Solve Q = 144 p y f o r y Q=\frac{144p}{y}\ \ \ \ \ \ for\ y Q = y 1 4 4 p f o r y

y = 144 p Q y=\frac{144p}{Q} y = Q 1 4 4 p

y = Q 144 p y=\frac{Q}{144p} y = 1 4 4 p Q

y = 144 p Q y=144pQ y = 1 4 4 p Q

y = 144 Q p y=\frac{144Q}{p} y = p 1 4 4 Q

If f ( x ) = 3 x + 1 2 f\left(x\right)=\frac{3x+1}{2} f ( x ) = 2 3 x + 1 then f − 1 ( x ) = f^{-1}\left(x\right)= f − 1 ( x ) =

2 x − 1 3 \frac{2x-1}{3} 3 2 x − 1

2 y − 1 3 \frac{2y-1}{3} 3 2 y − 1

3 y + 1 2 \frac{3y+1}{2} 2 3 y + 1

3 x + 2 1 \frac{3x+2}{1} 1 3 x + 2

Solve R = l + 3 w 2 f o r w R=\frac{l+3w}{2}\ \ \ \ \ \ \ \ for\ w R = 2 l + 3 w f o r w

w = 2 R − l 3 w=\frac{2R-l}{3} w = 3 2 R − l

2 R − l + 3 = w 2R-l+3=w 2 R − l + 3 = w

l = 2 R − 3 w l=2R-3w l = 2 R − 3 w

w = R 2 l − 3 w=\frac{R}{2l}-3 w = 2 l R − 3

Solve a x + b y + c = 0 ax+by+c=0\ a x + b y + c = 0 for y y y

y = − ( a x + c ) b y=-\frac{\left(ax+c\right)}{b} y = − b ( a x + c )

y = a x + c b y=\frac{ax+c}{b} y = b a x + c

x = − b y − c a x=\frac{-by-c}{a} x = a − b y − c

y = − a x − b y y=-ax-by y = − a x − b y

Given that sin ⁡ θ = 0.95 \sin\theta=0.95 sin θ = 0 . 9 5

θ = 71.8 ° \theta=71.8\degree θ = 7 1 . 8 °

θ = 55 ° \theta=55\degree θ = 5 5 °

θ = π r a d i a n s \theta=\pi\ radians θ = π r a d i a n s

θ = 0 ° \theta=0\degree θ = 0 °

D = 11 5 ( P − 15 ) f o r P D=\frac{11}{5}\left(P-15\right)\ for\ P D = 5 1 1 ( P − 1 5 ) f o r P Solve

P = 5 D 11 + 15 P=\frac{5D}{11}+15 P = 1 1 5 D + 1 5

P = 5 D + 15 11 P=\frac{5D+15}{11} P = 1 1 5 D + 1 5

P = 5 D − 11 15 P=\frac{5D-11}{15} P = 1 5 5 D − 1 1

P = 5 D 11 − 15 P=\frac{5D}{11}-15 P = 1 1 5 D − 1 5

( 2 x + 3 ) ( x − 5 ) ( x + 1 ) = 0 \left(2x+3\right)\left(x-5\right)\left(x+1\right)=0 ( 2 x + 3 ) ( x − 5 ) ( x + 1 ) = 0 The roots of

− 3 2 , 5 , − 1 -\frac{3}{2},\ 5,\ -1 − 2 3 , 5 , − 1

3 2 , − 5 , 1 \frac{3}{2},\ -5,\ 1 2 3 , − 5 , 1

− 2 3 , 5 , − 1 -\frac{2}{3},\ 5,\ -1 − 3 2 , 5 , − 1

− 3 , 5 , − 1 -3,\ 5,\ -1 − 3 , 5 , − 1

Two angles that are coterminal with 70 ° 70\degree\ 7 0 ° are:

430 ° a n d − 290 ° 430\degree\ and\ -290\degree 4 3 0 ° a n d − 2 9 0 °

70 ° a n d 110 ° 70\degree\ and\ 110\degree 7 0 ° a n d 1 1 0 °

430 ° a n d 610 ° 430\degree\ and\ 610\degree 4 3 0 ° a n d 6 1 0 °

110 ° a n d − 250 ° 110\degree\ and\ -250\degree 1 1 0 ° a n d − 2 5 0 °

Algebra 3 EOC

Authored by Arthur McIntosh

Mathematics

12th Grade

CCSS covered

Used 39+ times

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Solve
$Q=\frac{144p}{y}\ \ \ \ \ \ for\ y$

