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Serga [27]
4 years ago
10

Theorems! Please help. Problem in picture.

Mathematics
1 answer:
beks73 [17]4 years ago
6 0
I would review this back again later.
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Ill give brainliest - please help ASAP<br><br> WITH THE WORK THO PLEASE?
ohaa [14]

Answer:

5.

we have.

11x=½(16x+16+50)

22x=16x+66

22x-16x=66

6x=66

x=66/6

x=11

6.

118°=½(6x-11+181)

236=6x+170

6x=236-170

x=66/6

x=11°

7.

34°=½(arcAK-arc LN)

34×2=(110-arc LN)

arc LN =110-68

arc LN=42°

8.

<L=½(arc EN-arc KM)

<L=½(139°-73°)

<L=33

4 0
3 years ago
I need help plz will give brainliest
lesya [120]

Answer:

$10,000

Step-by-step explanation:

Cross multiply and divide by 55.

Hope this helps!

5 0
3 years ago
Read 2 more answers
Which equation is the inverse of y=x^2+16
stealth61 [152]
The inverse of the equation is the square root of x-16. In order to find inverse switch the y and x values and try to isolate y
5 0
3 years ago
Which shows the expressions rewritten with the least common denominator?
baherus [9]

Answer:

  D.  (14x-4)/(8x^2) and (x^2-x)/(8x^2)

Step-by-step explanation:

The least common denominator will be 8x^2, the product of 2 and x to the highest of their powers in either of the denominators.

  \left\{\dfrac{7x-2}{4x^2},\ \dfrac{x-1}{8x}\right\}=\left\{\dfrac{7x-2}{4x^2}\cdot\dfrac{2}{2},\ \dfrac{x-1}{8x}\cdot\dfrac{x}{x}\right\}\\\\=\left\{\dfrac{14x-4}{8x^2},\ \dfrac{x^2-x}{8x^2}\right\}

7 0
2 years ago
Graph the image of this figure after a dilation with a scale factor of 3 centered at (-7,-6)
Yuri [45]
Given that the triangle is dilated by factor 3, the image will be found as follows;
The object is at:
A(-7,-3), B(-3,-2), C(-4,-5)
when the image was enlarged the new coordinates will be:
A'=3(-7,-3)=(-21,-9)
B'=3(-3,-2)=(-9,-6)
C'=3(-4,-5)=(-12,-15)
since the image is centered at the point (-7,-6), the final point will be at:
A"=[(-21+-7),(-9+-6)=(-28,-15)
B''=[(-9+-7),(-6+-6)]=(-16,-12)
C''=[(-12+-7),(-15+-6)]=(-19,-17)
thus the coordinates of the final image are:
A''(-28,-15),B''(-16,-12),C''(-19,-17)
3 0
3 years ago
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