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3 years ago

Simplify 5/6i please help

Mathematics

2 answers:

Leni [432]3 years ago

6 0

Answer:

5/6 can´t be simplified

Step-by-step explanation:

avanturin [10]3 years ago

4 0

Answer:

5i/6

Step-by-step explanation:

$\frac{5}{6}i\\\\\mathrm{Multiply\:fractions}:\quad \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c}\\=\frac{5i}{6}$

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What is 3359085 rounded to the nearest ten thousand

user100 [1]

Answer:

where is the point?

if it is .3359085 the answer is .3359

if it's 3.359085 the answer is 3.3591

if it's 33.59085 the answer is 33.5909

if it's 335.9085 the answer is 335.9085

if there is no point the answer is 3360000

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

HELP HELP QUICK WILL GET MARKED BRAINLIEST

jeyben [28]

I believe you're right with what you have
the price decrease being 5% (the .95) and the increase by 15% (the 1.15)

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4 years ago

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mr Goodwill [35]

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[ 4 ] First substitution
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4 years ago

Christian and Tanae both leave Disneyland at the same time. Christian travels north at 65 mph. Tanae travels south at 55 mph. Ho

liq [111]

Answer:

Step-by-step explanation:

B looks like it would work.

You add speeds * time when you are travelling in opposite directions.

I don't know why you would add or subtract 1 as in A and C

120 * t = 540

t = 540/120

t = 4.5 hours.

So after 4.5 hours they are 540 miles apart.

5 0

3 years ago

Yuri thinks that 3/4 is a root of the following function.

sineoko [7]

Given:

The polynomial function is

$q(x)=6x^3+19x^2-15x-28$

Yuri thinks that $\dfrac{3}{4}$ is a root of the given function.

To find:

Why $\dfrac{3}{4}$ cannot be a root?

Solution:

We have,

$q(x)=6x^3+19x^2-15x-28$

If $\dfrac{3}{4}$ is a root, then the value of the function at $\dfrac{3}{4}$ is 0.

Putting $x=\dfrac{3}{4}$ in the given function, we get

$q(\dfrac{3}{4})=6(\dfrac{3}{4})^3+19(\dfrac{3}{4})^2-15(\dfrac{3}{4})-28$

$q(\dfrac{3}{4})=6(\dfrac{27}{64})+19(\dfrac{9}{16})-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=3(\dfrac{27}{32})+\dfrac{171}{16}-\dfrac{45}{4}-28$

$q(\dfrac{3}{4})=\dfrac{81}{32}+\dfrac{171}{16}-\dfrac{45}{4}-28$

Taking LCM, we get

$q(\dfrac{3}{4})=\dfrac{81+342-360-896}{32}$

$q(\dfrac{3}{4})=\dfrac{-833}{32}\neq 0$

Since the value of the function at $\dfrac{3}{4}$ is not equal to 0, therefore, $\dfrac{3}{4}$ is not a root of the given function.

4 0

3 years ago