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Sonbull [250]
3 years ago
11

Plz help me thank you

Mathematics
1 answer:
svet-max [94.6K]3 years ago
7 0

Answer:

the third option is the answer

Step-by-step explanation:

( {7}^{ \frac{1}{3} } ) {}^{3}

i) to raise a power to another power, multiply the exponents

{7}^{ \frac{1}{3}  \times 3}

{7}^{1}

ii) any power raised to the power of 1 equals itself

7 {}^{1}  = 7

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Justin travels at 70 km/h on his motorcycle. If the speed limit is 50 mi./hr., is he speeding? Why or why not?
balu736 [363]

Answer:

Justin is <em>not </em>speeding because there is 1.609 km in a mile. So if you compare it to 70 km/h to 50 mi/h it would 70 km and 43.50 miles per hour. So like I said, no he is not because he is going below 50 mi/h.

6 0
3 years ago
Read 2 more answers
What is the solution to the system of equations? y = A system of equations. Y equals StartFraction 2 over 3 EndFraction x plus 3
Sauron [17]

Answer:

The solution to the given system of equations is (-2,\frac{5}{3})

Therefore the values of x and y are x=-2 and y=\frac{5}{3}

Step-by-step explanation:

Given equations can be written as

y=\frac{2}{3}x+3\hfill (1)

x=-2x+3x=-2\hfill (2)

Solving equation(2) we get

x=-2

Substitute x=-2 in equation (1) we get

y=\frac{2}{3}(-2)+3

=-\frac{4}{3}+3

=\frac{-4+9}{3}

=\frac{5}{3}

Therefore the values of x and y are  x=-2 and y=\frac{5}{3}

The solution to the given system of equations is (-2,\frac{5}{3})

5 0
3 years ago
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1. Fill in the blanks for the rules for working with exponents:
natima [27]
1.a)add
B) substract
C) multiply
D) one

2) 1002=1.002×10^3

3)2.8×10^-3 < 2.5×10^-2 < 1.2×103 <4×103
5 0
3 years ago
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Kay was walking along the bridge above the river. She accidentally lost her grip on her camera when she was trying to get it out
nikklg [1K]

Answer:

5 seconds it will hit the water.

Step-by-step explanation:

y=-16t²+400

add -16t² to both sides

-16t²=-400

divide both sides by 16

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t=√25

t=5

8 0
3 years ago
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H = (4 x + 3 y) + to make x the subject​
Likurg_2 [28]
X=h/4 -3y/4 hope that helped
3 0
3 years ago
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