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2 years ago

Please help me this is due in an hour!

Mathematics

2 answers:

lawyer [7]2 years ago

8 0

Answer:

117

Step-by-step explanation:

it makes alternate exterior with the 177 angle

Lisa [10]2 years ago

4 0

Answer:

117°

Step-by-step explanation:

y° = 180° - 48° - 44°

y° = 88°

since opposite angles are equal,

2z° + 2y° = 360°

2z + (2*88)° = 360°

2z° = 360°-176°

2z° = 184°

z° = 92°

Now, check the picture attached.

44° + a° + 87° = 180°

a = 49°

Now, a° + b° + z ° = 180°

b° = 180° - 49° - 92°

b = 39°

a quadrilateral has four angles that sums up to 360°

thus,

117° + (39° + 48°) + (44° + 49°) + (c° + 39°) = 360°

c° + 39° = 63°

c° = 24°

Finally,

c° + b° + x° = 180°

x° = 180° - 24° - 39°

x° = 117°

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Answer:

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3 years ago

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Which equations are related equations to x + 4 = 15? Select each correct answer x=15 - 4 4= 15 - x (H) x = 15 + 4 cm x + 4 = 15

Mashutka [201]

Answer:

x=15-4

Step-by-step explanation:

because when you are doing a solving a equation you need to solved for x and

15 -4= x

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3 years ago

How to do this ?!!!!

kykrilka [37]

I actually forgot what a trapezoid was and had to look it up.
Anyway, in an isosceles trapezoid the two legs are the same length, so we now have 3x-9=7x-10. Now it's simply a problem where you solve for x.
3x-9=7x-10
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1=4x
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Hope this helps :)

4 0

3 years ago

A curve is traced by a point P(x, y) which moves such that its distance from the point A(-1,1) is three times its distance from

docker41 [41]

Answer:

$8x^2+8y^2+43-38x+20y=0$

Step-by-step explanation:

Let $A\left ( x_1,y_1 \right )$ and $B\left ( x_2,y_2 \right )$ be two points then distance AB is equal to $AB=\sqrt{\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2}$

Here, a curve is traced by a point $P(x,y)$ which moves such that its distance from the point $A(-1,1)$ is three times its distance from the point $B(2,-1)$ i.e $AP=3BP$

Using distance formula,

$AP=\sqrt{(-1-x)^2+(1-y)^2}$

$BP=\sqrt{(2-x)^2+(-1-y)^2}$

$AP=3BP\\\sqrt{(-1-x)^2+(1-y)^2}=3\sqrt{(2-x)^2+(-1-y)^2}$

On squaring both sides, we get

$(-1-x)^2+(1-y)^2=9\left [ (2-x)^2+(-1-y)^2 \right ]\\1+x^2+2x+1+y^2-2y=9\left ( 4 +x^2-4x+1+y^2+2y\right )\\1+x^2+2x+1+y^2-2y=36+9x^2-36x+9+9y^2+18y\\8x^2+8y^2+43-38x+20y=0$

So, equation of curve is $8x^2+8y^2+43-38x+20y=0$

5 0

3 years ago