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Vera_Pavlovna [14]

4 years ago

13

Pine Bluff Middle School is having its annual Spring Fling dance, which will cost $400. The student treasurer reported that the

dance fund has $75 left over from last year. Each Ticket to the dance cost $4. Write an inequality to determine how many tickets must be sold to pay for this years's dance

1 answer:

Pavel [41]4 years ago

3 0

4x + 75 > 400
( I meant to write the more than, equal to sign but I can't do that, so it shows as the more than sign, sorry about that.)

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Quinn is building an enclosed pen in his backyard he wants the perimeter to be no more than 50 feet he also wants the length to

ycow [4]

Answer:

Option B width = 5 feet and length = 12 feet

The solution in the attached figure

Step-by-step explanation:

The options of the questions are

Which combination of width and length will meet Quinn’s requirements for the pen?

A. width = 7 feet and length = 20 feet

B. width = 5 feet and length = 12 feet

C. width = 15 feet and length = 10 feet

D. width = 11 feet and length = 15 feet

Let

x -----> the length of the enclosed pen in feet

y-----> the width of the enclosed pen in feet

we know that

The perimeter is equal to

$P=2(x+y)$

In this problem

$2(x+y)\leq 50$

Simplify

$(x+y)\leq 25$ ----> inequality A

$x\geq y+5$ ---> inequality B

using a graphing tool

The solution is the triangular shaded area

see the attached figure N 1

Remember that

The values of x and y cannot be a negative number

If a ordered pair is a solution of the system of inequalities, then the ordered pair must satisfy both inequality

<u><em>Verify each case</em></u>

case A) width = 7 feet and length = 20 feet

so

For y=7, x=20

<em>Check inequality A</em>

$(20+7)\leq 25$

$(27)\leq 25$ ----> is not true

therefore

This combination of width and length will not meet Quinn’s requirements for the pen

case B) width = 5 feet and length = 12 feet

so

For y=5, x=12

<em>Check inequality A</em>

$(12+5)\leq 25$

$(17)\leq 25$ ----> is true

<em>Check inequality B</em>

$12\geq 5+5$

$12\geq 10$ -----> is true

therefore

<u><em>This combination of width and length will meet Quinn’s requirements for the pen </em></u>

case C) width = 15 feet and length = 10 feet

so

For y=15, x=10

<em>Check inequality A</em>

$(10+15)\leq 25$

$(25)\leq 25$ ----> is true

<em>Check inequality B</em>

$10\geq 15+5$

$10\geq 20$ -----> is not true

therefore

This combination of width and length will not meet Quinn’s requirements for the pen

case D) width = 11 feet and length = 15 feet

so

For y=11, x=15

<em>Check inequality A</em>

$(15+11)\leq 25$

$(26)\leq 25$ ----> is not true

therefore

This combination of width and length will not meet Quinn’s requirements for the pen

Note If the ordered pair is a solution of the system of inequalities, then the ordered pair must lie on the shaded area

see the attached figure N 2 to better understand the problem

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4 years ago

f is a quadratic function whose graph is a parabola opening upward and has a vertex on the x-axis. The graph of the new function

Dmitriy789 [7]

<span> Answer: ( - infinity , 2]
No need to spam, it makes you lose your hard-earned points!</span>

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4 years ago

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Given an equation y= -8x -4. Identify the slope and y-intercept

Naddika [18.5K]

Answer:

Slope = - 8

y-intercept = - 4

Step-by-step explanation:

$y= -8x -4 \\ \\ equating \: it \: with \\ \\ y = mx + b \\ \\ slope \: (m) = - 8 \\ \\ y - intercept \: (b) = - 4$

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3 years ago

Use division to determine whether the ratios form a proportion.

galben [10]

No they do not form a proportion. If i'm wrong i'm sorry. Maybe I did my math wrong but I don't think they form a proportion .

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3 years ago

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A. A 5in radius ball is in a

abruzzese [7]

Answer:

52.33%

Step-by-step explanation:

Since, all six sides of the box are touching the ball.

So, all the three dimensions of the box are equal.

Hence, it is a cubical box.

Side length of the box (l)

= diameter of the ball

= 2*radius of the ball

= 2*5

= 10 In.

$V_{ball} = \frac{4}{3} \pi r^3$

$V_{ball} = \frac{4}{3} \times 3.14(5)^3$

$V_{ball} = \frac{4}{3} \times 3.14\times 125$

$V_{ball} = \frac{1,570}{3}\: In^3$

$V_{box} =l^3$

$V_{box} =(10)^3$

$V_{box} =1000\: In^3$

Percentage of the space in the box occupied by the ball

$= \frac{V_{ball}}{V_{box} } \times 100$

$= \frac{\frac{1,570}{3}}{1000} \times 100$

$= \frac{1570}{3000} \times 100$

$= \frac{157}{3}$

$= 52.33\%$

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3 years ago