X squared over 5 gives you what?

Mathematics

1 answer:

il63 [147K]3 years ago

5 0

Answer:(x^(2))/(5)

Step-by-step explanation

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Witch one is bigger 3/5, 1/2, or 99/100

Solnce55 [7]

We need to make them all equivalent, so when all denominators are equal, we can see. 1/2 is 50/100, 3/5 is 60/100, and 99/100 stays the same.
The biggest is 99/100.

8 0

3 years ago

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Molly tried to evaluate 93\times5193×5193, times, 51 using partial products. Her work is shown below.

Mrac [35]

Since molly's solution tally's with the given solution, hence <em>Molly's solution is correct.</em>

Given the working on a partial product of 93 and 51 carried out by Molly as shown:

$\begin{array}{llrr} &&93 \\ &&\underline{{}\times51} \\ &\blueD{\text{Step 1}}&\blueD{3}& \blueD{1\times3\text{ ones}}\\ &\greenD{\text{Step 2}}&\greenD{90}& \greenD{1\times 9\text{ tens}}\\ &\maroonD{\text{Step 3}}&\maroonD{150}& \maroonD{50\times 3\text{ ones}}\\ &\goldE{\text{Step 4}}&\underline{{}+\goldE{ 4{,}500}}& \goldE{50\times 9\text{ tens}}\\ &\purpleD{\text{Step 5}}&\purpleD{4{,}743}& \end{array}$

This partial product can also be solved as shown below:

$93 \times 51 = (90+3)\times (50+1)$

Applying the distributive law:

$93 \times 51 = 90(50) + 90(1) + 3(50) + 3(1)\\93 \times 51 =4500 + 90 + 150 + 3\\93 \times 51 =4500+240+3\\93 \times 51 =4740+3\\93 \times 51 =4743$

Since molly's solution tally's with the given solution, hence <em>Molly solution is correct.</em>

Learn more about partial product at: brainly.com/question/24716925

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2 years ago

A ball is dropped from a certain height. The function below represents the height f(n), in feet, to which the ball bounces at th

Fittoniya [83]

We are given a function of the bouncing of the ball expressed as f(n) = 9(0.7)n in which n is an integer as the number of times the ball has dropped. 9 represents the initial height of the ball and 0.7 is the percent of which the height is reserved

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Come on help me pretty plz

m_a_m_a [10]

Answer:

im pretty sure its pink

Step-by-step explanation:

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3 years ago

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