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2 years ago

PLEASE HELP I WILL MARK BRAINLIEST

Mathematics

2 answers:

FrozenT [24]2 years ago

6 0

Answer: the answer is G

Step-by-step explanation:

Harlamova29_29 [7]2 years ago

3 0

Answer:

Step-by-step explanation:

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Pierce swims 10 laps in a pool in 8 minutes. He spent the same amount of time on each lap. How much time did each lap take him?

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Answer:

48 sec.

Step-by-step explanation:

8 devided by 10.

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What is 4,967 divided by 68 with a remainder no decimals

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The answer would be 73 remainder 0 if rounded to one decimal place

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Express the following as a fraction in its simplest form.

pychu [463]

Given:

Consider the given expression is:

$\dfrac{h+f}{3}+\dfrac{f+k}{2}+\dfrac{4h-k}{5}$

To find:

The simplified form of the given expression.

Solution:

We have,

$\dfrac{h+f}{3}+\dfrac{f+k}{2}+\dfrac{4h-k}{5}$

Taking LCM, we get

$=\dfrac{10(h+f)+15(f+k)+6(4h-k)}{30}$

$=\dfrac{10h+10f+15f+15k+24h-6k}{30}$

$=\dfrac{34h+25f+9k}{30}$

Therefore, the required simplified fraction for the given expression is $\dfrac{34h+25f+9k}{30}$ .

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2 years ago

Which of the following are solutions to the equation below 4x^2+20x+25=49

KATRIN_1 [288]

( 2x +5)^2 =4

2x+5=2

2x=-3

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3 years ago

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Which is the simplified form of the expression ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript nega

AlekseyPX

Answer:

The option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Step-by-step explanation:

Given expression is ((2 Superscript negative 2 Baseline) (3 Superscript 4 Baseline)) Superscript negative 3 Baseline times ((2 Superscript negative 3 Baseline) (3 squared)) squared

The given expression can be written as

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

To find the simplified form of the given expression :

$[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2$

$=(2^{-2})^{-3}(3^4)^{-3}\times (2^{-3})^2(3^2)^2$ ( using the property $(ab)^m=a^m.b^m$ )

$=(2^6)(3^{-12})\times (2^{-6})(3^4)$ ( using the property $(a^m)^n=a^{mn}$

$=(2^6)(2^{-6})(3^{-12})(3^4)$ ( combining the like powers )

$=2^{6-6}3^{-12+4}$ ( using the property $a^m.a^n=a^{m+n}$ )

$=2^03^{-8}$

$=\frac{1}{3^8}$ ( using the property $a^{-m}=\frac{1}{a^m}$ )

Therefore $[(2^{-2})(3^4)]^{-3}\times [(2^{-3})(3^2)]^2=\frac{1}{3^8}$

Therefore option "StartFraction 1 Over 3 Superscript 8" is correct

That is $\frac{1}{3^8}$ is correct answer

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3 years ago

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