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agasfer [191]
3 years ago
10

3/4 of the circle is shaded. What is the numerator ?

Mathematics
2 answers:
EleoNora [17]3 years ago
5 0

Answer: 3

Step-by-step explanation:

The numerator in the fraction 3/4 would be on top, so that would be 3.

ankoles [38]3 years ago
5 0

Answer:

the numerator is 3 for the shaded area, and 1 for the blank area

Step-by-step explanation:

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If h(x) = 5x − 3 and j(x) = −2x, solve h[j(2)] and select the correct answer below.
Sati [7]
Sub 2 for x in j(x) and evalueate
then sub that result for x in h(x)

j(2)=-2(2)
j(2)=-4

now sub-4 for  x in h(x)

h(-4)=5(-4)-3
h(-4)=-20-3
h(-4)=-23

answer is -23
8 0
3 years ago
A circle with a diameter of 2 inches and a square with 2-inch sides have the same center. Find the area of the region that is in
n200080 [17]
Area of square = 4 square inches
Area of circle = PI * radius ^2
Area of circle = 3.14 * 1^2
Area of circle = 3.14 square inches
Area inside the square and outside the circle =
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answer is .9 square inches


5 0
3 years ago
Find the point of intersection:<br> 3x=y-2<br> 6x=4-2y
IgorC [24]
Answer: (0, 2)

3x = y - 2
6x = 4 - 2y

3x = y - 2
- y - y
-y + 3x = - 2
- 3x - 3x
-y = - 3x - 2
/-1 /-1
y = 3x + 2


6x = 4 - 2y
+ 2y + 2y
2y + 6x = 4
- 6x - 6x
2y = - 6x + 4
/2 /2
y = -3x + 2


y = 3x + 2
y = -3x + 2


3x + 2 = -3x + 2
- 2 - 2
3x = -3x
/3 /3
x = 0


y = 3x + 2
y = 3(0) + 2
y = 2

(x, y) --> (0, 2)

Therefore, the two equations intersect at (0, 2)



Hope this helps!
7 0
3 years ago
Chad bought a box of raisins. he gave 1/6 of it to one brother, 1/4 to another brother, and he kept the rest for himself. what f
fenix001 [56]
Greetings!

This can be solved by creating a simple equation.
1/4+1/6+x=1

Solve for x.
1/4+1/6+x=1
Find the LCM.
3/12+2/12+x=1
5/12+x=1
Add -5/12 to both sides.
(5/12+x)+(-5/12)=(1)+(-5/12)
Simplify.
x=1/1-5/12
Find the LCM.
x=12/12-5/12
x=7/12

He kept 7/12 of the raisins for himself.

Hope this helps.
-Benjamin
4 0
3 years ago
I’m not sure how to do this
seraphim [82]
\bf \cfrac{\frac{x+4}{3}+\frac{1}{x}}{5+\frac{15}{x}}\qquad \cfrac{\impliedby \frac{LCD}{3x}}{\impliedby \stackrel{LCD}{x}}\implies \cfrac{\quad \frac{x(x+4)+3(1)}{3x}\quad }{\frac{x(5)+(1)15}{x}}\implies \cfrac{\quad \frac{x^2+4x+3}{3x}\quad }{\frac{5x+15}{x}}

\bf \cfrac{x^2+4x+3}{3\underline{x}}\cdot \cfrac{\underline{x}}{5x+15}\implies \cfrac{x^2+4x+3}{3(5x+15)}\implies \cfrac{x^2+4x+3}{15x+45}&#10;\\\\\\&#10;\cfrac{(x+3)(x+1)}{15x+45}\implies \cfrac{\underline{(x+3)}(x+1)}{15\underline{(x+3)}}\implies \cfrac{x+1}{15}
7 0
3 years ago
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