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2 years ago

Line a is parallel to line b, m<2 = 4x+44 , And m<6=6x+36. Find the value of x

Mathematics

1 answer:

Savatey [412]2 years ago

3 0

Answer:

x = 4

Step-by-step explanation:

<2 and <6 are equal because of the corresponding angles thm so:

4x + 44 = 6x + 36

4x - 6x = 36 - 44

-2x = -8

x = 4

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Tan 12 degrees = 10/x

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I need help can you?

gizmo_the_mogwai [7]

Answer:

Step-by-step explanation:

Statements Reasons

1). m∠QPS = m∠TPV 1). Given

2). m∠QPS = m∠1 + m∠3 2). Angle addition postulate

m∠TPV = m∠2 + m∠4

3). m∠1 + m∠3 = m∠2 + m∠4 3). Given

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2 years ago

Rachel budget $76.50 for school clothes. This is 0.45 of her total budget. How much does Rachel have in her total budget?

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Oct 23, 2015 12:21 AM

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2 years ago

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What is the equivalent ratio for 70:55

GarryVolchara [31]

Answer:

Step-by-step explanation:

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3 years ago

What is the result when x^3+ 7x^2+ 9x - 8 is divided by x+ 2?

Ahat [919]

Answer:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$ .

Step-by-step explanation:

We could do this by synthetic division since the denominator is a linear factor in the form $x-c$ .

Since we are dividing by $x+2=x-(-2)$ , this is our setup for the synthetic division:

-2 | 1 7 9 -8

| -2 -10 2

______________

1 5 -1 -6

So the quotient is $1x^2+5x+-1$ and the remainder is $-6$ .

So $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ .

We can also do long division.

x^2+5x-1

____________________

x+2| x^3+7x^2+9x-8

-(x^3+2x^2)

-------------------

5x^2+9x-8

-( 5x^2+10x)

--------------------

-x-8

-(-x-2)

--------------

-6

So we see here we get the same quotient, $x^2+5x-1$ . and the same remainder, $-6$ .

Now let's check our result that:

$\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ .

So I'm going to rewrite the right hand side as a single fraction:

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)}{x+2}+\frac{-6}{x+2}$ .

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

Now let's focus on multiplying $(x+2)(x^2+5x-1)$ .

We are going to multiply the first term of the first ( ) to every term in the second ( ).

We are also going to multiply the second term of the first ( ) to every term in the second ( ).

$x(x^2)=x^3$

$x(5x)=5x^2$

$x(-1)=-x$

$2(x^2)=2x^2$

$2(5x)=10x$

$2(-1)=-2$

---------------------------Combine like terms:

$x^3+(5x^2+2x^2)+(-x+10x)+-2$

$x^3+7x^2+9x-2$

So let's go back where we were in our check of $\frac{x^3+7x^2+9x-8}{x+2}=x^2+5x-1+\frac{-6}{x+2}$ :

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{(x+2)(x^2+5x-1)-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-2-6}{x+2}$

$\frac{x^3+7x^2+9x-8}{x+2}=\frac{x^3+7x^2+9x-8}{x+2}$

We have the exact same thing on both sides so we did good.

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2 years ago

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