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3 years ago

Find the area of the rectangle with a length of 2x+2/x-4 and the width of 3x-12/x^2+6x+5

1 answer:

3 0

We're trying to evaluate a fractional expression here:
$\frac{2x + 2}{x -4} * \frac{3x - 12}{x^{2} + 6x + 5}$

The first thing to do, is factor everything we can. The quadratic factors to (x + 1)(x+5) and the expression above that factors to 3(x - 4). 2x + 2 factors to 2(x + 1).
$\frac{2(x + 1)}{x -4} * \frac{3(x - 4)}{(x+1)(x+5)}$

It looks a little complicated now, but if we combine the two we can cancel some stuff.

$\frac{2(x + 1) * 3(x - 4)}{(x -4)(x+1)(x+5)}$

Still looking complicated, but look, we have (x+1) and (x+4) on the top and the bottom, that means they don't matter and we can get rid of them:

$\frac{2 * 3}{x+5} = \frac{6}{x+5}$

Tada! The area is 6/(x+5). We'd need to know x to go any further!

Need any more explanation?

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Which undefined geometric term can be described as a one-dimensional set of points that has no beginning or end?

alexandr402 [8]

A line is a set of points that has no beginning and no end.

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3 years ago

Read 2 more answers

About Exercise 2.3.1: Proving conditional statements by contrapositive Prove each statement by contrapositive

Sladkaya [172]

Answer:

See proofs below

Step-by-step explanation:

A proof by counterpositive consists on assuming the negation of the conclusion and proving the negation of the hypothesis.

a) Assume that n is not odd. Then n is even, that is, n=2k for some integer k. Hence n²=4k²=2(2k²)=2t for some integer t=2k². Then n² is even, therefore n² is not odd. We have proved the counterpositive of this statement.

b) Assume that n is not even, then n is odd. Thus, n=2k+1 for some integer k. Now, n³=(2k+1)³=8k³+6k²+6k+1=2(4k³+3k²+3k)+1=2t+1 for the integer t=4k³+3k²+3k. Thus n³ is odd, that is, n³ is not even.

c) Suppose that n is not odd, that is, n is even. Now, n=2k for some integer k. Then 5n+3=10k+3=2(5k+1)+1, thus 5n+3 is odd, then 5n+3 is not even.

d) Suppose that n is not odd, then n is even. Now, n=2k for some integer k. Then n²-2n+7=4k²-4k+7=2(2k²-2k+3)+1. Hence n²-2n+7 is odd, that is, n²-2n+7 is not even.

e) Assume that -r is not irrational, then -r is rational. Since -1 is rational, then (-1)(-r)=r is rational. Thus r is not irrational.

f) Assume that 1/z is not irrational. Then 1/z is rational. Multiplucative inverses of rational numbers are rational, hence z is rational, that is, z is not irrational.

g) Suppose that z>y. We will prove that z³+zy²≤z²y+y³ is false, that is, we will prove that z³+zy²>z²y+y³. Multiply by the nonnegative number z² in the inequality z>y to get z³>z²y (here we assume z and y nonzero, in this case either z³>0=y³ is true or z³=0>y³ is true). On the other hand, multiply by z² (positive number) to get zy²>y³. Add both inequalities to obtain z³+zy²>z²y+y³ as required.

h) Suppose than n is even. Then n=2k, and n²=4k² is divisible by 4.

i) Assume that "z is irrational or y is irrational" is false. Then z is rational and y is rational. Rational numbers are closed under sum, then z+y is rational, that is, z+y is not irrational.

3 0

3 years ago

William is reading a book for class . The book is 94 pages long . For the first assignment , William read 25 pages . For the sec

rjkz [21]

The answer is 32, I hope this helps!!

7 0

3 years ago

-18+-6 please help

Verdich [7]

Answer:

-24

Step-by-step explanation:

If you are struggling with this here's a tip!

-18+-6 is what you are trying to solve

The 6 is negative, so get rid of the plus sign -18-6

2 negative numbers are just added like positive numbers

So add 18 and 6, you should get 24

Don't forget about the negative!

-24

6 0

3 years ago

An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15

NeTakaya

Answer:

D. 22,417 feet

Step-by-step explanation:

Fine the diagram in the attachment for proper elucidation. Using the SOH, CAH, TOA trigonometry identity to solve for the distance (x) from the plane (P) to the observer (O), the longest side x is the hypotenuse and the side facing the angle of elevation is the opposite.

Hypotenuse = x and Opposite = 15,000feet

According to SOH;

$sin 42^0 = \frac{Opposite}{Hypotenuse} \\\\Sin42^0 = \frac{15000}{x}\\ \\x = \frac{15000}{sin42^0}\\\\ \\$

$x = \frac{15000}{ 0.6691} \\\\x = 22,417 feet$

Hence the distance (x) from the plane P to the observer O is approximately 22,417 feet

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3 years ago