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natali 33 [55]
3 years ago
14

The least common denominator of the fractions 13/12x and 7/9x is

Mathematics
1 answer:
djverab [1.8K]3 years ago
8 0
It’s 36x!! hope this helps
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Please help, thank you!
maxonik [38]

Answer:

P = \frac{4}{25}\\

Step-by-step explanation:

Big circle area:

A=\pi r^{2} =\pi (10)^{2} =100\pi

Smaller circle área:

A=\pi (4)^{2} =16\pi

Probability:

P=\frac{16\pi }{100\pi } =\frac{4}{25}

Hope this helps

5 0
2 years ago
Math. You get 10 points
Mashcka [7]

Answer:

i think 55

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Please help, i’m not sure if i’m correct
hammer [34]

Answer:

Looks good to me! :)

Step-by-step explanation: Hope you have a great rest of your day!

7 0
3 years ago
What facts are true for the graph of the function listed below? Please check all that apply.
lana [24]

Answer:

E, B, A,

Step-by-step explanation:

C is wrong because it isn't decreasing

D is wrong because if X=0 the 5 to the power of zero would be 1 and 2*1= 2 meaning y goes lower than 5

F is wrong because  if X=0 the 5 to the power of zero would be 1 and 2*1= 2 meaning the y intercept is 2 not 5

6 0
3 years ago
Read 2 more answers
Passing through (2, - 2) and perpendicular to the line whose equation is y= 5x+2
jonny [76]

Step-by-step explanation:

Hey there!!!

Here,

Given, A line passes through point (2,-2) and is perpendicular to the y= 5x+2.

The equation of a straight line passing through point is,

(y - y1) = m1(x - x1)

Now, put all values.

(y  + 2) = m1(x - 2)

It is the 1st equation.

Another equation is;

y = 5x +2........(2nd equation).

Now, Comparing it with y = mx + c, we get;

m2=5

As per the condition of perpendicular lines,

m1×m2= -1

m1 × 5 = -1

Therefore, m2= -1/5.

Keeping the value of m1 in 1st equation.

(y + 2) =  \frac{ - 1}{5} (x - 2)

Simplify them.

5(y + 2) =  - x + 2

5y + 10 =  - x + 2

x + 5y + 8 = 0

Therefore the required equation is x+5y+8= 0.

<em><u>Hope it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

7 0
3 years ago
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