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Levart [38]
3 years ago
15

PLEASE HELP ME QUESTION IS AT THE BOTTOM OF THE PICTURE

Mathematics
1 answer:
ivolga24 [154]3 years ago
4 0

Answer:

A 28 and 45

Step-by-step explanation:

So keep in mind that columns must add up to the total numbers, and rows must add up to their total numbers.

We know that what goes in the box canoeing + swimming is

x + 43 = 71.

It needs to add up to 71!

x = 71 - 43 = 28

We know that what goes in the box no swimming + no canoeing is

34 + y = 79

It needs to add up to 79.

y = 19 - 34 = 45

So the answer is <u>A. 28 and 45.</u>

Now, I did this with the columns, but you could easily do this with the rows and get the same result!

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The perimeter of a triangle is 82 feet. One side of the triangle is 2 times the second side. The third side is 2 feet longer tha
andrezito [222]

Answer:

first side=40

second side=20

third side=22

Step-by-step explanation:

LET THE SIDES OF THE TRIANGLES BE y

Perimeter=82

first side=2 x y = 2y

second side=y

third side=2 +y

2y + y + 2 + y=82

4y+2=82

4y= 82-2

4y=80

y=80/4

y=20

first side=2y=2(20)=40

second side=y=20

third side= 2 + y= 2 + 20 = 22

3 0
3 years ago
Please help me on this I’m so tired and I can’t sleep until my progress is at 80 and this is the end of my year it’s at 79%
Elena L [17]

Answer:

The answer is the bottom left

4 0
3 years ago
Find the x-coordinate of the vertex, y-coordinate of the vertex, and the x-intercepts.
galina1969 [7]

Answer:

X=-1

Y=-9

Step-by-step explanation:

x^2+2x-8

x^2+2x +1 =8+1

(x+1)-9

3 0
3 years ago
7.
Anuta_ua [19.1K]

Answer:

A. y ≥ 2x – 2

Step-by-step explanation:

4 0
3 years ago
What is the square root of <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B163%7D%20" id="TexFormula1" title=" \sqrt{163} " a
crimeas [40]

Hi,

\sqrt{163} = 12.76

There are no other simple steps to reduce the square root of one hundred and sixty-three.

The actual square root value is ≈12.767145334803704

But we have taken only significant figures and it is as 12.76.


Hope it helps! :)

5 0
3 years ago
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