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2 years ago

James has $32 and earns $10 per week for his allowance. What is the initial value for the scenario described?

Mathematics

2 answers:

Verdich [7]2 years ago

5 0

This answer is confirmed!

stealth61 [152]2 years ago

3 0

Your answer should be B.

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Help? I’ll mark u brainliest

andrew-mc [135]

Answer:

11 because adjacent is basically next to and NKL is 12

Step-by-step explanation:

6 0

2 years ago

D (3) (-9) = .<br> H-24 = 4 =

Softa [21]

Answer:

r u trying to look for the values of D and H?

D (3)(-9) = 0

D -27 = 0

D = 27

H - 24 = 4

H = 4 + 24

H = 28

7 0

3 years ago

3. The box-and-whisker plot below shows the heights,in inches, of the students in a 7th grade class.What percentage of the heigh

Fantom [35]

(3) 62.5% cuz u take the total number of the students and u divide it by the number of students between 60 to 65

7 0

2 years ago

In how many ways can the letters of the word ``COPYRIGHT'' be arranged?

PilotLPTM [1.2K]

Answer:

362,880 ways

Step-by-step explanation:

There are 9 letters so 9!

And none of them are repeated so 9!/0!

9! = 362,880

I hope this helps, and plz mark me brainliest!!

4 0

3 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.
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7 0

3 years ago