5488÷32 = a. 171R16 b. 176R18 c. 176R10 d. 171R8

A right triangle has a side with length 12 in and a hypotenuse with length 20 in. find the length of the missing leg. (round to

OLEGan [10]

Let the unknown side be "z" inches
Through pythagoras:
${12}^{2} + {z}^{2} = {20}^{2}$
Because the square of the two sides add up to the square of the hypotenuse in a right triangle.

This means that
${z }^{2} = {20}^{2} - {12}^{2}$
${z}^{2} = 400 - 144 = 256$
$z = \sqrt{256} = 16$
So the missing side is 16 inches

Hope this helped

6 0

3 years ago

The sum of two numbers is 15. four times the smaller number is 60 less than twice the larger number. what is the larger number?

kirza4 [7]

X = smaller number, y = larger number
x + y = 15.....x = 15 - y
4x = 2y - 60

4(15 - y) = 2y - 60
60 - 4y = 2y - 60
60 + 60 = 2y + 4y
120 = 6y
120/6 = y
20 = y <==== larger number is 20

x + y = 15
x + 20 = 15
x = 15 - 20
x = - 5

4 0

3 years ago

dog brought a new jet ski for $299 down in 14 monthly payments are $57 how much did Doug pay for the jet ski total

Effectus [21]

If he paid $57 monthly for 14 months, the total amount paid is:

$57\times14=798$

He paid $798 in total

8 0

7 months ago

Find the partial fraction decomposition of the rational expression with prime quadratic factors in the denominator

SpyIntel [72]

$\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}$
$\implies 5x^4-7x^3-12x^2+6x+21=a_1(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(a_4x+a_5)(x-3)$

When $x=3$ , you're left with

$147=49a_1\implies a_1=\dfrac{147}{49}=3$

When $x=\sqrt2$ or $x=-\sqrt2$ , you're left with

$\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}$

Adding the two equations together gives $-10=2a_5$ , or $a_5=-5$ . Subtracting them gives $2\sqrt2=2\sqrt2a_4$ , $a_4=1$ .

Now, you have

$5x^4-7x^3-12x^2+6x+21=3(x^2-2)^2+(a_2x+a_3)(x-3)(x^2-2)+(x-5)(x-3)$
$5x^4-7x^3-12x^2+6x+21=3x^4-11x^2-8x+27+(a_2x+a_3)(x-3)(x^2-2)$
$2x^4-7x^3-x^2+14x-6=(a_2x+a_3)(x-3)(x^2-2)$

By just examining the leading and lagging (first and last) terms that would be obtained by expanding the right side, and matching these with the terms on the left side, you would see that $a_2x^4=2x^4$ and $a_3(-3)(-2)=6a_3=-6$ . These alone tell you that you must have $a_2=2$ and $a_3=-1$ .

So the partial fraction decomposition is

$\dfrac3{x-3}+\dfrac{2x-1}{x^2-2}+\dfrac{x-5}{(x^2-2)^2}$

7 0

3 years ago

Domain and range of the inequality y<√x+3 +1

Margaret [11]

The domain of the given inequality is y>1.

<h3>What is Inequality?</h3>

An inequality is a relation which makes a non-equal comparison between two numbers or other mathematical expressions.

Here, given inequality:

y < $\sqrt{x+3}+1$

Related equation:

y = $\sqrt{x+3}+1$

The equation defined as,

x+3 > 0

x > -3

In the given inequality, the sign of inequality is <, it means the points on the boundary line are not included in the solution set. Thus, -3 is not included in the domain.

So, the domain of the given inequality is x>-3.

We know that,

$\sqrt{x+3} \geq 0$