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3 years ago

6

Solve one step and two step problems

1 answer:

Nastasia [14]3 years ago

5 0

1. First would be 9-4+7=13
2. Second one is 4+6-2=8

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Can u pls help me with this thanks

aleksandrvk [35]

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3 years ago

Chas and Tyler have been collecting video games. Chas

marshall27 [118]

Answer:

40

Step-by-step explanation:

m+25=65

65-25=40

m=40

4 0

3 years ago

How many bowls of cereal did he eat pls and thx

miv72 [106K]

36 bowls - since with one gallon he can eat 12 bowls (as demonstrated by 1/12), you multiply 12 by 3 to get 36 bowls.

8 0

3 years ago

Read 2 more answers

Givers the linear sequence,<br> -2,7,X,Y, 2... Find the value of<br> Z.

Harrizon [31]

Answer:

Z = 34

Step-by-step explanation:

From the series, we have that:

a = -2

a + d = 7

Hence, -2 + d = 7

∴ d = 7 + 2 = 9

a + 2d = X; a + 3d = Y and a + 4d = Z

-2 + 4(9) = Z

∴ Z = 36 - 2 = 34

7 0

2 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