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bekas [8.4K]
3 years ago
12

Can you please find the missing side these are similar figures.

Mathematics
1 answer:
storchak [24]3 years ago
8 0

Answer:

1.108

2.4

3.6

4.28

5.80

6. 10.5

hope this will help you

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Andrew [12]
Let's call the number x. So four times the number minus five would look like
4x-5
Twice the number plus three would look like
2x+3
Now make the two equations equal
4x-5=2x+3
I'm going to subtract 2x from each side
2x-5=3
Now I'm going to add five to each side
2x=8
Now divide both sides by 2
x=4
To double check, just plug back into the original equations
4(4)-5=11
2(4)+3=11
They are equal, so the answer is x=4.
7 0
3 years ago
Read 2 more answers
the question says write an equation for the nth term of the geometric sequence. Then find the fifth term in the sequence. hopefu
Ne4ueva [31]

Step-by-step explanation:

u can u sue math_way just take out the under score

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3 years ago
√secA+tanA/√secA-tanA × √cosecA-1/√cosecA+1=1
Snowcat [4.5K]

Use:\\\\\sec A=\dfrac{1}{\cos A}\\\\\tan A=\dfrac{\sin A}{\cos A}\\\\\csc A=\dfrac{1}{\sin A}\\\\\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{\dfrac{a}{b}}=\dfrac{\sqrt{a}}{\sqrt{b}}\\---------------------------------\\\\\sec A+\tan A=\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}=\dfrac{1+\sin A}{\cos A}\\\\\sec A-\tan A=\dfrac{1-\sin A}{\cos A}\\\\\csc A-1=\dfrac{1}{\sin A}-\dfrac{\sin A}{\sin A}=\dfrac{1-\sin A}{\sin A}\\\\\csc A+1=\dfrac{1+\sin A}{\sin A}


\dfrac{\sqrt{\sec A+\tan A}}{\sqrt{\sec A-\tan A}}\cdot\dfrac{\sqrt{\cos A-1}}{\sqrt{\cos A+1}}=1\\\\L_s=\sqrt{\dfrac{\sec A+\tan A}{\sec A-\tan A}\cdot\dfrac{\cos A-1}{\cos A+1}}=\sqrt{\dfrac{\frac{1+\sin A}{\cos A}}{\frac{1-\sin A}{\cos A}}\cdot\dfrac{\frac{1-\sin A}{\sin A}}{\frac{1+\sin A}{\sin A}}}\\\\=\sqrt{\dfrac{1+\sin A}{\cos A}\cdot\dfrac{\cos A}{1-\sin A}\cdot\dfrac{1-\sin A}{\sin A}\cdot\dfrac{\sin A}{1+\sin A}}\\\\\text{Everything are simplified}\\\\=\sqrt{1}=1=R_s

7 0
3 years ago
Lines k and p are perpendicular, neither is vertical and p passes through the origin. Which is greater? A. The product of the sl
AlladinOne [14]
What do we know about those two lines?

They are perpendicular, meaning they have the same slope.
We know the slope of both is not zero (neither is vertical).
Therefore either
1) Both slopes are positive and therefore the product is positive
2) Both slopes are negative and therefore the product is positive (minus by a minus is a plus)

For the y intercepts, we know that the line P passes through the origin.
Therefore its Y intercept is zero.
[draw it if this is not obvious and ask where does it cross the y axis]
Therefore the Y intercept of line K and line P is zero.
[anything multiplied by a zero is a zero]

So we know that the product of slopes is positive, and we know that the product of Y intercepts is zero.
So the product of slopes must be greater.
Answer A
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3 years ago
DEFG is a rhombus with m∠EFG=28. What is m∠FGE?<br> 76<br> 152<br> 14
Katen [24]
14 adverage between numbers but i might be wrong
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3 years ago
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