All categories

3 years ago

How to factor a polynomial in the form p^2-m^2 HELP!!

1 answer:

6 0

Answer: if you are going to put it like p^2-m^2 no one going to have that answer so I need the p and the m like for example the p=5 and the m=4

Step-by-step explanation: so it going to be like

Evaluate for m=4,p=5

5^2−4^2

So I can’t answer how you put it

You might be interested in

WILL GIVE BRAINLIEST! 15 POINTS! ASAP!

gogolik [260]

Answer:

12/7 = 1 5/7

Step-by-step explanation:

divide 12 by 7 to get 1 with a remainder of 5 and take the denominator and put it under the remainder 5/7 the answer is 1 5/7

7 0

3 years ago

A sunflower grows to be 200cm tall. After four weeks, it was 30% of this height. How tall

hichkok12 [17]

Answer:

60cm

Step-by-step explanation:

200*3/10=20*3=60

6 0

2 years ago

On average, the moon is 384,400 kilometers from the Earth. How is this distance written in

KiRa [710]

Answer:

Step-by-step explanation:

hope this helps with the work

5 0

3 years ago

Read 2 more answers

Need help with this one!!!

DaniilM [7]

$\bf (2x^2)^4(5x^6)\impliedby \textit{distributing the exponent}
\\\\\\
2^4x^{2\cdot 4}5x^6\implies 2^4\cdot 5x^{2\cdot 4}x^6\implies 16\cdot 5x^8x^6\implies 80x^{8+6}
\\\\\\
80x^{14}$

8 0

3 years ago

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l

tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

Step(ii):-

The Empirical rule

$Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }$

Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = P(Z≥2.383)

= 1- P( Z<2.383)

= 1-( 0.5 -+A(2.38))

= 0.5 - A(2.38)

= 0.5 -0.4913

= 0.0087

Final answer:-

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0

2 years ago