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Serhud [2]
2 years ago
10

The graph of y = 4 – is shown.

Mathematics
1 answer:
Ratling [72]2 years ago
7 0
Rather than try to select the correct answer without doing work, I'd rather do the work first and the matching later.

Given y = 4 - x, or    y = 4 - 1x,

1) the y-intercept is (0,4).
2) this being a linear function, it has neither maximum nor minimum.
3) the x-intercept is found by letting y = 0 and solving for x:  0 = 4 - x; x - 4.  The x-intercept is (4,0).

Which of the given statements is true?
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Trigonometry
Leona [35]

Answer:

7 units

Step-by-step explanation:

These points lie on the same horizontal line.

No vertical distance

Only horizontal distance of:

15-8 = 7 units

6 0
3 years ago
Read 2 more answers
I Need Help With My Work <br><br>x-9=-5
Ede4ka [16]
Add 9 to both sides to get the x-value alone

x-9=-5
x=4
6 0
3 years ago
ANOTHER ONE, The graph of a linear function is shown.Which word descibes the slope?
Trava [24]

This is a negative graph because the graph go to the left with a negative slope. Y=-1/2x
5 0
3 years ago
Find the length of the diagonal of a square with perimeter of 24
julsineya [31]

Answer:

d = 6sqrt(2) or 8.4853

Step-by-step explanation:

<u><em>Formula</em></u>

P = 4*s

s^2 + s^2 = d^2 where d is the diagonal and s is the side.

<u><em>Givens</em></u>

P = 24

<u><em>Solution</em></u>

P = 4s         Substitute for s

24 = 4*s      Divide by 4

24/4 = s

s = 6

================

d^2 = s^2 + s^2

d^2 = 6^2 + 6^2

d^2 = 36 + 26

d^2 = 72

d = sqrt(72)

Factors of 72

72: 6 * 6 * 2

<em><u>Rule</u></em>: Every pair of = factors allows you to take one of them outside the sqrt sign and throw the other a way. If there are no pairs, whatever you started with stays under the root sign.

sqrt(6*6*2) = 6sqrt(2)

  • The diagonal length is either
  • d = 6*sqrt(2) or
  • d = 8.4853

8 0
3 years ago
A rectangular mat has a length of 12 in. and a width of 4 in. Drawn on the mat are three circles. Each circle has a radius of 2
Andru [333]
The answer is 0.79!! You're welcome
4 0
2 years ago
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