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Arlecino [84]
3 years ago
11

How do I do the substitution method?

Mathematics
2 answers:
Lina20 [59]3 years ago
5 0

Step-by-step explanation:

1: Solve one Equation for one of the variables

2: Substitute (plug-in) this expression into the other equation and solve

3: Resubtitue the value into the original equation to find the corresponding variable

alekssr [168]3 years ago
4 0

Answer:

1. Solve one equation for one of the variables.

2. Substitute (plug-in) this expression into the other equation and solve.

3. Resubstitute the value into the original equation to find the corresponding variable.

Now at first glance, this may seem complicated, but it'll get easier along the way. <3

Done by NeighborhoodDealer

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Helpp pleaseeeee and could u put an explanation of how you did it?
umka21 [38]

Answer: Option A

Step-by-step explanation:

To get from A to A', what is done?

A is at the point (2,6) and A' is at the point (2,-1)

The x value of the two points stayed constant (didn't change) meaning that options C and D are invalid (because those two are transformations to the x-value)

Now let us take a look at options A and B which both concern the y value

The y-value of A is 6 and A' is -1

To get from 6 to -1, do you

a) subtract 7

or b) = multiply by -1

Let us test it out:

a) 6 - 7 = -1 (yes, it equals -1)

b) 6 * -1 = -6 (no, it does not equal the intended value of -1)

Thus correct choice is A, we subtract 7

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3 years ago
Help!<br> Find the value of x for which ABCD must be a parallelogram.
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In a parallelogram, the diagonals (the lines inside from corner to corner) bisect each other. That means the two halves of the diagonal are equal to each other.
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3 years ago
Help with Algebra 2: 1. <img src="https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B7%5D%7Bx%5E5%7D%20%7D%7B%5Csqrt%5B4%5D%7Bx%5E2%7D%
Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

1. Given the expression

\dfrac{\sqrt[7]{x^5} }{\sqrt[4]{x^2} }

Note that

\sqrt[7]{x^5}=x^{\frac{5}{7}} \\ \\\sqrt[4]{x^2}=x^{\frac{2}{4}}=x^{\frac{1}{2}}

When dividing \sqrt[7]{x^5} by \sqrt[4]{x^2}, we have to subtract powers (we cannot subtract 4 from 7, because then we get another expression), so

\dfrac{5}{7}-\dfrac{2}{4}=\dfrac{5}{7}-\dfrac{1}{2}=\dfrac{5\cdot 2-1\cdot 7}{14}=\dfrac{3}{14}

and the result is x^{\frac{3}{14}}=\sqrt[14]{x^3}

2. Given equation 3\sqrt[4]{(x-2)^3} -4=20

Add 4:

3\sqrt[4]{(x-2)^3} -4+4=20+4\\ \\3\sqrt[4]{(x-2)^3}=24

Divide by 3:

\sqrt[4]{(x-2)^3} =8

Rewrite the equation as:

(x-2)^{\frac{3}{4}}=8\\ \\(x-2)^{\frac{3}{4}}=2^3

Hence,

\left((x-2)^{\frac{3}{4}}\right)^{\frac{4}{3}}=(2^3)^{\frac{4}{3}}\\ \\x-2=2^{3\cdot \frac{4}{3}}\\ \\x-2=2^4\\ \\x-2=16\\ \\x-2+2=16+2\\ \\x=18

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It's C or A if it's C that's right if it's A I'm wrong be pretty sure it is C

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