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2 years ago

PLEASE HELP ME WITH MY PROBLEM!!! 50 POINTS!!!

Mathematics

1 answer:

Molodets [167]2 years ago

5 0

Well he started off right by solving for y=y to find the abscissa of the intersection points between the two points,but then he assumed that their ordinate (y coordinates) are both zero,which is incorrect obviously. To find their ordinate,all we have to do is plug the x values into any of the two equations, since they both will pass through that point ੴ

$y = 4 x+ 1 \\ y = 4(3) + 1 \\ y = 12 + 1 \\ y = 13$

$y = 4x + 1 \\ y = 4( - 2) + 1 \\ y = - 8 + 1 \\ y = - 7$

I chose to plug the values in the second linear equation, since it's easier to compute

<h2>Points:</h2>

You might be interested in

Simplify each rational expression to lowest terms, specifying the values of xx that must be excluded to avoid division

k0ka [10]

Answer:

(a) $\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}$ . The domain of this function is all real numbers not equal to -2 or 5.

(b) $\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}$ . The domain of this function is all real numbers not equal to 0, $-1+\sqrt{2}$ or $-1+\sqrt{2}$ .

(c) $\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}$ .The domain of this function is all real numbers not equal to 2 or -4.

(d) $\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}$ . The domain of this function is all real numbers not equal to -2.

(e) $\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$ . The domain of this function is all real numbers.

Step-by-step explanation:

To reduce each rational expression to lowest terms you must:

(a) For $\frac{x^2-6x+5}{x^2-3x-10}$

$\mathrm{Factor}\:x^2-6x+5\\\\x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)\\x^2-6x+5=x\left(x-1\right)-5\left(x-1\right)\\\\\mathrm{Factor\:out\:common\:term\:}x-1\\x^2-6x+5=\left(x-1\right)\left(x-5\right)$

$\mathrm{Factor}\:x^2-3x-10\\\\x^2-3x-10=\left(x^2+2x\right)+\left(-5x-10\right)\\x^2-3x-10=x\left(x+2\right)-5\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\x^2-3x-10=\left(x+2\right)\left(x-5\right)$

$\frac{x^2-6x+5}{x^2-3x-10}=\frac{\left(x-1\right)\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x-5\\\\\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}$

The denominator in a fraction cannot be zero because division by zero is undefined. So we need to figure out what values of the variable(s) in the expression would make the denominator equal zero.

To find any values for x that would make the denominator = 0 you need to set the denominator = 0 and solving the equation.

$x^2-3x-10=\left(x+2\right)\left(x-5\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+2=0\\x=-2\\\\x-5=0\\x=5$

The domain is the set of all possible inputs of a function which allow the function to work. Therefore the domain of this function is all real numbers not equal to -2 or 5.

(b) For $\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3+3x^2+3x+1\mathrm{\:and\:the\:divisor\:}x^3+2x^2-x\mathrm{\::\:}\frac{x^3}{x^3}=1$

Quotient = 1

$\mathrm{Multiply\:}x^3+2x^2-x\mathrm{\:by\:}1:\:x^3+2x^2-x$

$\mathrm{Subtract\:}x^3+2x^2-x\mathrm{\:from\:}x^3+3x^2+3x+1\mathrm{\:to\:get\:new\:remainder}$

Remainder = $x^2+4x+1}$

$\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}$

The domain of this function is all real numbers not equal to 0, $-1+\sqrt{2}$ or $-1+\sqrt{2}$ .

$x^3+2x^2-x=0\\\\x^3+2x^2-x=x\left(x^2+2x-1\right)=0\\\\\mathrm{Solve\:}\:x^2+2x-1=0:\quad x=-1+\sqrt{2},\:x=-1-\sqrt{2}$

(c) For $\frac{x^2-16}{x^2+2x-8}$

$x^2-16=\left(x+4\right)\left(x-4\right)$

$x^2+2x-8= \left(x-2\right)\left(x+4\right)$

$\frac{x^2-16}{x^2+2x-8}=\frac{\left(x+4\right)\left(x-4\right)}{\left(x-2\right)\left(x+4\right)}\\\\\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}$

The domain of this function is all real numbers not equal to 2 or -4.

$x^2+2x-8=0\\\\x^2+2x-8=\left(x-2\right)\left(x+4\right)=0$

(d) For $\frac{x^2-3x-10}{x^3+6x^2+12x+8}$

$\mathrm{Factor}\:x^2-3x-10\\\left(x^2+2x\right)+\left(-5x-10\right)\\x\left(x+2\right)-5\left(x+2\right)$

$\mathrm{Apply\:cube\:of\:sum\:rule:\:}a^3+3a^2b+3ab^2+b^3=\left(a+b\right)^3\\\\a=x,\:\:b=2\\\\x^3+6x^2+12x+8=\left(x+2\right)^3$

$\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)^3}\\\\\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}$

The domain of this function is all real numbers not equal to -2

$x^3+6x^2+12x+8=0\\\\x^3+6x^2+12x+8=\left(x+2\right)^3=0\\x=-2$

(e) For $\frac{x^3+1}{x^2+1}$

$\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$

The domain of this function is all real numbers.

$x^2+1=0\\x^2=-1\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-1},\:x=-\sqrt{-1}$

4 0

3 years ago

-21+70n=77<br>please help me Idk the answer

asambeis [7]

Add 21 to both sides: 70n=98
Divide each side by 70: n=98/70
Reduce the fraction: n=7/5

7 0

4 years ago

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PLEASE HURRY!! <br> simplify completely<br> show all work for full credit

Irina18 [472]

Answer:

$4x^{3}$

Step-by-step explanation:

<u>Given:</u>

$(\frac{192x^{12}}{3x^{3}} )^{\frac{1}{3}}$

<u>Apply exponent rule of distribution:</u>

$\frac{(192x^{12})^{\frac{1}{3}}}{(3x^{3})^{\frac{1}{3}}}$

<u>Simplify the numerator:</u>

$\\\\\frac{4 * 3^{\frac{1}{3}}x^{4}}{(3x^{3})^{\frac{1}{3}}}$

<u>Simplify the denominator:</u>

$\\\frac{4 * 3^{\frac{1}{3}}x^{4}}{3^\frac{1}{3}x}}$

<u>Simplify:</u>

$4x^{3}$

-> To explain this party since it is a bigger jump, $3^{\frac{1}{3}}x$ is on the top and the bottom, so it becomes a one. We are left with a four on the top, and using properties of exponents 4 - 1 = 3, explaining why we have $x^{3}$ leftover too.

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

3 0

3 years ago

What value is added to both sides of the equation x2 − 4x = −2 in order to solve by completing the square? 4, 2 , −2 , −4

Fiesta28 [93]

X^2 - 4 x = -2
Add 4 to both sides:
x^2 - 4 x + 4 = 2
Write the left-hand side as a square:
(x - 2)^2 = 2
Take the square root of both sides:
x - 2 = sqrt(2) or x - 2 = -sqrt(2)
Add 2 to both sides:
x = 2 + sqrt(2) or x - 2 = -sqrt(2)
Add 2 to both sides:
Answer: x = 2 + sqrt(2) or x = 2 - sqrt(2)

6 0

3 years ago

The ratio of height between Joey and his girlfriend is 8:7. If Joey is 72 inches tall, how tall is his girlfriend in inches

olganol [36]

Ratio is Joey/girlfriend = 8/7=8x/7x.
8x is height of Joey.
8x=72
x=72/8=9

7x is height of Joey's girlfriend.
7x=7*9=63 in.
Answer: 63 in.

7 0

3 years ago

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