Which inequality shows the region represented by the
inequality in vertex form?
Oy2(x-50)² +625
O yz-(x +50)² +625
O y2(x +50)² +625
O yz-(x-50)² +625

The instructor chose to use the highest score now mode, hence the score of the second attempt would be used that is 10/10.

In statistics, the value that consistently appears in a particular set is referred to as the mode. The mode or modal value in a data set is sometimes referred to as the value or number that occurs most frequently in the data set. In addition to mean and median, it is one of the three measures of central tendency.

The scores at different attempts are:

First attempt: 2/10 = 0.2

Second attempt: 10/10 = 1

Third attempt: 3/10 = 0.3

The instructor chose to use the highest score now mode, hence the score of the second attempt would be used that is 10/10.

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Answer: 245 yds

Step-by-step explanation:

61 1/4 times 4 = 245

The weight of the robot is within the error limit of 0.3%.

Given is that a robot building club organizes competitions by weight categories called classes each class permit a percent error of 0.3%. In the 5 kg class one robot weighs 4.925 kg.

Let -

{x} = 0.3% of 5

{x} = 0.3/100 x 5

{x} = 3/1000 x 5

{x} = 15/1000

{x} = 3/200

{x} = 0.015                                                                            .... Eq { 1 }

Now -

For the weight of the robot to be under the weight limit -

4.925 ≤ (5 - x)

4.925 ≤ (5 - 0.015)

4.925 ≤ 4.985                                                       ..... Eq { 2 }

which is true.                              

Therefore, the weight of the robot is within the error limit of 0.3%.

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The people that are eating record toys are 48

From the question, we have the following parameters that can be used in our computation:

People at the movies = 80

People are eating popcorn = 2/5

Using the above as a guide, we have the following:

Record toys = (1 - People are eating popcorn) * People at the movies

Substitute the known values in the above equation, so, we have the following representation

Record toys = (1 - 2/5) * 80

Evaluate

Record toys = 48

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Answer: 115%

Step-by-step explanation:

100% = 1

10%=0.1

5%=0.05

100%+10%+5%=115%

1+0.1+0.05=1.15

Answer: 15

Step-by-step explanation:

x+x*15/100

x(1+.15)

1.15x

The equation that can be used to solve the problem is 66/x + 40/x = 5.3

gage rate of travel on the mountain roads is 20 mph and on the country roads is 48 mph.

A mathematical equation is a formula that uses the equals sign to represent the equality of two expressions. Finding the values of the variables that result in the equality is the first step in solving an equation with variables.

Given that:

In mountain Road

distance = 66 miles ; rate = x mph

⇒ time = d/r = 66/x hrs

In country Road

distance = 96 miles ; rate = 2.4x mph

⇒ time = d/r = 96/(2.4x)

⇒ time = d/r = 40/x hrs.

Equation:

time + time = 5.3 hrs

⇒ 66/x + 40/x = 5.3

⇒ 106/x = 5.3

⇒ 5.3x = 106

⇒ x = 20 mph on mountain Roads

⇒ 2.4x = 2.4*20 mph on country Roads

⇒ 48 mph on country Roads

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Answer: -1

Step-by-step explanation:

When the graph g hits g=2, the y is -1

Answer:

So, Donna served 9, Alonzo served 19 and kevin served 57 orders.

Step-by-step explanation:

Let orders be :

Donna - x, Alonzo - y, Kevin - z

x = y - 10 - eqn i

z = 3y - eqn ii

Now, By the question,

x + y + z = 85

we know from eqn i and eqn ii,

y - 10 + y + 3y = 85

or, 5y = 95

so, y = 19

Now,

z = 3(19) = 57

x = 19 - 10 = 9

Check the picture below.

The following numbers are the result:

22 play trumpet, 11 play trombone, and 8 play saxophone.

Two algebraic expressions having the same value and symbol '=' in between are called an equation.

Given:

There are 41 students in group 2.

Twice as many students play the trumpet as play the trombone,

but 8 students play the saxophone.

Let x be the number of students playing the trombone.

So, according to the question,

x + 2x + 8 = 41

3x = 33

x = 11

Hence, 22 play trumpet, 11 play trombone, and 8 play saxophone.

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The polynomial  n² + 6n - 1 can be factored as,

(n + 3 + √10)(n - 3 + √10)(3n - 2).

