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aalyn [17]
3 years ago
13

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

Mathematics
1 answer:
cricket20 [7]3 years ago
6 0
<span>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. </span>64c^15-a^27
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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

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2 years ago
A little help?!? Pleaseee
Ksju [112]

Answer:

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3 years ago
Prove :<br>sin²θ + cos²θ = 1<br><br><br>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 !

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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

0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)

Thus

\phi(y) = 1-0.1 = 0.9

by looking at the table, we find that y = 1.28, therefore

\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

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

Download pdf
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