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3 years ago

12

Pls help I’ll brainlest

1 answer:

Makovka662 [10]3 years ago

8 0

Answer:

D looks to be the only good solution.

Step-by-step explanation:

19.8 is roughly 20

9.8 is roughly 10, so 1.1 represents 1.10 which is increased by 10%

5.1 is still around 5, & 1.05 is 1 increased by 5%

Very rough estimate will be given.

Hope this helps. :0)

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Find the product of 51 and -2.5.

yaroslaw [1]

The answer is C because 51×-2=-102 and 51×-0.5=-25.5

So it -102+(-25.5) or C

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3 years ago

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Complementary of supplementary? Find the value of x

prohojiy [21]

Answer: x=6

The two angles shown in each are complementary because they add up to 90°.

10 & 12 would be supplementary to one another because they would add up to 180°.

Step-by-step explanation:

We know that on both 10 & 12 the angles add up to equal 90° so...

10. 8x+7x=90

15x=90

x=6

12. it's the same in pic as 10

The two angles shown in each are complementary because they add up to 90°.

10 & 12 would be supplementary to one another because they would add up to 180°.

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3 years ago

What is the exponent on 10 when you right 5.3 in scientific notation?

salantis [7]

It is zero

the answer is 5.3 x 10^0
because 10^1 would force you to move the decimal place to make it 53

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3 years ago

Divide. Write the quotient in lowest terms. 3 3/8 ÷ 9

Troyanec [42]

3 3/8 = 27/8
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4 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.

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3 years ago

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