We know that if x = a is one of the roots of a given polynomial x - a = 0 is a factor of the given polynomial.

To confirm if x - a = 0 is a factor of a polynomial we replace f(x) with f(a) and if the remainder is zero then it is confirmed that x - a = 0 is a factor.

Given, A product of two polynomials (n² + 6n - 1)(3n - 2).

Now, The quadratic expression n² + 6n - 1 can be factored as,

n² + 6n - 1.

[- 6 ± √(36 + 4)]/2.

= [- 6 ± √40]/2.

= [- 6 ± 2√10]/2.

= - 3 ± √10.

Therefore, (n² + 6n - 1)(3n - 2)

= (n + 3 + √10)(n - 3 + √10)(3n - 2).

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Answer:

Step-by-step explanation:

Answer:

Around 11:15 am

Step-by-step explanation:

Let h represents the number of hours the plumber works

Since the plumber charges $65 fee plus $75 per hour, the bill can be calculated by:

→ bill = 65 + 75h

Since the bill is $195.25

→ 195.25 = 65 + 75h

→ 75h = 130.25

→ h = 1.74

1.74 implies that the plumber worked almost 1 hour and 45 minutes. Since the plumber started at 9:30, the repair was completed around 11:15 am.

The percent change of an object is the ratio of its final value to its initial value.

The percentage of an item's final value versus its starting value is its percent change. By dividing the difference in data by the starting value, you can get the percent change.

"Rate" simply refers to the quantity divided by another number, typically 100, 1,000, or another multiple of 10. A rate per 100 is known as a percentage.

Percentage Change is the difference that results from deducting the old value from the new value, dividing by the old value, and multiplying the result by 100 to represent it as a percentage. In most cases, multiplying a fraction by 100 results in a percent conversion.

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The amplitude of subsequent oscillations is 0.093 m.

The spring stiffness is quantified by the spring constant, or k. For various springs and materials, it varies. The stiffer the spring is and the harder it is to stretch, the bigger the spring constant.

Given that,

Mass of a block, m = 1.2 kg

Spring constant of the spring, k = 14.0 N/m

While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 32.0 cm/s or 0.32 m/s

We need to find the amplitude of the subsequent oscillations.

We can use the conservation of energy here. Initially, the kinetic energy of the block is maximum and then it gets converted to the potential energy of the spring.

Mathematically,

[tex]\frac{1}{2} mv^2=\frac{1}{2}kv^2[/tex]

A is the amplitude of subsequent oscillations.

[tex]A=\sqrt{(\frac{mv^2}{k} )} \\\\A=\sqrt{\frac{(1.2)(0.32)^2}{14} } \\\\A=0.093m[/tex]

So, the amplitude of subsequent oscillations is 0.093 m.

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The solution to the system of equations is (13/3, 16/3).

The graph is given below.

An equation is a mathematical statement that is made up of two expressions connected by an equal sign.

Example:

2x + 4 = 8 is an equation.

We have,

y = 14x - 2 _____(1)

y = x + 1 ______(2)

Now,

The graph of the equation is given below.

Now,

From (1) and (2).

14x - 2 = x + 1

14x - x = 1 + 2

13x = 3

x = 13/3

And,

y = x + 1

y = 13/3 + 1

y = 16/3

So,

The solution is (13/3, 16/3)

Thus,

The solution is (13/3, 16/3).

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Step-by-step explanation:

a probability is always the ratio

desired cases / totally possible cases.

if 2 non-overlapping events should occur, the probability of both events happening is the product of both individual probabilities.

e.g. the probability to roll two 6 with 2 dice (or in 2 consecutive rolls with 1 die) is 1/6 × 1/6 = 1/36

if 1 of 2 possible non-overlapping events should occur, the probability of one of the events happening is the sum of the individual probabilities.

e.g. the probability to roll a 1 or a 2 with a die is

1/6 + 1/6 = 2/6 = 1/3

both students are dual-enrolled.

in our case, for the first selection we have the totally possible cases of 27.

the desired cases (dual-enrolled) are 17.

so, the probabilty to select a dual-enrolled student is

17/27

for the second selection (without replacement of the first pull) the totally possible cases are now 26 (because we pulled one student from the general pool of candidates in the first selection).

the desired cases are now 16 (because for our case we pulled a dual-enrolled student from the general pool).

