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Otrada [13]
3 years ago
13

As a motivational speaker, Bree speaks at school assemblies, charging a school

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
4 0

Answer:

<em>Bree spoke at 55 assemblies in this district.</em>

Step-by-step explanation:

Bree speaks at school assemblies charging a school district a total cost C modeled by the equation:

C(a) = 50a+1,500

Being a the number of assemblies attending the meeting.

It's given Bree charged the Escambia City School District $4,250  to speak at school assemblies.

We can find the number of assemblies by solving the equation:

50a+1,500=4,250

Subtracting 1,500:

50a=4,250-1,500=2,750

Dividing by 50;

a = 2,750 / 50=55

Bree spoke at 55 assemblies in this district.

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Answer:

it equals 91

Step-by-step explanation:

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3 years ago
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The area of a trapezoid is 60 square inches. One of the bases is 8 inches long. The height is 6 inches long. What is the length
Mashcka [7]

Answer:

12 in

Step-by-step explanation:

The area (A) of a trapezoid is calculated as

A = \frac{1}{2} h(b₁ + b₂ )

where h is the height and b₁, b₂ the parallel bases

Given h = 6, b₁ = 8 and A = 60 , then

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8 + b₂ = 20 ( subtract 8 from both sides )

b₂ = 12

The length of the second base is 12 inches

6 0
3 years ago
Julie is digging a square area in her backyard a for a vegtable garden. If the area is 52 square feet, what is the approximate l
AURORKA [14]

Answer:

<em>The correct option will be:   (b) 7 feet.</em>

Step-by-step explanation:

Suppose, the length of the garden is  x feet.

As the <u>shape of the garden is like a square, then the area of the garden</u> will be: (length)^2 = x^2 feet²

Given that, the <u>area is 52 square feet</u>. So, the equation will be.......

x^2= 52\\ \\ x=\sqrt{52} = 7.211.... \approx 7

So, the approximate length of the garden will be 7 feet.

4 0
3 years ago
Which equation is parallel to the line LaTeX: y=\frac{1}{2}x+3y = 1 2 x + 3and passes through the point (10, -5)?
sukhopar [10]

Answer:

Equation\ of\ line:\ y=\frac{1}{2}x-10

Step-by-step explanation:

Let\ the\ required\ equation\ is\ y=mx+c\\\\where\ m\ is\ the\ slope\ of\ the\ equation\ and\ c\ is\ y-intercept\\\\It\ is\ parallel\ to\ the\ equation\ y=\frac{1}{2}x+3\\\\Hence\ slope\ of\ these\ two\ lines\ will\ be\ same.\\\\Slope\ of\ y=\frac{1}{2}x+3\ is\ \frac{1}{2}\\\\Hence\ slope\ of\ y=mx+c\ is\ \frac{1}{2}\Rightarrow m=\frac{1}{2}\\\\Equation:y=\frac{1}{2}x+c\\\\Line\ passes\ through\ (10,-5).\ Hence\ this\ point\ satisfies\ the\ equation\ of\ line.\\\\-5=\frac{1}{2}\times 10+c

-5=-5+c\\\\c=-10

Equation\ of\ line:\ y=\frac{1}{2}x-10

8 0
3 years ago
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Answer:

See if this will help,

https://www.pmschools.org/site/handlers/filedownload.ashx?moduleinstanceid=701&dataid=3612&FileName=2015%20triangle%20proofs%20answers.pdf



8 0
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