 onions is $ 1.20.

Step-by-step explanation:

To determine what is the cost, in dollars, of 16 onions if 3 onions weigh 1.5 lb and the price of onions is 33 cents per kilogram, the following calculation must be performed:

1.5 pounds = 0.68 kilos

0.68 / 3 = 0.22666 kilos each onion

16 x 0.22666 = 3.626 kilos

0.33 x 3.626 = 1.20

Therefore, the cost of 16 onions is $ 1.20.

Step-by-step explanation:

this is the correct answer for the question

Answer:

5,5,6,6,13

Step-by-step explanation:

Mode means most often.  The 5 numbers has 2 modes 5 and 6

This means that 4 of the numbers must be 5,5,6,6

Median means the middle number must be 6

5,5,6,6,x is the only way to to get the middle number to be 6

We need to average to 7

(5+5+6+6+x) /5 = 7

(5+5+6+6+x) /5  *5= 7*5

(5+5+6+6+x) =35

22+x = 35

x = 35-22

x = 13

The other number is 13

Answer:

[tex]x = 77[/tex]

Step-by-step explanation:

Given

[tex]First \to Second[/tex]

[tex]35 \to x[/tex]

[tex]20 \to 44[/tex]

[tex]20 \to 44[/tex]

Required

Find x by SSS

Represent the triangle sides as a ratio

[tex]35 : x = 20 : 44[/tex]

Express as fraction

[tex]\frac{x}{35} = \frac{44}{20}[/tex]

Multiply by 35

[tex]x = \frac{44}{20} * 35[/tex]

[tex]x = \frac{44 * 35}{20}[/tex]

[tex]x = \frac{1540}{20}[/tex]

[tex]x = 77[/tex]

Answer:

ccccccccccccccccccccccccccc

Step-by-step explanation:

Answer:

Step-by-step explanation:

(0, -1)  & (3 ,1)

[tex]Slope=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\=\frac{1-[-1]}{3-0}\\\\=\frac{1+1}{3}\\\\=\frac{2}{3}[/tex]

Answer:

[tex]44[/tex]

Step-by-step explanation:

The dimensions of the garden is 12 by 8. If we have a walkway that surrounds the garden, the dimensions of the walkway is 2. Since it surrounds the rectangle all sides add 2 to each of the dimensions so now the dimensions of the garden and walkway is 14×10.

The area of the garden is 96 square ft.

The area of the garden and walkway is 140 so let subtract the area of the garden from the total area of both the garden and walkway.

[tex]140 - 96 = 44[/tex]

The area is 44.

Answer:

120 square feet

Step-by-step explanation:

(8+2*2)

(18+2*2) - 8*18 = 120 square feet.

Answer:

4

Step-by-step explanation:

The mean is the average of a data set. It can be found by adding up all of the values in a data set and then dividing it by the number of values in the data set.

The values in this data set;

[tex]7,5,5,3,2,2[/tex]

The number of values in this data set,

[tex]6[/tex]

Find the mean;

[tex]\frac{sum\ of\ vlaues}{number\ of\ values}[/tex]

[tex]=\frac{7+5+5+3+2+2}{6}\\\\=\frac{24}{6}\\\\=4[/tex]

Answer:

[tex]x = 5600 + 5600 * 0.042 * 1[/tex]

Step-by-step explanation:

Given

[tex]P = 5600[/tex] -- Principal

[tex]R = 4.2\%[/tex] -- Rate

[tex]T = 1[/tex] -- Time

Required

The amount (x) to be paid

This is calculated as:

[tex]x = P + I[/tex]

Where:

[tex]I = PRT[/tex]

So, we have:

[tex]x = 5600 + 5600 * 4.2\% * 1[/tex]

Express percentage as decimal

[tex]x = 5600 + 5600 * 0.042 * 1[/tex]

(c) is correct

Answer: x=4, y=-6

Step-by-step explanation:

Answer:

C $13,220

Step-by-step explanation:

Using a linear approximation, the estimated cube root of 217 is  6.00925.

Given that the number is,

The cube root of 217

Now, for the cube root of 217 using a linear approximation, use differentials.

So, the derivative of the function [tex]f(x) = x^{(1/3)[/tex] at a known point.

Taking the derivative of [tex]f(x) = x^{(1/3)[/tex], we get:

[tex]f'(x) = (\dfrac{1}{3} )x^{-2/3[/tex]

Now, we can choose a point near 217 to evaluate the linear approximation.

Let's use x = 216, which is a perfect cube.

