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meriva
3 years ago
5

Write the following word phase as a mathematical equation. The sum of 7 and twice a number, squared.

Mathematics
1 answer:
agasfer [191]3 years ago
4 0

Answer:

(7+(2n))^2

You don't technically need the ( ) around the 2n, but I included them to avoid confusion.

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At the beginning of spring, Arianna planted a small sunflower in her backyard. When
zhannawk [14.2K]

Answer:

3 weeks: 21.5 inches

w weeks: 20+0.5w inches

Step-by-step explanation:

Equation: 20+0.5w.

20 was when it was first planted and 0.5 is the amount it grows per week.

6 0
3 years ago
Look at the relationship between a and b.
tensa zangetsu [6.8K]

Answer:

42 is the answer to life, the universe, and the amount of pizza i need. The 0 is the amount of braincells i have and the 6 is because I have 6 working teeth. The 9 is because it is almost 10

i hope this helped hehe

Step-by-step explanation:

well its all up there i legit have no brain cells

3 0
3 years ago
Helppp!!!! please!!!
Vitek1552 [10]

Answer:

d. 15 square yard

Step-by-step explanation:

area \: of \: shape \\  =  \frac{1}{2}  \times base \times height \\  \\  =  \frac{1}{2}  \times 10 \times 3 \\  \\  =  \frac{1}{2}  \times 30 \\  \\  = 15 \:  {yd}^{2}

4 0
3 years ago
Read 2 more answers
What’s the answer to 10a - 13a being factored
solmaris [256]

Answer:

a(10 - 13)

Step-by-step explanation:

The only common factor between these numbers is a, as 10 and 13 do not share factors (13 is prime). We can only factor out a.

7 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
4 years ago
Read 2 more answers
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