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Karo-lina-s [1.5K]
3 years ago
10

Which of the following would be equivalent to 3^2 x 3^5

Mathematics
1 answer:
Goshia [24]3 years ago
5 0
The first one... 3^10/3^3
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Please help. <br> also, i hope you enjoyed you holidays !
lions [1.4K]

Answer:

1. angle AXB

2. 52

Step-by-step explanation:

1.supplementary is 180⁰

2.90-38=52

6 0
3 years ago
3 square root 4 • square root 3
Anton [14]

Answer:

6 square root 3

Step-by-step explanation:

3 square root 4 .square root 3= 6 square root 3

8 0
3 years ago
HELLLPP ME PLEASSEEE!!!!!
antoniya [11.8K]

The similar circles P and Q can be made equal by dilation and translation

  • The horizontal distance between the center of circles P and Q is 11.70 units
  • The scale factor of dilation from circle P to Q is 2.5

<h3>The horizontal distance between their centers?</h3>

From the figure, we have the centers to be:

P = (-5,4)

Q = (6,8)

The distance is then calculated using:

d = √(x2 - x1)^2 + (y2 - y1)^2

So, we have:

d = √(6 + 5)^2 + (8 - 4)^2

Evaluate the sum

d = √137

Evaluate the root

d = 11.70

Hence, the horizontal distance between the center of circles P and Q is 11.70 units

<h3>The scale factor of dilation from circle P to Q</h3>

We have their radius to be:

P = 2

Q = 5

Divide the radius of Q by P to determine the scale factor (k)

k = Q/P

k = 5/2

k = 2.5

Hence, the scale factor of dilation from circle P to Q is 2.5

Read more about dilation at:

brainly.com/question/3457976

8 0
2 years ago
Find the area of the following figure. Explain how you found your answer.
iragen [17]

Answer:

Below

Step-by-step explanation:

Let's say that each square on the grid represents 1 cm

First find the area of the two triangles

For the larger one: A = bh/2

A = 3 x 3 / 2

  = 4.5 cm^2

For the smaller one : A = bh/2

A = 2 x 2 / 2

   = 2 cm^2

For the rectangle : A = lw

A = 4 x 5

   = 20 cm^2

Add them all up to get the area

4.5 + 2 + 20 = 26.5 cm^2

Hope this helps!

5 0
3 years ago
Read 2 more answers
Tan^2x+sec^2x=1 for all values of x. <br> true or false?
amid [387]

This is not true.

\tan^2x+\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac1{\cos^2x}=\dfrac{1-\cos^2x+1}{\cos^2x}=2\sec^2x-1

2\sec^2x-1=1\implies\sec^2x=1\implies\sec x=\pm1\implies x=n\pi

where is n is any integer. So suppose we pick some value of x other than these, say x=\dfrac\pi4. Then

\tan^2\dfrac\pi4+\sec^2\dfrac\pi4=1+2=3\neq1

5 0
3 years ago
Read 2 more answers
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