Ask question

Ask question

All categories

ivanzaharov [21]

3 years ago

5

What am i supposed to do??

1 answer:

miskamm [114]3 years ago

5 0

Exterior angle theorem
16x - 7 = 8x + 2 + 10x - 19 ===> 16x -7 = 18x - 17
16x - 16x - 7 = 18x - 16x - 17 ===> -7 = 2x - 17
-7 + 17 = 2x - 17 + 17 ===> 10 = 2x ======> 5 = x
Plug x in to each equation.
Angle P: 8(5) + 2 = 42 deg, Angle Q: 10(5) - 19 = 31
The sum of interior angles = 180. 180 - 42 -31 = 107 deg (angle of PRQ)
To check that line other angle : 16(5) -7 = 73 deg + 107 = 180 deg

You might be interested in

Solve for m.<br> 7<br> D= (m - 64)<br> 2<br> ==

DedPeter [7]

=(D-64)2

We move all terms to the left:

-((D-64)2)=0

We calculate terms in parentheses: -((D-64)2), so:

(D-64)2

We multiply parentheses

2D-128

Back to the equation:

-(2D-128)

We get rid of parentheses

-2D+128=0

We move all terms containing D to the left, all other terms to the right

-2D=-128

D=-128/-2

D=+64

7 0

4 years ago

H(n)=-2n^2+4; find h(4)

IceJOKER [234]

Answer: h(4) = 36

Step-by-step explanation:

All you have to do is plug in 4 to the equation.

So, h(4) = 2(4)^2+4

=2(16)+4

=32+4

=36

5 0

3 years ago

Given: m arc AB = 4x, m arc BC = 3x<br> m arc CD = 2x, m arc DA = 3x<br> Find: m∠APB

Vitek1552 [10]

Answer:

a > m/bp

Step-by-step explanation:

flip the equation

abp > m

divide both sides with bp

abp/bp > m/bp

a > m/bp

3 0

4 years ago

Read 2 more answers

Find the Perimeter and area

Maslowich

Perimeter is 36. This is because the radius and side lengths are the same in a regular hexagon.

Are is roughly 93.53.

6 0

4 years ago

Find x so that x, X +2, X'+3 are the first three terms of a geometric sequence. Then find the 5" term of the

velikii [3]

Answer:

The fifth term is -1/4.

Step-by-step explanation:

We know that the first three terms of the geometric sequence is <em>x, x</em> + 2, and <em>x </em>+ 3.

So, our first term is <em>x</em>.

Then our second term will be our first term multiplied by the common ratio <em>r</em>. So:

$x+2=xr$

And our third term will be our first term multiplied by the common ratio <em>r</em> twice. Therefore:

$x+3=xr^2$

Solve for <em>x</em>. From the second term, we can divide both sides by <em>x: </em>

$\displaystyle r=\frac{x+2}{x}$

Substitute this into the third equation:

$\displaystyle x+3=x\Big(\frac{x+2}{x}\Big)^2$

Square:

$\displaystyle x+3 = x\Big( \frac{(x+2)^2}{x^2} \Big)$

Simplify:

$\displaystyle x+3=\frac{(x+2)^2}{x}$

We can multiply both sides by <em>x: </em>

$x(x+3)=(x+2)^2$

Expand:

$x^2+3x=x^2+4x+4$

Isolate the <em>x: </em>

$-x=4$

Hence, our first term is:

$x=-4$

Then our common ratio <em>r</em> is:

$\displaystyle r=\frac{(-4)+2}{-4}=\frac{-2}{-4}=\frac{1}{2}$

So, our first term is -4 and our common ratio is 1/2.

Then our sequence will be -4, -2, -1, -1/2, -1/4.

You can verify that the first three terms indeed follow the pattern of <em>x</em>, <em>x</em> + 2, and <em>x</em> + 3.

So, our fifth term is -1/4.

4 0

3 years ago