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castortr0y [4]
3 years ago
9

What's 2x+3y=6 and x-3y=9?

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
4 0
Set up a system of equations:

2x+3y=6
x-3y=9

You can solve this with elimination by adding the two equations together:

3x+0y = 15
3x=15
x=5

Then, plug this value back into either of the original equations to find the y value:

5-3y=9
-3y=4
y= -4/3

The point of intersection is (5,-4/3)
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Answer:

6m - 3

Step-by-step explanation:

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Dividing integers
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3 years ago
Which expressions have a value of 16/81? Check all that apply.
lana [24]

Which expression have a value of 16/81? check all that apply. (2/3)^4, (16/3)^4, (4/81)^2, and (4/9)^2

Answer:

First and last option is correct.

(\frac{2}{3})^{4}=\frac{16}{81}

(\frac{4}{9})^{2}=\frac{16}{81}

Step-by-step explanation:

Given:

There are four options.

(2/3)^4, (16/3)^4, (4/81)^2, and (4/9)^2

We need to check all given options for value of 16/81.

Solution:

Using rule.

(\frac{a}{b})^{n}=\frac{a^{n}}{b^{n}}

Solve for option (\frac{2}{3})^4.

(\frac{2}{3})^{4}=\frac{2^4}{3^4}=\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3}=\frac{16}{81}

Solve for option (\frac{16}{3})^4.

(\frac{16}{3})^{4}=\frac{16^4}{3^4}=\frac{16\times 16\times 16\times 16}{3\times 3\times 3\times 3}=\frac{65536}{81}

Solve for option (\frac{4}{81})^2.

(\frac{4}{81})^{2}=\frac{4^2}{81^2}=\frac{4\times 4}{81\times 81}=\frac{16}{6561}

Solve for option (\frac{4}{9})^2.

(\frac{4}{9})^{2}=\frac{4^2}{9^2}=\frac{4\times 4}{9\times 9}=\frac{16}{81}

Therefore, expression (\frac{2}{3})^4 and (\frac{4}{9})^2 have a value of  \frac{16}{81}.

6 0
3 years ago
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Sergeeva-Olga [200]

Answer:

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Step-by-step explanation:

for A, none of the x's repeat which makes it a FUNCTION

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3 years ago
Solve for the missing item in the following. (Do not round intermediate calculations. Round your answer to the nearest cent.)
iogann1982 [59]

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Answer:

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Step-by-step explanation:

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The principal amount was $1904.76.

7 0
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