Substituting x = 216 into the derivative, we get:

[tex]f'(216) = (\dfrac{1}{3} )(216)^{-2/3[/tex]

            [tex]= 0.00925[/tex]

Next, use the linear approximation formula:

Δy ≈ f'(a)Δx

Since our known point is a = 216 and we want to estimate the cube root of 217,

since 217 - 216 = 1

Hence, Δx = 1

Δy ≈ f'(216)

Δx ≈ 0.00925 × 1

      ≈ 0.00925

Finally, add this linear approximation to the known value at the known point to get our estimate:

Estimated cube root of 217 ;

f(216) + Δy = 6 + 0.00925

                 = 6.00925

Therefore, the estimated cube root of 217 is 6.00925.

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9514 1404 393

Answer:

Step-by-step explanation:

a) The starting height is h(0) = 7.5 feet, the constant in the quadratic function.

The irrigation system is positioned 7.5 feet above the ground

__

b) The axis of symmetry for quadratic ax^2 +bx +c is x = -b/(2a). For this quadratic, that is x=-10/(2(-1)) = 5. This is the horizontal distance to the point of maximum height. The maximum height is ...

  h(5) = (-5 +10)(5) +7.5 = 32.5 . . . feet

The spray reaches a maximum height of 32.5 feet at a horizontal distance of 5 feet from the sprinkler head.

__

c) The maximum distance will be √32.5 + 5 ≈ 10.7 ft.

The spray reaches the ground at about 10.7 feet away.

Answer:

7.5

32.5

5

maximum

10.7

Step-by-step explanation:

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

Given:

First lamp rating

Power (P) = 100W

Voltage (V) = 110V

Second lamp rating

Power (P) = 100W

Voltage (V) = 220V

Source

Voltage = 220V

i. Get the resistance of each lamp.

Remember that power (P) of each of the lamps is given by the quotient of the square of their voltage ratings (V) and their resistances (R). i.e

P = [tex]\frac{V^2}{R}[/tex]

Make R subject of the formula

⇒ R = [tex]\frac{V^2}{P}[/tex]             ------------------(i)

For first lamp, let the resistance be R₁. Now substitute R = R₁, V = 110V and P = 100W into equation (i)

R₁ = [tex]\frac{110^2}{100}[/tex]

R₁ = 121Ω

For second lamp, let the resistance be R₂. Now substitute R = R₂, V = 220V and P = 100W into equation (i)

R₂ = [tex]\frac{220^2}{100}[/tex]

R₂ = 484Ω

ii. Get the equivalent resistance of the resistances of the lamps.

Since the lamps are connected in series, their equivalent resistance (R) is the sum of their individual resistances. i.e

R = R₁ + R₂

R  = 121 + 484

R = 605Ω

iii. Get the current flowing through each of the lamps.

Since the lamps are connected in series, then the same current flows through them. This current (I) is produced by the source voltage (V = 220V) of the line and their equivalent resistance (R = 605Ω). i.e

V = IR [From Ohm's law]

I = [tex]\frac{V}{R}[/tex]

I = [tex]\frac{220}{605}[/tex]

I = 0.36A

iv. Get the power consumed by each lamp.

From Ohm's law, the power consumed is given by;

P = I²R

Where;

I = current flowing through the lamp

R = resistance of the lamp.

For the first lamp, power consumed is given by;

P = I²R           [Where I = 0.36 and R = 121Ω]

P = (0.36)² x 121

P = 15.68W

For the second lamp, power consumed is given by;

P = I²R           [Where I = 0.36 and R = 484Ω]

P = (0.36)² x 484

P = 62.73W

Therefore;

The power consumed in the lamp marked 100W - 110V is 15.68W

The power consumed in the lamp marked 100W - 220V is 62.73W

(1) Recall that

sin(x - y) = sin(x) cos(y) - cos(x) sin(y)

sin²(x) + cos²(x) = 1

Given that α lies in the third quadrant, and β lies in the fourth quadrant, we expect to have

• sin(α) < 0 and cos(α) < 0

• sin(β) < 0 and cos(β) > 0

Solve for cos(α) and sin(β) :

cos(α) = -√(1 - sin²(α)) = -3/5

sin(β) = -√(1 - cos²(β)) = -5/13

Then

sin(α - β) = sin(α) cos(β) - cos(α) sin(β) =  (-4/5) (12/13) - (-3/5) (-5/13)

==>   sin(α - β) = -63/65

(2) In the second identity listed above, multiplying through both sides by 1/cos²(x) gives another identity,

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

==>   tan²(x) + 1 = sec²(x)

Rewrite the equation as

3 sec²(x) tan(x) = 4 tan(x)

3 (tan²(x) + 1) tan(x) = 4 tan(x)