$y=\frac{Q}{144p}$

$y=144pQ$

$y=\frac{144p}{Q}$

$y=\frac{144Q}{p}$

Tags

CCSS.HSA.CED.A.4

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

When I divide $\left(2x^2-9x-5\right)$ by $\left(x-5\right)$ the remainder is:

-7

Tags

CCSS.HSA.APR.B.2

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$The\ quotient\ of\ \left(2x^2-9x-5\right)\div\left(2x+1\right)\ is$

$x-5$

$0$

$x+5$

$2x-5$

Tags

CCSS.HSA.APR.D.6

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Calculate the value of h

Tags

CCSS.8.G.B.7

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

The measure of x is equal to:

21.7

19.8

22.6

Tags

CCSS.HSG.SRT.C.8

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

From a geometric perspective, the inverse of a function is:

The reflection of it's graph in the y-axis as the mirror

The reflection of it's graph in the x-axis as the mirror

The reflection of it's graph in the line y=x as the mirror

The reflection of it's graph in the line y=-x as the mirror

Tags

CCSS.HSF-BF.B.4C

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

If $f\left(x\right)=\frac{3x+1}{2}$ then $f^{-1}\left(x\right)=$

$\frac{2x-1}{3}$

$\frac{2y-1}{3}$

$\frac{3y+1}{2}$

$\frac{3x+2}{1}$

Tags

CCSS.HSF-BF.B.4A

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Solve $R=\frac{l+3w}{2}\ \ \ \ \ \ \ \ for\ w$

$2R-l+3=w$

$w=\frac{2R-l}{3}$

$l=2R-3w$

$w=\frac{R}{2l}-3$

Tags

CCSS.HSA.CED.A.4

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Algebra 3 EOC

Solve
$Q=\frac{144p}{y}\ \ \ \ \ \ for\ y$

When I divide $\left(2x^2-9x-5\right)$ by $\left(x-5\right)$ the remainder is:

$The\ quotient\ of\ \left(2x^2-9x-5\right)\div\left(2x+1\right)\ is$

Calculate the value of h

The measure of x is equal to:

From a geometric perspective, the inverse of a function is:

If $f\left(x\right)=\frac{3x+1}{2}$ then $f^{-1}\left(x\right)=$

Solve $R=\frac{l+3w}{2}\ \ \ \ \ \ \ \ for\ w$

Which law would you use to answer this question?

To the nearest whole number, c is equal to

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Algebra 3 EOC

Solve Q=144py for yQ=\frac{144p}{y}\ \ \ \ \ \ for\ yQ=y144p​ for y

When I divide (2x2−9x−5)\left(2x^2-9x-5\right)(2x2−9x−5) by (x−5)\left(x-5\right)(x−5) the remainder is:

The quotient of (2x2−9x−5)÷(2x+1) isThe\ quotient\ of\ \left(2x^2-9x-5\right)\div\left(2x+1\right)\ isThe quotient of (2x2−9x−5)÷(2x+1) is

Calculate the value of h

The measure of x is equal to:

From a geometric perspective, the inverse of a function is:

If f(x)=3x+12f\left(x\right)=\frac{3x+1}{2}f(x)=23x+1​ then f−1(x)=f^{-1}\left(x\right)=f−1(x)=

Solve R=l+3w2 for wR=\frac{l+3w}{2}\ \ \ \ \ \ \ \ for\ wR=2l+3w​ for w

Which law would you use to answer this question?

To the nearest whole number, c is equal to

Access all questions and much more by creating a free account

Similar Resources on Wayground

Popular Resources on Wayground

Solve
$Q=\frac{144p}{y}\ \ \ \ \ \ for\ y$

When I divide $\left(2x^2-9x-5\right)$ by $\left(x-5\right)$ the remainder is:

$The\ quotient\ of\ \left(2x^2-9x-5\right)\div\left(2x+1\right)\ is$

If $f\left(x\right)=\frac{3x+1}{2}$ then $f^{-1}\left(x\right)=$

Solve $R=\frac{l+3w}{2}\ \ \ \ \ \ \ \ for\ w$