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Anestetic [448]
3 years ago
6

Under his cell phone plan, Ian pays a flat cost of $59.50 per month and $5 per gigabyte. He wants to keep his bill under $80 per

month. Which inequality can be used to determine xx, the maximum number of gigabytes Ian can use while staying within his budget?
Btw this is Delta math Inequalities in Context (MC)
Mathematics
1 answer:
inessss [21]3 years ago
3 0
You would start by saying his budget is $80 so you start with 80= he pays an initial fee of $59.50 every month plus $5 per gigabyte so you would set it up like
80=59.50+5x
X being how many gigabytes he uses then you solve 80-59.50 is 20.5 divided by 5 gives you 4.1 but you round down cause you don’t want to pass the limit so therefor the most gigabytes he can use is 4
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If the sum of a number and eight is doubled, the result is three less than the number. Find the number.
nydimaria [60]

Answer:

n = -19

Step-by-step explanation:

2(n+8) = n - 3

2n + 16 = n - 3

2n = n - 19

n = - 19

8 0
2 years ago
A training field is formed by joining a rectangle and two semicircles, as shown below. The rectangle is 86m long and 64m wide. F
KIM [24]

To solve this problem, we have to find the area of the semi circles and the area of the rectangle and add them up together.

The area of the field is equal to 8719.36m^2

<h3>Area of the Field</h3>

Let us start by finding the area of the rectangle.

Data;

  • width = 86m
  • length = 64m

Area of a rectangle is given as

a = length * width \\a = 86 * 64\\a = 5504m^2\\

The area of the rectangle is 5504m^2

Area of the semicircle

The formula of a semi circle is given as

A = \frac{1}{2} \pi r^2

But since we have two semi circles here, we can simply multiply it by two and the formula becomes

A = 2 * \frac{1}{2} \pi r^2\\A = \pi r^2

But in the question, we have the diameter of the semi circle as the length of the rectangle.

radius = \frac{diameter}{2} \\radius = \frac{64}{2} \\radius = 32m

The radius of the semicircle is 32m

Area of the semicircle is

A = \pi r^2\\A = 3.14 * 32^2\\A = 3215.36m^2

The area of the field is

area of field = 3215.36 + 5504\\area = 8719.36m^2

The area of the field is equal to 8719.36m^2

Learn more on area of rectangle and semi circle here;

brainly.com/question/2906652

#SPJ1

7 0
2 years ago
Rewrite the equation below in slope-intercept form.<br><br> 2y−8x=18
zavuch27 [327]

2y - 8x = 18

2y = 18 + 8x

y = 8x/2 + 18/2

y = 4x + 9

5 0
2 years ago
Read 2 more answers
7+4×-9=-6<br>please show the work
VikaD [51]

7+4=11 * -9= -99  hope this helped        

3 0
2 years ago
1/3 cup of flour is used to make 5 dinner rolls.
RoseWind [281]

we know that

\frac{1}{3} cup of flour is used to make 5 dinner rolls.

Part a) How much flour is needed to make one dinner roll?

by using proportion

\frac{\frac{1}{3}}{5}  =\frac{x}{1} \\ \\ x=\frac{\frac{1}{3}}{5}\\ \\ x=\frac{1}{15}

therefore

the answer Part a) is

To make one dinner roll is needed \frac{1}{15} cup of flour

Part b) How many cups of flour are needed to make 3 dozen dinner rolls?

Multiply the value obtained in part a) by 36

so

\frac{1}{15} *36=\frac{36}{15} \\ \\ =\frac{12}{5}

\frac{12}{5} =2\frac{2}{5}

therefore

the answer Part b) is

To make 3 dozen dinner rolls are needed 2\frac{2}{5} cup of flour

Part c) How many rolls can you make with 5 2/3 cups of flour?

5\frac{2}{3} =\frac{17}{3}

by using proportion

\frac{\frac{1}{3}}{5}  =\frac{\frac{17}{3}}{x} \\ \\ \frac{x}{3} =\frac{85}{3} \\ \\ x=85

therefore

the answer part c) is

85 rolls

6 0
3 years ago
Read 2 more answers
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