3 tan³(x) + 3 tan(x) = 4 tan(x)

3 tan³(x) - tan(x) = 0

tan(x) (3 tan²(x) - 1) = 0

Solve for x :

tan(x) = 0   or   3 tan²(x) - 1 = 0

tan(x) = 0   or   tan²(x) = 1/3

tan(x) = 0   or   tan(x) = ±√(1/3)

x = arctan(0) + nπ   or   x = arctan(1/√3) + nπ   or   x = arctan(-1/√3) + nπ

x = nπ   or   x = π/6 + nπ   or   x = -π/6 + nπ

where n is any integer. In the interval [0, 2π), we get the solutions

x = 0, π/6, 5π/6, π, 7π/6, 11π/6

(3) You only need to rewrite the first term:

[tex]\dfrac{\sin(x)}{1-\cos(x)} \times \dfrac{1+\cos(x)}{1+\cos(x)} = \dfrac{\sin(x)(1+\cos(x))}{1-\cos^2(x)} = \dfrac{\sin(x)(1+\cos(x)}{\sin^2(x)} = \dfrac{1+\cos(x)}{\sin(x)}[/tex]

Then

[tex]\dfrac{\sin(x)}{1-\cos(x)}+\dfrac{1-\cos(x)}{\sin(x)} = \dfrac{1+\cos(x)+1-\cos(x)}{\sin(x)}=\dfrac2{\sin(x)}[/tex]

Answer:

B: Ones.

Step-by-step explanation:

Because this number is 4.09, and the decimal is right next to the 4, that means that it is in the ones place. Decimals are always adjacent on the right to the ones place.

Let's see

[tex]\\ \tt\leadsto A=P(1+r/n)^{nt}[/tex]

[tex]\\ \tt\leadsto A=1200(1+0.055)^t[/tex]

[tex]\\ \tt\leadsto A=1200(1.055)^t[/tex]

Answer:

[tex]\sf A=1200(1.055)^t[/tex]

Step-by-step explanation:

Annual Compound Interest Formula

[tex]\large \text{$ \sf A=P\left(1+r\right)^{t} $}[/tex]

where:

Given:

Substitute the given values into the equation:

[tex]\implies \sf A=1200(1+0.055)^t[/tex]

[tex]\implies \sf A=1200(1.055)^t[/tex]

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Answer:

Following are the response to the given points:

Step-by-step explanation:

For question 5.11:

For point a:

For all the particular circumstances, it was not an appropriate sampling strategy as each normal distribution acquired is at a minimum of 30(5) = 150 or 2.5 hours for a time. Its point is not absolutely fair if it exhibits any spike change for roughly 10 minutes.

For point b:

The problem would be that the process can transition to an in the state in less than half an hour and return to in the state. Thus, each subgroup is a biased selection of the whole element created over the last [tex]2 \frac{1}{2}[/tex] hours. Another sampling approach is a group.

For question 5.12:

This production method creates 500 pieces each day. A sampling section is selected every half an hour, and the average of five dimensions can be seen in a [tex]\bar{x}[/tex]line graph when 5 parts were achieved.

This is not an appropriate sampling method if the assigned reason leads to a sluggish, prolonged uplift. The difficulty would be that gradual or longer upward drift in the procedure takes or less half an hour then returns to a controlled state. Suppose that a shift of both the detectable size will last hours [tex]2 \frac{1}{2}[/tex] . An alternative type of analysis should be a random sample of five consecutive pieces created every [tex]2 \frac{1}{2}[/tex] hour.

Answer:

DL/dt = 529 miles/h

Step-by-step explanation:

The radio station (point A) the point just up the radio station ( point B), and the variable position of the plane ( at specif t  point C) shape a right triangle wich hypothenuse L is:

L²  =  d² + x²

d is the constant distance between the plane and the ground

Then differentiation with respect to time on both sides of the equation

2*L*dL/dt  = 2*d* Dd/dt  + 2*x*dx/dt

But   Dd/dt  =  0

L*dL/dt  =  x*dx/dt

x =  5 miles       dx/dt  =  570 m/h        L  = √ d² + x²    L  √ (5)² + (2)²

L = √29      L  = 5.39 m

5.39 *DL/dt  =  5*570 m/h

DL/dt  =   5*570/5.39   miles/h

DL/dt  =  528.76  miles/h

DL/dt = 529 miles/h

Answer:

34n=306

Step-by-step explanation:

Use inverse operation to find it, 306÷34= 9, check again 34(9)=306, so it's correct!

Answer:

x = -2; y = 1

Step-by-step explanation:

See picture below.

We are told matrices B is the inverse of matrix A.

