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3 years ago

What is the simplified prime factorization of 320?

Mathematics

2 answers:

Butoxors [25]3 years ago

8 0

Answer:I think it would be d (2x6)x5

Step-by-step explanation:

320-160x2-80x2-40x2-20x2-10x2-5x2

2x2x2x2x2x2x5 2 to the 6th power x 5 so (2x6)5 sounds accurate

katen-ka-za [31]3 years ago

6 0

Answer:2*2*2*2*2*2*5

Step-by-step explanation: thats what I found

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Solve<br>3 tan 8 <-1 for 0 5 0521.

BlackZzzverrR [31]

Answer:

Substitute the given value into the function and evaluate.

3 tan ( 8 ) < − 1 = 0.4216225 < − 1

0 = 0.4216225 < − 1

5 = 0.4216225 < − 1

521 = 0.4216225 < − 1

Step-by-step explanation:

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3 years ago

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Yuki888 [10]

You can pay in check

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3 years ago

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The answer is p = -2/3 and q= -2

igomit [66]

$y=\dfrac{p}{q+x}\\\\\dfrac{1}{y}=\dfrac{1}{\frac{p}{q+x}}=\dfrac{q+x}{p}$

We have x and y intercept:

(2, 0) and (0, 3).

Substitute the coordinates of points to the equation:

$(2,\ 0)\\\\0=\dfrac{q+2}{p}\to q+2=0\to q=-2\\\\(0,\ 3)\\3=\dfrac{q+0}{p}\to3p=q\to3p=-2\to p=-\dfrac{2}{3}$

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3 years ago

Neptune is an average distance of 4.5 × 10^9 km from the Sun. Estimate the length of the Neptunian year using the fact that the

zlopas [31]

Answer:

164.32 earth year

Step-by-step explanation:

distance of Neptune, Rn = 4.5 x 10^9 km

distance of earth, Re = 1.5 x 10^8 km

time period of earth, Te = 1 year

let the time period of Neptune is Tn.

According to the Kepler's third law

T² ∝ R³

$\left ( \frac{T_{n}}{T_{e}} \right )^{2}=\left ( \frac{R_{n}}{R_{e}} \right )^{3}$

$\left ( \frac{T_{n}}{1} \right )^{2}=\left ( \frac{4.5\times10^{9}}}{1.5\times10^{8}}} \right )^{3}$

Tn = 164.32 earth years

Thus, the neptune year is equal to 164.32 earth year.

Step-by-step explanation:

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4 years ago

Consider the system of inequalities and its graph. y ≤ –0.75x y ≤ 3x – 2 In which section of the graph does the actual solution

AlekseyPX

The answer to your question is 4

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3 years ago

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