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marin [14]
4 years ago
9

Multiply 2×-5/7×-3/-8​

Mathematics
1 answer:
solniwko [45]4 years ago
6 0

Hey Kat! :)

Glad to see you on Brainly.

Look below and watch how we solve this problem.

Ask any questions if you need it! I'll respond when you need it!

This is a bit of a hard, one. So please ask questions.

------

Please use the rule a/b x c/d = ac/bd while solving.

After using that equation, it should look like this:

2 x -5 x -3 / 7 x -8

Simplify 2 x -5 to -10.

Now, it should look like -10 x -3 / 7 x -8

30/7 x -8

30/-56

- 30/56 (the negative is front of the WHOLE equation.)

-15/28 is our final answer.

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5. Point U is the midpoint of ty. Find the value of x. Round to the nearest tenth if necessary.
sattari [20]

The value of x is 8.5.

Solution:

U is the mid-point of TV.

TU = 6

UV = 2x - 11

TU = UV

6 = 2x - 11

Add 11 on both sides.

6 + 11 = 2x - 11 + 11

17 = 2x

Divide by 2 on both sides.

$\frac{17}{2} =\frac{2x}{2}

8.5 = x

Switch the sides.

x = 8.5

Option B is the correct answer.

7 0
3 years ago
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lesya692 [45]
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6 0
3 years ago
Read 2 more answers
Here is a division equation: 4/5 ÷ 2/3 = ? Write a multiplication equation that corresponds to the division equation.
Finger [1]

Answer:

4/5 * 3/2

Step-by-step explanation:

fraction division = multiplying the first fractions by the reciprocal of the second fractions (the tops are bottoms are flipped).

5 0
3 years ago
What is the estimated temperature if the dissolved salt is 98g?<br><br> 54<br> 61<br> 161<br> 171
madreJ [45]

Ths solubility curve is used to show the amount of salt dissolved (solubility).

<h3>What is the solubility curve?</h3>

The solubility curve is a depiction of the solubility of a substance plotted against its temperature. It can be used most times to show the solubility of a susbtance at different temperatures. This question is incomplete hence so we can not be able to obtain the solubility of the salt at this temperature.

If the solubility curve has been shown in the graph, then we can be able to obtain the solubility of the salt from the data shown on the plot.

Learn more about solubility curve: brainly.com/question/9537462

7 0
2 years ago
which of the following is equivalent to 3 sqrt 32x^3y^6 / 3 sqrt 2x^9y^2 where x is greater than or equal to 0 and y is greater
Nutka1998 [239]

Answer:

\frac{\sqrt[3]{16y^4}}{x^2}

Step-by-step explanation:

The options are missing; However, I'll simplify the given expression.

Given

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} }

Required

Write Equivalent Expression

To solve this expression, we'll make use of laws of indices throughout.

From laws of indices \sqrt[n]{a}  = a^{\frac{1}{n}}

So,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } gives

\frac{(32x^3y^6)^{\frac{1}{3}}}{(2x^9y^2)^\frac{1}{3}}

Also from laws of indices

(ab)^n = a^nb^n

So, the above expression can be further simplified to

\frac{(32^\frac{1}{3}x^{3*\frac{1}{3}}y^{6*\frac{1}{3}})}{(2^\frac{1}{3}x^{9*\frac{1}{3}}y^{2*\frac{1}{3}})}

Multiply the exponents gives

\frac{(32^\frac{1}{3}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

Substitute 2^5 for 32

\frac{(2^{5*\frac{1}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})}

From laws of indices

\frac{a^m}{a^n} = a^{m-n}

This law can be applied to the expression above;

\frac{(2^{\frac{5}{3}}x*y^{2})}{(2^\frac{1}{3}x^{3}*y^{\frac{2}{3}})} becomes

2^{\frac{5}{3}-\frac{1}{3}}x^{1-3}*y^{2-\frac{2}{3}}

Solve exponents

2^{\frac{5-1}{3}}*x^{-2}*y^{\frac{6-2}{3}}

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}}

From laws of indices,

a^{-n} = \frac{1}{a^n}; So,

2^{\frac{4}{3}}*x^{-2}*y^{\frac{4}{3}} gives

\frac{2^{\frac{4}{3}}*y^{\frac{4}{3}}}{x^2}

The expression at the numerator can be combined to give

\frac{(2y)^{\frac{4}{3}}}{x^2}

Lastly, From laws of indices,

a^{\frac{m}{n} = \sqrt[n]{a^m}; So,

\frac{(2y)^{\frac{4}{3}}}{x^2} becomes

\frac{\sqrt[3]{(2y)}^{4}}{x^2}

\frac{\sqrt[3]{16y^4}}{x^2}

Hence,

\frac{\sqrt[3]{32x^3y^6}}{\sqrt[3]{2x^9y^2} } is equivalent to \frac{\sqrt[3]{16y^4}}{x^2}

8 0
3 years ago
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