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ch4aika [34]
4 years ago
11

GEOMETRY. image attached

Mathematics
1 answer:
Inessa [10]4 years ago
3 0
\bf \measuredangle DFC=\cfrac{1}{2}(\widehat{BG}+\widehat{CD})\implies 
\measuredangle DFC=\cfrac{\widehat{BG}+\widehat{CD}}{2}

you know what the arcCD is, and what the angle DFC is, so
solve for the arcBG
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What is 34827 x 7 =<br><br> Jasper over here..
kirill [66]

Answer:

243789

Step-by-step explanation:

(34827) 7 = 243789

Hope this helps!

3 0
2 years ago
Read 2 more answers
How would you solve this? help.
Eddi Din [679]

Answer:

<em>Center: (3,3)</em>

<em>Radius: </em>2\sqrt{5}<em />

Step-by-step explanation:

<u>Midpoint and Distance Between two Points</u>

Given two points A(x1,y1) and B(x2,y2), the midpoint M(xm,ym) between A and B has the following coordinates:

\displaystyle x_m=\frac{x_1+x_2}{2}

\displaystyle y_m=\frac{y_1+y_2}{2}

The distance between both points is given by:

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Point (5,7) is the center of circle A, and point (1,-1) is the center of the circle B. Given both points belong to circle C, the center of C must be the midpoint from A to B:

\displaystyle x_m=\frac{5+1}{2}=\frac{6}{2}=3

\displaystyle y_m=\frac{7-1}{2}=\frac{6}{2}=3

Center of circle C: (3,3)

The radius of C is half the distance between A and B:

d=\sqrt{(1-5)^2+(-1-7)^2}

d=\sqrt{16+64}=\sqrt{80}=\sqrt{16*5}=4\sqrt{5}

The radius of C is d/2:

r =4\sqrt{5}/2 = 2\sqrt{5}

Center: (3,3)

Radius: 2\sqrt{5}

8 0
3 years ago
If using the method of completing the square to solve the quadratic equation x^2 + 9 x + 7 = 0 , which number would have to be a
FinnZ [79.3K]

Answer:

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Step-by-step explanation:

6 0
3 years ago
Can help me on the question 17!?
dsp73

Answer:

18 feet

Step-by-step explanation:

5 years to 30 years is 6

6x5=30

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7 0
3 years ago
) dy 2x<br> ------ = ---------------<br> dx yx2 + y
larisa86 [58]

Step-by-step explanation:

\dfrac{dy}{dx} = \dfrac{2x}{y(x^2 + 1)}

Rearranging the terms, we get

ydy = \dfrac{2xdx}{x^2 + 1}

We then integrate the expression above to get

\displaystyle \int ydy = \int \dfrac{2xdx}{x^2 + 1}

\displaystyle \frac{1}{2}y^2 = \ln |x^2 +1| + k

or

y = \sqrt{2\ln |x^2 + 1|} + k

where I is the constant of integration.

6 0
3 years ago
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