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Fittoniya [83]
4 years ago
6

HEEEEEEEEEEEEEEEEEEEEEEEEELP!!!!!!!!!!!!!!

Mathematics
2 answers:
klemol [59]4 years ago
7 0
1,883 divided by 14 is 134.5
Jet001 [13]4 years ago
6 0

Answer: Maybe it's 138,5 of the remaining...

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What is 10^-3x*10^x= 1/10
densk [106]
Since 1/10=10^-1, and multiplying exponents with the same base just adds the actual exponents at the top, x=2
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3 years ago
I need help please i will give brainly ​
Arisa [49]

Answer:

C

Step-by-step explanation:

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3 years ago
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Find the product. (-2x^2 )^3 ·3 x
jeka94

Assignment: \bold{Solve \ \left(-2x^2\right)^3\cdot \:3x}

<><><><><>

Answer: \boxed{\bold{-24x^7}}

<><><><><>

Explanation: \downarrow\downarrow\downarrow

<><><><><>

[ Step One ] Rewrite \bold{\left(-2x^2\right)^3}

\bold{-2^3\left(x^2\right)^3}

[ Step Two ] Rewrite Equation

\bold{-2^3\cdot \:3x^6x}

[ Step Three ] Apply Exponent Rule

Note: \bold{Exponent \ Rule \ \:a^b\cdot \:a^c=a^{b+c}: \ x^6x=\:x^{6+1}=\:x^7}

\bold{-2^3\cdot \:3x^7}

[ Step Four ] Refine

\bold{-24x^7}

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\bold{\rightarrow Rhythm \ Bot \leftarrow}

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3 years ago
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Hi, I am taking PSAT and I am working on the math section. I need help with answering this math type of question. Please give th
ValentinkaMS [17]

By using the concept of uniform rectilinear motion, the distance surplus of the average race car is equal to 3 / 4 miles. (Right choice: A)

<h3>How many more distance does the average race car travels than the average consumer car?</h3>

In accordance with the statement, both the average consumer car and the average race car travel at constant speed (v), in miles per hour. The distance traveled by the vehicle (s), in miles, is equal to the product of the speed and time (t), in hours. The distance surplus (s'), in miles, done by the average race car is determined by the following expression:

s' = (v' - v) · t

Where:

  • v' - Speed of the average race car, in miles per hour.
  • v - Speed of the average consumer car, in miles per hour.
  • t - Time, in hours.

Please notice that a hour equal 3600 seconds. If we know that v' = 210 mi / h, v = 120 mi / h and t = 30 / 3600 h, then the distance surplus of the average race car is:

s' = (210 - 120) · (30 / 3600)

s' = 3 / 4 mi

The distance surplus of the average race car is equal to 3 / 4 miles.

To learn more on uniform rectilinear motion: brainly.com/question/10153269

#SPJ1

4 0
1 year ago
Rational number word problems
Temka [501]

Answer:

1583.90

Step-by-step explanation:

Khan academy

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