All categories

2 years ago

Please help this is too hard for me

Mathematics

1 answer:

makvit [3.9K]2 years ago

8 0

Answer:

The answer would be 10 companies.

Step-by-step explanation:

All the numbers on the stem-lesf plot are under 60.

You might be interested in

A regular polygon has sides all the same length, and angles all the same measure. Find the pelmeter of the regular polygon below

Ede4ka [16]

The perimeter of the regular polygon is 70 inches

<h3>How to determine the perimeter of the regular polygon?</h3>

The sides of the regular polygon is given as:

Side = 10 in

The regular polygon has 7 sides

So, the perimeter of the polygon is calculated as:

P = Side lengths * Number of sides

This gives

P = 10 inches * 7

Evaluate

P = 70 inches

Hence, the perimeter of the regular polygon is 70 inches

Read more about perimeter at

brainly.com/question/24571594

#SPJ1

6 0

10 months ago

Jamiah bought two 125 ounce bags of flour for baking. He used up 32 ounces of flour to bake a batch of brownies. How much flour

V125BC [204]

Answer: its 218

Step-by-step explanation: you just have to subtract 250 minus 32

7 0

1 year ago

Find the slope of the function f ( x ) = 5x/2 + 3, by the definition of limit. Express answer as a fraction using the "/" key as

musickatia [10]

$\huge{\boxed{\frac{5}{2}}}$

Slope-intercept form is $f(x)=mx+b$ , where $m$ is the slope and $b$ is the y-intercept.

$\frac{5x}{2}=\frac{5}{2}x$ , so change the equation to represent this. $f(x)=\frac{5}{2}x+3$

Great! Now the function is in slope-intercept form, so we can just see that $m$ , or the slope, is equal to $\boxed{\frac{5}{2}}$

3 0

2 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
</span>

7 0

2 years ago

-7 x -8 = 56 True or False

dalvyx [7]

Answer:

true negative ×negative =positive

8 0

2 years ago

Read 2 more answers