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3 years ago

Which is the factored form of 64p3 + 125?

Mathematics

2 answers:

xz_007 [3.2K]3 years ago

8 0

Answer:

q .no d

Step-by-step explanation:

64p^3+125

4p^3+ 5^3

(4p+ 5 ) ( 4p^2- 4p *5 + 5^2)

(4p+5) (16p- 20p+ 25)

Schach [20]3 years ago

3 0

Answer:

The answer is

Step-by-step explanation:

64p³ + 125

Write the number in exponential form of 3

That's

4³p³ + 5³

Multiply the terms

That's

(4p)³ + 5³

Using the formula

Factor the expression

That's

( 4p + 5 )[ (4p)² - 4p(5) + 5²) ]

Solve the terms in the bracket

We have the final answer as

Hope this helps you

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A chain weighs 10 pounds per foot. How many ounces will 3 inches weigh?

KengaRu [80]

Answer:

40 ounces

Step-by-step explanation:

One foot = 12 inches

3 inches is 1/4 of a foot

Divide 10 pounds by 4 to get 2.5 pounds

Convert to ounces

3 0

3 years ago

Before noon today, the shipping department at a factory has sent 36 boxes. this is 15% of the total number of boxes in a special

HACTEHA [7]

36/15=2.4
2.4*100=240 boxes

3 0

4 years ago

2. Suppose 27 blackberry plants started growing in a yard. Absent constraint, the blackberry plants will spread by 80% a month.

Marta_Voda [28]

Explanation

The question indicates we should use a logistic model to estimate the number of plants after 5 months.

This can be done using the formula below;

$\begin{gathered} P(t)=\frac{K}{1+Ae^{-kt}};A=\frac{K-P_{0_{}}}{P_0}_{} \\ \text{From the question} \\ P_0=\text{ Initial Plants=27} \\ K=\text{Carrying capacity =140} \end{gathered}$

Workings

Step 1: We would need to get the value of A using the carrying capacity and initial plants that started growing in the yard.

This gives;

$\begin{gathered} A=\frac{140-27}{27} \\ A=\frac{113}{27} \end{gathered}$

Step 2: Substitute the value of A into the formula.

$P(t)=\frac{140}{1+\frac{113}{27}e^{-kt}}$

Step 3: Find the value of the constant k

Kindly recall that we are told that the plants increase by 80% after each month. Therefore, after one month we would have;

$\begin{gathered} P(1)=27+(\frac{80}{100}\times27) \\ P(1)=\frac{243}{5} \end{gathered}$

We can then have that after t= 1month

$\begin{gathered} \frac{140}{1+\frac{113}{27}e^{-k\times1}}=\frac{243}{5} \\ Flip\text{ the equation} \\ \frac{1+\frac{113}{27}e^{-k}}{140}=\frac{5}{243} \\ 243(1+\frac{113}{27}e^{-k})=700 \\ 243+1017e^{-k}=700 \\ 1017e^{-k}=700-243 \\ 1017e^{-k}=457 \\ e^{-k}=\frac{457}{1017} \\ -k=\ln (\frac{457}{1017}) \end{gathered}$

Step 4: Substitute -k back into the initial formula.

$\begin{gathered} P(t)=\frac{140}{1+\frac{113}{27}e^{\ln (\frac{457}{1017})t}} \\ =\frac{140}{1+\frac{113}{27}(e^{\ln (\frac{457}{1017})})^t} \\ P(t)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^t} \\ \end{gathered}$

The above model is can be used to find the population at any time in the future.

Therefore after 5 months, we can estimate the model to be;

$\begin{gathered} P(5)=\frac{140}{1+\frac{113}{27}(\frac{457}{1017}^{})^5} \\ P(5)=\frac{140}{1.07668} \\ P(5)=130.029\approx130 \end{gathered}$

Answer: The estimated number of plants after 5 months is 130 plants.

8 0

1 year ago

How many times greater is 1,560,000 than 260,000

77julia77 [94]

6 times greater djdjjdjd

5 0

3 years ago

Can someone help me real quick? I’ll give brainliest

blsea [12.9K]

Answer:

Number 4 will be Quadrant 3

Number 5 will be Quadrant 1

Number 6 will be 3 -4

Step-by-step explanation:

For number 6 it says a point so that is what I did sorry if i`m wrong

8 0

3 years ago

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