All categories

3 years ago

If y=-8 when x=-2,find x when y =32

1 answer:

6 0

So when y=32 it is 4x more than the original y which means they've multiplied it by four. Whatever you do to one you have to do to the other side so you multiply 2 by 4 to get 8 so x=8

You might be interested in

every year, a food company shares 1/5 of its earnings with its employees as a bonus. last year the total bonus was 220,000. how

NNADVOKAT [17]

Answer:

Step-by-step explanation:

1/5 of earnings = 220,000

Let earning be x

1/5 of x = 220,000

$\frac{1}{5}*x=220000\\\\x=220000*5\\\\x=1,100,000$

5 0

3 years ago

Read 2 more answers

Factor the polynomial completely

aliina [53]

We have $2x^2-8$ .

First let us factor the 2 to obtain,

$2(x^2-4)$

We can now see clearly that the expression inside the parentheses is a difference of two squares.

We now write the 4 as 2². So that we obtain the expression,

$2(x^2-2^2)$

Recall that

$a^2-b^2=(a+b)(a-b)$

Hence our expression becomes,

$2(x+2)(x-2)$

Therefore, when factored completely,

$2x^2-8=2(x+2)(x-2)$

5 0

3 years ago

Please Answer! Thanks. <br><br> $45,838.22 / 12=

kondaur [170]

the answer is 3819.85

6 0

3 years ago

Read 2 more answers

Use triangle ABC drawn below & only the sides labeled. Find the side of length AB in terms of side a, side b & angle C o

Brrunno [24]

Answer:

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

Step-by-step explanation:

Given:

The above triangle

Required

Solve for AB in terms of a, b and angle C

Considering right angled triangle BOC where O is the point between b-x and x

From BOC, we have that:

$Sin\ C = \frac{h}{a}$

Make h the subject:

$h = aSin\ C$

Also, in BOC (Using Pythagoras)

$a^2 = h^2 + x^2$

Make $x^2$ the subject

$x^2 = a^2 - h^2$

Substitute $aSin\ C$ for h

$x^2 = a^2 - h^2$ becomes

$x^2 = a^2 - (aSin\ C)^2$

$x^2 = a^2 - a^2Sin^2\ C$

Factorize

$x^2 = a^2 (1 - Sin^2\ C)$

In trigonometry:

$Cos^2C = 1-Sin^2C$

So, we have that:

$x^2 = a^2 Cos^2\ C$

Take square roots of both sides

$x= aCos\ C$

In triangle BOA, applying Pythagoras theorem, we have that:

$AB^2 = h^2 + (b-x)^2$

Open bracket

$AB^2 = h^2 + b^2-2bx+x^2$

Substitute $x= aCos\ C$ and $h = aSin\ C$ in $AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = h^2 + b^2-2bx+x^2$

$AB^2 = (aSin\ C)^2 + b^2-2b(aCos\ C)+(aCos\ C)^2$

Open Bracket

$AB^2 = a^2Sin^2\ C + b^2-2abCos\ C+a^2Cos^2\ C$

Reorder

$AB^2 = a^2Sin^2\ C +a^2Cos^2\ C + b^2-2abCos\ C$

Factorize:

$AB^2 = a^2(Sin^2\ C +Cos^2\ C) + b^2-2abCos\ C$

In trigonometry:

$Sin^2C + Cos^2 = 1$

So, we have that:

$AB^2 = a^2 * 1 + b^2-2abCos\ C$

$AB^2 = a^2 + b^2-2abCos\ C$

Take square roots of both sides

$AB = \sqrt{a^2 + b^2-2abCos\ C}$

6 0

3 years ago

I’m not sure how to do this please help

Nikolay [14]

Answer:

B :step 2 she didnt collect all the like terms and calculate

Step-by-step explanation:

first rewrite remove all ( )Parentheses

2nd collect all like terms and calculate

a^4 + 7a -16 -12^a^3 + 5a -3

a^4 + 12a - 19 - 12a^3 ( Like terms are 7a +5a and -16 +-3)

so she skipped the second step

3 0

2 years ago