The product of a matrix and its inverse is the identity matrix.

Answer:

6/11, 11/2, 5.77, 5.772

Step-by-step explanation:

Answer:

sin A = 4/5

cos A = 3/5

Step-by-step explanation:

SOHCAHTOA

cot A = 1/tan A

tan A = opp/adj

cot A = 1/tan A = adj/opp

cot A = 3/4

adj = 3; opp = 4

adj^2 + opp^2 = hyp^2

3^2 + 4^2 = hyp^2

9 + 16 = hyp^2

hyp = 5

sin A = opp/hyp

sin A = 4/5

cos A = adj/hyp

cos A = 3/5

Answer:

sin A = 4/5

cos A = 3/5

1km = 0.621371miles

195 km= ?

cross multiplication

= 121.167 miles

25 miles= 1hour

121.167miles = ?hours

121.167=25x

divide by 25x both sides

=4.84 hours

approx 5hours

She must ride for 5 hours if she wants to bike 195 km.

Average speed is defined as the ratio of the total distance traveled by a body to the total time taken for the body to reach its destination.

Given that cyclist rides at an average speed of 25 miles per hour.

Since 1 km = 0.621371 miles

So 195 km = 121.167 miles

The speed of the cyclist (s)  = 24 miles per hour.

Distance covered by the rider = 195 km

Distance covered by the rider (d) = 121.167 miles

By using the formula,  time taken by a body, we calculate the time,

⇒ t = d/s

Substitute the value of d and s in above the equation

⇒ t = 121.167/ 24

Apply the division operation,

⇒ t = 5

Hence, she must ride for 5 hours if she wants to bike 195 km.

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Answer:

y - 1 =  -1/4(x+5)

Step-by-step explanation:

Answer:

Step-by-step explanation:

There are 2 different ways to do this: calculus and by completing the square. In this particular instance, calculus is WAY easier, and since I don't know for what class you are doing this, I'll do both ways. First the calculus way. We know the position equation, and the first derivative of the position is velocity. We also know that when the velocity is equal to 0 is when the object is at its max height. So we'll find the derivative first, then solve it for t:

If [tex]s(t)=-3t^2+20t+2[/tex] then the first derivative is

v(t) = -6t + 20 Solving for t requires that we set the velocity equal to 0 (again, this is where the object is at its max height), so

0 = -6t + 20 and

-20 = -6t so

t = 3.3 seconds. Now that we know that at 3.3 seconds the object is at its highest point, we sub that time into the position function to see where it is at that time:

s(3.3) = [tex]-3(3.3)^2+20(3.3)+2[/tex] and

s(3.3) = 35.3 meters.

Now onto the more difficult way...completing the square. Begin by setting the position function equal to 0 and then move over the constant to get:

[tex]-3t^2+20t=-2[/tex] Since the leading coefficient is not a 1 (it's a 3), we have to factor out the 3, leaving us with:

[tex]-3(t^2-\frac{20}{3}t)=-2[/tex] Now the rule is to take half the linear term, square it, and add it to both sides. Our linear term is [tex]\frac{20}{3}[/tex] and half of that is [tex]\frac{20}{6}[/tex]. Squaring that:

[tex](\frac{20}{6})^2=\frac{400}{36}=\frac{100}{9}[/tex]. We will add that in to both sides. On the left it's easy, but on the right we have to take into account that we still have that -3 sitting out front, refusing to be ignored. So we have to multiply it in when we add it to the right. Doing that gives us:

[tex]-3(t^2-\frac{20}{3}t+\frac{100}{9})=-2-\frac{100}{3}[/tex] We will clean this up a bit now. The reason we do this is because on the left we have created a perfect square binomial which will give us the time we are looking for to answer this question. Simplifying the right and at the same time writing the perfect square binomial  gives us:

[tex]-3(t-\frac{20}{6})^2=-\frac{106}{3}[/tex] Now the last step is to move the constant back over and set the quadratic back equal to y:

[tex]y=-3(t-\frac{20}{6})^2+\frac{106}{3}[/tex].  The vertex of this quadratic is

[tex](\frac{20}{6},\frac{106}{3})[/tex] where

[tex]\frac{20}{6}=3.3[/tex] as the time it takes for the ball to reach its max height of

[tex]\frac{106}{3}=35.3[/tex] meters.

I'd say if you plan on taking calculus cuz you're not there yet, you'll see that many of these types of problems become much simpler when you know it!

Answer:4778.3 cm^3

Step-by-step explanation: The formula for volume of a cone is V=1/3h pi r^2. By plugging in the height and the radius we get our answer.

Answer:

4778.4 :)

Step-by-step explanation: