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zheka24 [161]
2 years ago
6

A side of the triangle below has been extended to form an exterior angle of 62°. Find the value of xx.

Mathematics
1 answer:
umka21 [38]2 years ago
3 0

Answer:

Photo?

Step-by-step explanation:

If there is a photo linked I dont see it

Also: Dont click the stpid link below its malware

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X - (-20) = 5 _________________
tamaranim1 [39]

X - (-20) = 5

When you subtract a negative, change it to addition:

X + 20 = 5

Subtract 20 from both sides:

X = -15

8 0
3 years ago
Read 2 more answers
Solve for x <br><br> (5x-1) 8x
trapecia [35]

Answer:

x = 7

Step-by-step explanation:

Given is the figure of a kite.

Opposite sides are equal... (given)

Since, measures of the angles opposite to equal sides are equal.

Also, diagonals of a kite interests at right angles.

Therefore,

8x \degree + (5x - 1) \degree + 90 \degree = 180 \degree \\  \\  (8x + 5x - 1) \degree   = 180 \degree -  90 \degree\\  \\(13x - 1) \degree   = 90 \degree\\  \\ 13x - 1 = 90 \\  \\ 13x = 90 + 1 \\  \\ 13x = 91 \\  \\ x =  \frac{91}{13}  \\  \\ x = 7

Therefore,

5 0
3 years ago
1) After a dilation, (-60, 15) is the image of (-12, 3). What are the coordinates of the image of (-2,-7) after the same dilatio
BARSIC [14]

Answer:

a) k = 5; (-10, -35)

Step-by-step explanation:

Given:

Co-ordinates:

Pre-Image = (-12,3)

After dilation

Image = (-60,15)

The dilation about the origin can be given as :

Pre-Image(x,y)\rightarrow Image(kx,ky)

where k represents the scalar factor.

We can find value of k for the given co-ordinates by finding the ratio of x or y co-ordinates of the image and pre-image.

k=\frac{Image}{Pre-Image}

For the given co-ordinates.

Pre-Image = (-12,3)

Image = (-60,15)

The value of k=\frac{-60}{-12}=5

or k=\frac{15}{3}=5

As we get k=5 for both ratios i.e of x and  y co-ordinates, so we can say the image has been dilated by a factor of 5 about the origin.

To find the image of (-2,-7), after same dilation, we will multiply the co-ordinates with the scalar factor.

Pre-Image(-2,-7)\rightarrow Image((-2\times5),(-7\times 5))

Pre-Image(-2,-7)\rightarrow Image(-10,-35) (Answer)

7 0
3 years ago
Please help me .... Thank you ​
Verizon [17]

\\ \sf\longmapsto \dfrac{4}{6x+2}

\\ \sf\longmapsto \dfrac{5x+3y}{16-8y}

\\ \sf\longmapsto \dfrac{2}{4x-3}

\\ \sf\longmapsto \dfrac{2ab+c^2}{3n-2m}

\\ \sf\longmapsto \dfrac{a^2-16}{7x}

3 0
2 years ago
I’ll give brainliest to first correct answer.
kirza4 [7]
Ok but u forgot to add the picture
4 0
3 years ago
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