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3 years ago

GIVING BRAINLIEST!

Mathematics

1 answer:

alexira [117]3 years ago

5 0

Answer:

a= 8x+6x+48x+1

Step-by-step explanation:

Area of large rectangle is represented by a

a= 8x+6x+48x+1 - This is expanded form

Can be simplified to a= 62x+1

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Are these correct? Im having trouble

Gennadij [26K]

Answer:

yes

Step-by-step explanation:

it is correct

3 0

2 years ago

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Geometry:

Eddi Din [679]

Answer:

<AEB = obtuse angle

<AED = straight angle

<BEA = obtuse angle

<BEC = straight angle

<CDE = not angle

<CEA = acute angle

<DEA = straight angle

<DEB = acute angle

<DEC = obtuse angle

Step-by-step explanation:

3 0

2 years ago

Write the inequality:<br> 3 less than n is no more than -8

rusak2 [61]

Answer:

n - 3 ≤ -8

Step-by-step explanation:

n - 3 ≤ -8

3 less than n: n - 3

is no more than: ≤ - 8

5 0

2 years ago

Using f(x) = 4x + 6 with a domain of {-1, 0, 2 }, find the range.

lianna [129]

Answer:

{2, 6, 14}

Step-by-step explanation:

Using f(x) = 4x + 6 with a domain of {-1, 0, 2 }, find the range.

To get the range, we will substitute the values of the domain into the given function as shown;

when x = -1

f(-1) = 4(-1)+6

f(-1) = -4+6

f(-1) = 2

when x = 0

f(0) = 4(0)+6

f(0) = 0+6

f(0) = 6

when x = 2

f(2) = 4(2)+6

f(2) = 8+6

f(2) = 14

Hence the required range are {2, 6, 14}

6 0

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

Read 2 more answers