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Lerok [7]
3 years ago
7

Can u find a it please

Mathematics
1 answer:
blsea [12.9K]3 years ago
6 0

Answer:

3 + 3 + 1

Step-by-step explanation:

The doubles is 3+3 which is 6 plus 1 is 7

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What’s the lcd for 1/10& 7/45
NeTakaya
The lowest common denominator is 9/90 and 14/90, multiplying 1/10 by 9/9 and 7/45 by 2/2. I am assuming you know how to solve for LCD? If you don't I can explain in comments.
8 0
4 years ago
What is the slope of the line that passes<br> through<br> 0(-4, 8) and (5, 2)?
zlopas [31]

Answer:

(-8,-4) and (-2,5). at reflect on x axis

4 0
3 years ago
10g - 4 (g - 2) + 11 = 1
sergij07 [2.7K]

Answer:

g = -3

Step-by-step explanation:

10g-4(g-2)+11=1

1) 10g-4g+8+11=1  Combine like terms

2) 6g+8+11=1

3) 6g+19=1

4) 6g=-18   To continue the process of getting the variable by itself, we have to get rid of 19. So we are going to subtract 19 from 19 to get 0, and whatever we do to the other side, so we subtract 19 from 1 to get -18.

5)6g=-18    To get the variable by itself, instead of mutiplying, we will divide by 6 on both sides to get... g=-3! Hope this helps!

5 0
3 years ago
What is the equation of this graph? Please answer this I am soo stuck.
OLEGan [10]

Answer:

y = 8x + 50

Step-by-step explanation:

Use (0,50) and (10,130)

Slope: m = (130-50)/(10-0) = 8

Y-intercept is 50

y = 8x + 50

3 0
3 years ago
Karina is designing a logo on a coordinate plane. The logo is a square whose area is 200 square units. She places a circle in th
Katarina [22]

Answer:

D

Step-by-step explanation:

To find the percentage we must first find the area of the circle:

We can find the radius of the circle using the distance formula;

⇒ d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}

⇒ r = \sqrt{(6.3 - 0)^{2} + (1.6 - 0)^{2}}

⇒ r = 6.5

Now we calculate the area;

⇒ A = \pi r^{2}

⇒ A = \pi (6.5)^{2}

⇒ A = \frac{169}{4}\pi

To get the percentage we divide the area of the circle by the square’s are and multiply by 100;

⇒ Percentage = \frac{42.25\pi }{200} * 100

⇒ Percentage = 66 (2sf)

5 0
2 years ago
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