the probabilty to select a dual-enrolled student in the second selection is

16/26 = 8/13

the overall probability to select two dual-enrolled students in 2 selections is

17/27 × 8/13 = 136/351 = 0.387464387... ≈ 0.387

the first student is dual-enrolled, the second is not.

for the first selection the totally possible cases are 27.

the desired cases are again 17.

the probability of the first dejection is again

17/27

for the second selection the totally possible cases are 26 (see above).

the desired cases are still 10.

the probably for this second selection is

10/26 = 5/13

the overall probability to select first a dual-enrolled student and then a not-dual-enrolled student is

17/27 × 5/13 = 85/351 = 0.242165242... ≈ 0.242

The expression 6j^2 - 4j 12 can be rewritten as [tex]$-24(j\frac{1}{2})^2 -12$[/tex]where , and [tex]$q=-12$[/tex]. Thus,[tex]$\frac{q}{p}=-24$[/tex].

The expression[tex]$6j^2 - 4j 12$[/tex] can be rewritten in the form [tex]$c(j p)^2 q$[/tex], where c, p, and q are constants. To do this, we can factor out a common factor from the expression. In this case, the common factor is j. Then, we can rewrite the expression as [tex]$j(6j - 12)$[/tex] which simplifies to [tex]$-24j^2 -12$[/tex]. Dividing both sides by j, we can rewrite the expression as [tex]$-24(j\frac{1}{2})^2 -12$[/tex] where [tex]$c=-24$[/tex], [tex]$p=\frac{1}{2}$[/tex] and [tex]$q=-12$[/tex]. Thus, [tex]$\frac{q}{p}=-24$[/tex]. To summarize, the expression [tex]$6j^2 - 4j 12$[/tex] can be rewritten in the form [tex]$c(j p)^2 q$[/tex] where c, p, and q are constants, and [tex]$\frac{q}{p}=-24$[/tex].

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Answer:

Step-by-step explanation:

The inequality 5x+12y ≤ 30 represents a shaded region in the xy-plane that includes all the points that satisfy the inequality. To graph this inequality, we can first graph the equation 5x+12y = 30, which is the boundary line of the shaded region.

To graph the boundary line, we can find its x- and y-intercepts:

x-intercept: Set y = 0 and solve for x: 5x + 12(0) = 30, which gives x = 6.

y-intercept: Set x = 0 and solve for y: 5(0) + 12y = 30, which gives y = 2.5.

So the boundary line passes through the points (6, 0) and (0, 2.5).

Next, we can determine which side of the boundary line to shade. One way to do this is to pick a test point that is not on the line and plug it into the inequality. For example, the point (0,0) is a convenient test point. Plugging in x=0 and y=0 into the inequality, we get:

5(0) + 12(0) ≤ 30

which simplifies to 0 ≤ 30. Since this is true, we know that the region containing the point (0,0) is part of the shaded region, and therefore the shaded region is the one that contains the origin.

Putting it all together, we can graph the inequality by drawing the line 5x+12y = 30 and shading the region below the line, as shown in the figure below:

    |

 3  |       x

    |      /

 2  |     /

    |    /

 1  |   /  

    |  /  

 0  | /    

    ------------

      0  6

The shaded region includes all the points below the line 5x+12y=30, including the points on the line.

The values of a = 17 and b = - 56

that make the following piece wise defined function both continuous and differentiable everywhere.

A continuous function in mathematics is one that doesn't change value unexpectedly due to discontinuities, or breaks in the function's continuity. A continuous function in mathematics is a function that causes the value of the function to continuously vary as a result of a continuous variation of the argument, or a change without a leap. Thus, there aren't any discontinuities—rapid changes in value.

conditions are:

1.

f(x) is continuous if,

[tex]\lim_{x \to \ 5^{-}}f(x)[/tex] = [tex]\lim_{x \to \ 5^{+}}f(x)[/tex] = f(5)

2.

differentiable if,

[tex]\lim_{x \to \ 5^{-}}f^'(x)[/tex] = [tex]\lim_{x \to \ 5^{+}}f^'(x)[/tex]

for 1st condition:

⇒ [tex]\lim_{x \to \ 5^{-}}f(x) = \lim_{x \to \ 5^{+}}f(x)[/tex]

