Someone please help me with this

Find x and find unknown angle measure. Please hellppppppp

Kay [80]

Answers:

x = 13

Question mark angle is 64 degrees

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Explanation:

Let y be the angle labeled as a question mark.

This angle y is adjacent to the exterior angle of measure 116 degrees

This means y and 116 must add to 180 to form a straight angle

y+116 = 180

y = 180-116

y = 64

So 64 degrees will replace the question mark.

Then we can find the angle x. Recall that all angles of a triangle add to 180

90+2x+y = 180

90+2x+64 = 180

2x+154 = 180

2x = 180-154

2x = 26

x = 26/2

x = 13

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Alternatively, you can find x through the remote interior angle theorem

The interior angles 90 and 2x add to the exterior angle 116

90+2x = 116

2x = 116-90

2x = 26

x = 26/2

x = 13

This method doesn't involve angle y.

6 0

3 years ago

I need help with this question i dont understand it

RUDIKE [14]

The perimeter, by definition, is the outside measure of that figure. MN and LM are the same length and LK and NK are the same length....we just need to find the lengths! Use the distance formula to find the distance between the 2 points:
$\sqrt{( x_{2} - x_{1} ) ^{2} +( y_{2}- y_{1} ) ^{2} }$
For the segment MN, use the coordinates of M as your x1, y1, and use the coordinates of N for x2, y2:
$\sqrt{(3-2) ^{2}+(4-3) ^{2} }$
which simplifies to
$\sqrt{(1 )^{2}+(1) ^{2} }$ which is $\sqrt{2}$
So that is the length of both MN and LM. So far our perimeter is $\sqrt{2} + \sqrt{2}=2 \sqrt{2}$
Now let's use the same formula to find out the length of one of the longer segments:
$\sqrt{(5-3) ^{2} +(3-2) ^{2} }$
which simplifies down to
$\sqrt{(2) ^{2} +(1) ^{2} }$
which is of course $\sqrt{5}$
Since we have 2 of those lengths, $\sqrt{5} + \sqrt{5}=2 \sqrt{5}$
So our perimeter is, in the end, $2 \sqrt{2}+2 \sqrt{5}$
That's the third choice down

7 0

3 years ago

Two points are located at (−9,−8) and (−6,−4).

Mrac [35]

Answer:

$l(AB)=5\ units$

is the distance between the point A and point B.

Step-by-step explanation:

Let

A ≡ (x₁, y₁) ≡ (-9 , -8 )

B ≡ (x₂, y₂) ≡ (-6 , -4 )

Now by Pythagoras Distance formula we have

a² + b² = c²

$l(AB) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}\\\\l(AB) = \sqrt{((-6--9)^{2}+(-4--8)^{2} )}\\\\l(AB)=\sqrt{(3)^{2}+(4)^{2} }\\\\l(AB) =\sqrt{25} \\\\l(AB) = 5\ units$