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alina1380 [7]
3 years ago
6

which is the rationalized form of the expression the square root of x divided by the square root of x +the square root of 7

Mathematics
1 answer:
Ilya [14]3 years ago
8 0

Answer:

I assume your goal is to rationalize the denominator -

So you need to multiply the denominator by its conjugate.

So we have:

\frac{\sqrt{x} }{\sqrt{x}-\sqrt{7} } *\frac{\sqrt{x} +\sqrt{7} }{\sqrt{x} +\sqrt{7}}

->

\frac{x-\sqrt{7}\sqrt{x} }{x-7}

gg

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Which fraction is not equivalent to 7 /30
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3 years ago
The graph of which function(s) will contain the point (4, 12)? y = x + 8 y = 3x y = 2x y = x + 6
Alex777 [14]

Answer:

y=x+8  and y=3x

Step-by-step explanation:

The Given point is (4,12)

now we can substitute the point in each of the function and check wether it lies on it or not.

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substitute y=12,x=4

12=4+8=12 which is true

B.  y=3x

substitute y=12,x=4

12=3*4=12 which is true

C.  y=2x

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4 0
3 years ago
Express answer in exact form.
bazaltina [42]
The answer is 5.13 in²

Step 1. Calculate the diameter of the circle (d).
Step 2. Calculate the radius of the circle (r).
Step 3. Calculate the area of the circle (A1).
Step 4. Calculate the area of the square (A2).
Step 5. Calculate the difference between two areas (A1 - A2) and divide it by 4 (because there are total 4 segments) to get <span>the area of one segment formed by a square with sides of 6" inscribed in a circle.
</span>
Step 1:
The diameter (d) of the circle is actually the diagonal (D) of the square inscribed in the circle. The diagonal (D) of the square with side a is:
D = a√2            (ratio of 1:1:√2 means side a : side a : diagonal D = 1 : 1 : √2)
If a = 6 in, then D = 6√2 in.
d = D = 6√2 in

Step 2.
The radius (r) of the circle is half of its diameter (d):
r = d/2 = 6√2 / 2 = 3√2 in

Step 3.
The area of the circle (A1) is:
A = π * r²
A = 3.14 * (3√2)² = 3.14 * 3² * (√2)² = 3.14 * 9 * 2 = 56.52 in²

Step 4.
The area of the square (A2) is:
A2 = a²
A2 = 6² = 36 in²

Step 5:
(A1 - A2)/4 = (56.52 - 36)/4 = 20.52/4 =  5.13 in²
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3 years ago
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