⇒ [tex]\lim_{x \to \ 5^{-}}(-3x -6) = \lim_{x \to \ 5^{+}}(-2x^2+ax+b)[/tex]

⇒ (-3 × (-5) - 6) = - 2(5)² + a(5) + b

⇒ - 21 = - 50 + 5a + b

⇒ 5a + b = 29 ................... (1)

For 2nd condition:

⇒ [tex]\lim_{x \to \ 5^{-}}f^'(x) = \lim_{x \to \ 5^{+}}f^'(x)[/tex]

⇒ [tex]\lim_{x \to \ 5^{-}} \frac{d}{dx} (-3x - 6) = \lim_{x \to \ 5^{+}} \frac{d}{dx}(-2x^2 + ax + b)[/tex]

⇒ - 3 = - 4x + a

⇒ - 3 = - 4 (5) + a

⇒ a = 17

Substitute a = 17 in equation (1) we get:

5(17) + b = 29

or, b = - 56

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The complete question is as follows:

Using the Divergence Theorem, we get the flux of F across the closed surface S is π/2 (27√2 - 27).

To apply the Divergence Theorem, we will first compute the divergence of F as follows -

div(F) = ∂Fx/∂x + ∂Fy/∂y + ∂Fz/∂z

= 1 + 1 + 1

= 3

Now we will calculate the flux of F across the closed surface S that completely surrounds T. By the Divergence Theorem, this is equal to the triple integral of the divergence of F over the region T and is given by -

∫∫S F•n dS = ∭T div(F) dV

We can describe the region T using cylindrical coordinates:

0 ≤ r ≤ 2

-π/2 ≤ θ ≤ π/2

0 ≤ z ≤ √(9 - r sin θ)

The bounds on r and θ come from the fact that the half cylinder is contained within the plane x = 2, and the plane y = ±3. The bounds on z come from the equation of the half cylinder.

Now we can write the triple integral as follows -

∭T div(F) dV = ∫0^2 ∫-π/2^π/2 ∫0^√(9 - r sin θ) 3 r dz dθ dr

Evaluating this integral, we get,

∭T div(F) dV = π/2 (27√2 - 27)

Therefore, the flux of F across the closed surface S is π/2 (27√2 - 27).

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Answer:

3?

Step-by-step explanation:

im not sure sorry

Answer: 0

Step-by-step explanation:

Using GEMDAS

- Grouping

- Exponents

- Multiplication

- Division

- Addition

- Subtraction

[5-6+1]

= [-1+1] because you add/subtract from left to right

= [0] because -1+1=0.

Answer is 0

Answer:

Yes, dilation does not change the angle measurements.

Step-by-step explanation:

Answer: The sample space of possible outcomes is:

{SM, SJ, MJ, MS, JS, JM}.

There are 6 possible outcomes in the sample space, and 1 of these outcomes results in the first student being a boy (Franklin, represented by "F").

Therefore, the probability of the first student being a boy is P(A) = 1/6.

Step-by-step explanation:

Answer:

Fourth option: √65/4

Step-by-step explanation:

Since the angles are similar, each of the angles of ΔEFG must be equal to each of the corresponding angles of ΔHIJ

Specifically, m ∠J in ΔHIJ = m∠G in Δ EFG

Let's first find ∠G

Since EFG is a right triangle,

tan(G) = EF/FG = √65/4

tan J must be the same as tan G  = √65/4

This is the fourth  option

The value of tanJ is in the right angled triangle HIJ√65 / 4

An equation is an expression that contains numbers and variables linked together by mathematical operations of addition, subtraction, multiplication, division and exponents. An equation can either  be linear, quadratic, cubic, depending of the degree of the variable.

Two triangles are similar if their corresponding angles are equal to each other.

Given that triangle EFG and triangle HIJ are similar, hence:

Angle G = Angle J

tanJ = tanG = √65 / 4

The value of tanJ is  √65 / 4

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Answer:

  x = 5

Step-by-step explanation:

You want the length of x in the figure showing similar triangles.

The bottom edge of the triangle is 2/3 of the left edge.

The similar triangles have the same ratio of side lengths. That means x is 2/3 of (3+4.5):

  (2/3)(7.5) = 5

The length x is 5 units.

Answer:

21+15+9=45/4=that the answer

Answer:

Top one is 3/8

Middle is 4/7

Bottom is 1/3

Step-by-step explanation: