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2 years ago

5

All the dimensions of a pentagon were multiplied by the same factor, and the area of the figure changed by a factor of 25. By wh

at factor were the dimensions of the pentagon multiplied?
12. 5

625

5

50

1 answer:

elena-s [515]2 years ago

8 0

5

reason is because if you’re pentagon has lines of 5 then the area would be 25

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What si the exponential form of 243

dezoksy [38]

2.43 x 10^2

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3 years ago

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Find X and Y and show please show calculations

torisob [31]

Answer:

Y = 90

X = 52

Step-by-step explanation:

Y is on a straight line. The sum of the angles on a straight line is 180.

The only other angle (that is marked) on this straight line, is a 90 degree one. So, <y + 90 = 180.

<y = 90.

Now <x + 38 = 90

so, x = 52

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3 years ago

How to solve for x where (2^x)*x =n?

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Tough call.
You may have to use Newton's Method or something like that.
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3 years ago

Help!! 50 points and brainliest!

Viktor [21]

Answer:

Second choice:

$x=2t$

$y=4t^2+4t-3$

Fifth choice:

$x=t+1$

$y=t^2+4t$

Step-by-step explanation:

Let's look at choice 1.

$x=t+1$

$y=t^2+2t$

I'm going to subtract 1 on both sides for the first equation giving me $x-1=t$ . I will replace the $t$ in the second equation with this substitution from equation 1.

$y=(x-1)^2+2(x-1)$

Expand using the distributive property and the identity $(u+v)^2=u^2+2uv+v^2$ :

$y=(x^2-2x+1)+(2x-2)$

$y=x^2+(-2x+2x)+(1-2)$

$y=x^2+0+-1$

$y=x^2$

So this not the desired result.

Let's look at choice 2.

$x=2t$

$y=4t^2+4t-3$

Solve the first equation for $t$ by dividing both sides by 2:

$t=\frac{x}{2}$ .

Let's plug this into equation 2:

$y=4(\frac{x}{2})^2+4(\frac{x}{2})-3$

$y=4(\frac{x^2}{4})+2x-3$

$y=x^2+2x-3$

This is the desired result.

Choice 3:

$x=t-3$

$y=t^2+2t$

Solve the first equation for $t$ by adding 3 on both sides:

$x+3=t$ .

Plug into second equation:

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

Expanding using the distributive property and the earlier identity mentioned to expand the binomial square:

$y=(x^2+6x+9)+(2x+6)$

$y=(x^2)+(6x+2x)+(9+6)$

$y=x^2+8x+15$

Not the desired result.

Choice 4:

$x=t^2$

$y=2t-3$

I'm going to solve the bottom equation for $t$ since I don't want to deal with square roots.

Add 3 on both sides:

$y+3=2t$

Divide both sides by 2:

$\frac{y+3}{2}=t$

Plug into equation 1:

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

This is not the desired result because the $y$ variable will be squared now instead of the $x$ variable.

Choice 5:

$x=t+1$

$y=t^2+4t$

Solve the first equation for $t$ by subtracting 1 on both sides:

$x-1=t$ .

Plug into equation 2:

$y=(x-1)^2+4(x-1)$

Distribute and use the binomial square identity used earlier:

$y=(x^2-2x+1)+(4x-4)$

$y=(x^2)+(-2x+4x)+(1-4)$

$y=x^2+2x+-3$

$y=x^2+2x-3$ .

This is the desired result.

3 0

3 years ago

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Given: <PQR is a right angle Prove: <PQS and <SQR are complemantary

amm1812

Answer:

see the explanation

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

m∠PQS+m∠SQR=m∠PQR ----> equation A (by Addition Angle Postulate)

we have that

m∠PQR=90° ----> equation B given problem (because is a right angle)

substitute equation B in equation A

m∠PQS+m∠SQR=90°

Remember that

Two angles re complementary is their sum is equal to 90 degrees (Definition of complementary angles)

therefore

m∠PQS and m∠SQR are complementary angles

3 0

3 years ago