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Serhud [2]
3 years ago
9

Can you please help me thanks

Mathematics
1 answer:
mash [69]3 years ago
6 0

Answer:

x = 9.6

Step-by-step explanation:

Before we can figure out what x is, we need to figure out what the unlabeled side is. To figure that out, multiply the hypotenuse (12) by the sine of the angle labeled (34)

12 * sin(34)

<em>sin(34) equals 0.559192903470747; I'll round it to 0.6 for convenience.</em>

12 * 0.6

<em>Now simply multiply 12 by 0.6 to get 7.2.</em>

The unlabeled side is approx. 7.2 units long.

Now we know what the unlabeled side is. Now, to find x, find the square root of 12 squared minus 7.2 squared.

x = √12² - 7.2²

<em>12 squared is 144; 7.2 squared is 51.84.</em>

x = √144 - 51.84

<em>Subtract 51.84 from 144 to get 92.16.</em>

x = √92.16

<em>The square root of 92.16 is 9.6 (on the spot!).</em>

x equals 9.6.

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8 0
3 years ago
DF=78, DE=5x-9, and EF-2x+10, find EF
lbvjy [14]

Answer:

As e is point between df. So we can write df as

df = de + ef

As given df = 78 , de = 5x - 9 , ef = 2x + 10.

So we can write

78 = 5x - 9 + 2x + 10

78 = 7x + 1

On subtracting 1 on booth side

78 - 1 = 7x +1 - 1

77 = 7x

On dividing both side by 7.

11 = x

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Step-by-step explanation:

8 0
3 years ago
PLEASE HELP.
krek1111 [17]

Answer:

Part a) The radii are segments AC and AD and the tangents are the segments CE and DE

Part b) DE=4\sqrt{10}\ cm

Step-by-step explanation:

Part a)

we know that

A <u>radius</u> is a line from any point on the circumference to the center of the circle

A <u>tangent</u> to a circle is a straight line which touches the circle at only one point. The tangent to a circle is perpendicular to the radius at the point of tangency.

In this problem

The radii are the segments AC and AD

The tangents are the segments CE and DE

Part b)

we know that

radius AC is perpendicular to the tangent CE

radius AD is perpendicular to the tangent DE

CE=DE

Triangle ACE is congruent with triangle ADE

Applying the Pythagoras Theorem

AE^{2}=AC^{2}+CE^{2}

substitute the values and solve for CE

14^{2}=6^{2}+CE^{2}

CE^{2}=14^{2}-6^{2}

CE^{2}=160

CE=\sqrt{160}\ cm

CE=4\sqrt{10}\ cm

remember that

CE=DE

so

DE=4\sqrt{10}\ cm

7 0
3 years ago
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