All categories

3 years ago

Which expression is a difference of cubes 9w^33-y^12 18p^15-q^21 36a^22-b^16 64c^15-a^27

1 answer:

6 0

A. 9w^33-y^12
B. 18p^15-q^21
C. 36a^22-b^16
D. 64c^15-a^27 -----> 64c^15 = (4c^5)^3; a^27 = (a^9)^3

Answer: D. 64c^15-a^27

You might be interested in

Please help 2х -3y =18

siniylev [52]

Answer:

x = 3(6 + y)/2

Step-by-step explanation:

Solving for x

Add 3y to both sides.

2x = 18 + 3y

Divide both sides by 2.

x = 18 + 3y/2

Factor out the common term 3.

x = 3(6 + y)/2

3 0

2 years ago

A little help?!? Pleaseee

Ksju [112]

Answer:

I think no. Because degree is the sum all variables in first equation degree. Is14 but second equation degree not 14.

4 0

3 years ago

Prove : sin²θ + cos²θ = 1 thankyou ~

Gnom [1K]

Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

$\sf\longrightarrow sin^2\theta + cos^2\theta = 1$

Imagine a right angled triangle with one of its acute angle as $\theta$ .

The side opposite to this angle will be perpendicular .
Also we know that ,

$\sf\longrightarrow sin\theta =\dfrac{p}{h} \\$

$\sf\longrightarrow cos\theta =\dfrac{b}{h}$

And by Pythagoras theorem ,

$\sf\longrightarrow h^2 = p^2+b^2 \dots (i)$

Where the symbols have their usual meaning.

Now , taking LHS ,

$\sf\longrightarrow sin^2\theta +cos^2\theta$

Substituting the respective values,

$\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\$

$\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\$

$\sf\longrightarrow \dfrac{p^2+b^2}{h^2}$

From equation (i) ,

$\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\$

$\sf\longrightarrow \bf 1 = RHS$

Since LHS = RHS ,

Hence Proved !

I hope this helps.

5 0

2 years ago

The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard devia

KengaRu [80]

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

$W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}$

The values of the cumulative distribution function of W, denoted by $\phi$ , can be found in the attached file. Lets call $y = \frac{z-70}{4}$ . We have