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VLD [36.1K]
3 years ago
9

Write the prime factorization of 500 in index form.

Mathematics
2 answers:
makvit [3.9K]3 years ago
8 0

Answer:

2^{2} × 5^{3}

Step-by-step explanation:

500

= 2 x 250

= 2 x 10 x 25

= 2 x 2 x 5 x 5 x 5

= 2^{2} × 5^{3}

vovangra [49]3 years ago
4 0

Answer:

#refer to the attachment!!

Factors \:of \:500

\\ \longmapsto 1, 2, 4, 5, 10, 20, 25, 50, 100, 125, 250, and 500.

Prime \:Factorization \:of\: 500

\\ \longmapsto 500=2^{2} × 5^{3}

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dybincka [34]
<h3>Given :-</h3>
  • Pranav ran 20.3 km more than Branda

  • Pravin ran 38.6 km

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To find :

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\\  \\

<h3>Let :</h3>

  • No. of kilometers ran by Brenda be x.

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<h3>Solution:</h3>

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Equation formed:-

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Total distance covered by Pravan = More distance covered by Pravan + distance covered by Brenda.

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\leadsto \sf38.6 = 20.3 + x

Write the equation

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\leadsto \sf38.6 - 20.3 =x

When we transfer 20.3 to left side the positive sign (+) will change into negative sign (–)

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\leadsto \sf x = 38.6 - 20.3

Arrange the equation because x is always represented at left side.

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\leadsto \boxed {\pmb{\sf x = 18.3}}\star

After subtracting 38.6 with 20.3 we will get result as 18.3 .

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\therefore \red{ \underline{  \pmb{\frak{Distance  ~ covered ~by ~Brenda~ is ~equal ~to~18.3~kilometers}}}}

7 0
3 years ago
2(x - 4) - 2 = 5x + 23<br> x= ?
crimeas [40]
X=-11 :) let me know if you need more help!
3 0
3 years ago
A factory makes 12 bikes in 3 hours. If it keeps making bikes at the same rate, how many bikes will it have made in 8 hours? (HI
Vilka [71]

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4 0
3 years ago
Use the arc length formula to find the length of the curve y = 2 − x2 , 0 ≤ x ≤ 1. Check your answer by noting that the curve is
Lina20 [59]

By using the arc length formula, we will see that the length of the curve is L = 1.48

<h3>How to use the arc length formula?</h3>

Here we have the curve:

y = 2 - x^2   with 0 ≤ x ≤ 1

And we want to find the length of the curve.

The arc length formula for a curve y in the interval [x₁, x₂] is given by:

L  =\int\limits^{x_2}_{x_1} {\sqrt{1 + \frac{dy}{dx}^2} } } \, dx

For our curve, we have:

dy/dx = -2x

And the interval is [0, 1]

Replacing that we get:

L  =\int\limits^{1}_{0} {\sqrt{1 + (-2x)^2} } } \, dx\\\\L  \int\limits^{1}_{0} {\sqrt{1 + 4x^2} } } \, dx\\

This integral is not trivial, using a table you can see that this is equal to:

L = (\frac{Arsinh(2x)}{4}  + \frac{x*\sqrt{4x^2 + 1} }{2})

evaluated from x = 1 to x = 0, when we do that, we will get:

L = \frac{Arsinh(2) + 2*\sqrt{5} }{4}  = 1.48

That is the length of the curve.

If you want to learn more about curve length, you can read:

brainly.com/question/2005046

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Write an equation that has ONE SOLUTION to the equation: y=5x+3
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Answer:(0,3)

Step-by-step explanation:

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