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3 years ago

Rewrite the equation below into standard form. -3/2x=-2/5y-10/3

Mathematics

2 answers:

SVETLANKA909090 [29]3 years ago

7 0

Answer:

Step-by-step explanation:

You want to rewrite the equation

Ax +By=C A,B,C are integers

30(-3/2x) = 30(-2/5y-10/3)

-45x = -12y - 100

-45x + 12y = -100

kaheart [24]3 years ago

4 0

Answer:

standard form is 45x -12y -100 = 0.

Step-by-step explanation:

Given the equation is -3/2x=-2/5y-10/3.

Finding common denominator, L.C.M.(2,5,3) = 30.

Multiplying both sides by 30.

30*(-3/2x) = 30*(-2/5y-10/3)

-45x = -12y -100

Moving all terms on left side, keeping right side equal to 0.

-45x +12y +100 = 0

Making x's coefficient positive by changing all signs.

45x -12y -100 = 0

Hence, standard form is 45x -12y -100 = 0.

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What is 713 divided by 9 and then the quotient estimated

Murljashka [212]

713/9

79 2/9 Decimal form: 79.222

713 divided by 9 is 79 with a quotient estimate of 2.

79 r 2

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3 years ago

Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

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3 years ago

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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cluponka [151]

Answer:

1.5 seconds

Step-by-step explanation:

h=-16t² – 16t+ 60

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