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2 years ago

Please help me I don't get this

Mathematics

1 answer:

patriot [66]2 years ago

3 0

A.0 because they are the same number there is no change from LMN to OPQ

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Simplify 4^7/5^2 all raised to the 3rd power<br><br> 4^10/5^5<br> 4^4/5<br> 4^21/5^6<br> 12^7/15^2

zepelin [54]

[4^7/5^2]³ = 4^21/5^6

When a power is raised to an other power, you multiply these 2 powers

Answer: 4^21/5^6

5 0

2 years ago

Mr. Case is placing boxes along one of the walls in his attic. If the wall is 11 1⁄3 feet long and each box is 2 1⁄2 feet long,

rosijanka [135]

Answer: he can fit 4 boxes in his attic

Step-by-step explanation:

4 0

2 years ago

Gamestop is having a one-day 35% off special. If Donald bought $124.50 worth of items, how much did he save? how much was his fi

Neko [114]

He save $43.58
His final bill would be $80.92

Solution #1:
35/100 × $124.50 = $43.575

When you round $43.575 to the nearest hundredths you will get $43.58

Solution #2:
$124.50 - $43.58 = $80.92

8 0

2 years ago

There are 31 fish in a tank. The fish are either orange or red. There are 7 more orange fish than the number of red fish. How ma

Lapatulllka [165]

Answer:

There are 12 red fish, 19 orange fish

Step-by-step explanation:

Let y rep amount of red fish

y + 7 = Amount of orange fish

y + y + 7 = 31

2y = 31 - 7

2y = 24

y = 12

Then

y + 7

12 + 7 = 19

4 0

2 years ago

The right end of a relaxed standard spring is at the origin; the left end is clamped at some point on the negative x-axis. Holdi

ella [17]

The work required to stretch the spring is 350J.

<h3>Hooke's Law</h3>

This law can be written by the formula: F=kx, where:

F= elastic force (N)

k= spring constant (N/m)

x= linear displacement (m)

<h3>Spring Work</h3>

For finding the spring work in J, you should apply the formula $W=\frac{k*(\Delta x)^2}{2}$ , where:

W= work (J)

k= spring constant (N/m)

x= difference between the linear displacements (m)

The question gives:

x=5 cm=0.05m requires a force of 1.4N

x=8 cm =0.08m

Step 1 - First, you should find the spring constant from Hooke's law, for x=5 cm and F=1.4N.

$F=kx\\ 1.4=k*0.05\\ \\ k=\frac{1.4}{0.05} =28\frac{N}{m}$

Step 2 - Now you can apply the formula for spring work.

$W=\frac{k*(\Delta x)^2}{2}\\ \\ W=\frac{28*(8-3)^2}{2} \\ \\ W=\frac{28*(5)^2}{2}\\ \\ W=\frac{28*25}{2}\\ \\ W=\frac{700}{2}=350J$

Read more about the spring work here:

brainly.com/question/3317535

5 0

2 years ago