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g100num [7]
3 years ago
13

GEOMETRY HELP WILL MARK BRAINLIEST

Mathematics
2 answers:
daser333 [38]3 years ago
8 0

Answer:

x = 5 and z = 108

Step-by-step explanation:

We have two intersection lines and since z and 108 lie opposite each other, they are called vertical angles. By definition, vertical angles are congruent to each other, so that means z = 108.

Now, notice that z and 6x + 42 lie on the same line. A line has a measure of 180 degrees, which means that these two will add up to 180 (they are supplementary angles). Then we can write:

z + (6x + 42) = 180

We already know that z = 108, so plug that in and solve for x:

108 + (6x + 42) = 180

6x + 150 = 180

6x = 30

x = 5

Thus, x = 5 and z = 108.

Marysya12 [62]3 years ago
7 0

Answer:

z=108

x=5

Step-by-step explanation:

First, let's find z.

z is opposite from an angle that measures 108 degrees. Since they are opposite, they are vertical angles. This means that they are congruent or equal.

z=108

Next, let's find x.

108 and 6x+42 are on a straight line, which means they are supplementary, and add to 180 degrees.

6x+42+108=180

Combine like terms by adding 42 and 108

6x+150=180

Now, we have to solve for x, by getting x by itself.

First, subtract 150 from both sides, because 150 is being added to 6x.

6x+150-150=180-150

6x=180-150

6x=30

Next, divide both sides by 6, because 6 and x are being multiplied.

6x/6=30/6

x=30/6

x=5

So, z=108 and x=5

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avanturin [10]
Hey!

The lower quartile is 13
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Hope this helps!
3 0
3 years ago
Read 2 more answers
Can you prove it??<br>it's hard,I tried but couldn't solve this.
BARSIC [14]

I'll abbreviate s=\sin\theta and c=\cos\theta, so the identity to prove is

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)

On the left side, we can simplify a bit:

\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}

\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}

Then

\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)

So the establish the original equality, we need to show that

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s

Combine the fractions:

\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}

We can rewrite the denominator as

c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1

then using the fact that c^2+s^2=\cos^2\theta+\sin^2\theta=1, we get

1-2s^2+2s-1=2s-2s^2

so that we have

\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s

as desired.

6 0
4 years ago
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Aleksandr [31]

Answer:

72 units

Step-by-step explanation:

There are two ways to solve the problems.

Count graphically, horizontally there are 9 units, vertically there are 8 units in the rectangle.  So the area is 9*8 = 72 units.

Alternatively, check the difference between two adjacent vertices,

1. between (4,-3) and (4,5), the difference is (0,8), or 8 units.

2. between (4,5) and (-5,5), the difference is (9,0), or 9 units.

So again, the area of the square is 8*9=72 units.

5 0
3 years ago
Write an<br> equation of a line with the given slope and y-intercept.<br> m=-2, b=3
GuDViN [60]

Answer:

y=-2x+3

Step-by-step explanation:

y=mx+b

m=-2,b=3

so, y=-2x+3

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