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seropon [69]
2 years ago
8

Someone pls help me with this

Mathematics
1 answer:
Vesna [10]2 years ago
5 0

Answer:

1 =125

2=55

3=95

...........

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Find an order pair (x, y) that is a solution to the equation <br> x-3y=3<br> (x,y) =
aniked [119]

Answer:

(6,1)

Step-by-step explanation:

If you plug these numbers in for x and y, the equation works.

4 0
2 years ago
Find the area of the following figure:
stellarik [79]

Answer:

12m

Step-by-step explanation:

You can break this shape down into a square and a trapezoid. Frist you can find the area of the square by multiplying 2 by 5 to get 10. Then you would find the area of a trapezoid by doing Area = 1/2height(base1+base2). We can find the height by subtracting 5 from 9. (We do this because we know the side of the square is 5m) Therefore the height would be 4m. We know the bases are 2 and 4. From there you can plug those numbers in to the formula. Area=1/2 * 4 (2 + 4)

Area= 1/2 * 4 (6)

Area = 2(6)

Area = 12m

4 0
3 years ago
Find the value of x. <br><br> A. 60<br><br> B. 50 <br><br> C. 40 <br><br> D. 30
garri49 [273]

We can see that two sides of that triangle area equal

so, their corresponding angle will also be equal

now, we know that

sum of all angles in any triangle is always 180

so, we can get equation as

x+100+x=180

now, we can solve for x

2x+100=180

2x=80

x=40

so, option-C..............Answer

8 0
3 years ago
Read 2 more answers
Which equation is equivalent to the given equation?
madam [21]

Answer:

x² - x - 12 = 0

Step-by-step explanation:

0 = -x² + x + 12

→ Add x² to both sides

x² = x + 12

→ Minus x from both sides

x² - x = 12

→ Minus 12 from both sides

x² - x - 12 = 0

4 0
2 years ago
Please Help!!
algol13
When you see questions of this nature, test the individual inequalities and look out for their intersection.

For
y < \frac{2}{3} x
Choose a point in the lower or upper half plane created by the line
y = \frac{2}{3} x
The above line is the one which goes through the origin.

Now testing (1,0) yields,

0 < \frac{2}{3} (1)
That is,

0 < \frac{2}{3}
This statement is true. So we shade the lower half of
y = \frac{2}{3} x

For
y \geqslant - x + 2
We test for the origin because, it is not passing through the origin.

0 \geqslant - (0) + 2
This yields
0 \geqslant 2
This statement is false so we shade the upper half.

The intersection is the region shaded in B. The top right graph
8 0
3 years ago
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