All categories

2 years ago

If g(x) = -4x^2 - 3x + 2, determine g(-2).

Mathematics

2 answers:

Solnce55 [7]2 years ago

8 0

Answer:

Step-by-step explanation:

Since -2 is replacing x, all the other Xs get changed to -2:

Input -2:

g(-2)=-4(-2)^2-3(-2)+2

P.E.M.D.A.S:

Simplify Parenthesis:

g(-2)=8^2+6+2

Simplify Exponents:

g(-2)=64+6+2

no m, d, or s so straight to A:

Add:

g(-2)=72

i hope this helps u pls give a brainliest and a thx ;)

Scorpion4ik [409]2 years ago

6 0

Answer:

yea is 72

Step-by-step explanation:

You might be interested in

What other points are on the line of direct variation through (5, 12)? Check all that apply. (0, 0) (2.5, 6) (3, 10) (7.5, 18) (

aalyn [17]

we know that

A relationship between two variables, x, and y, represent a direct variation if it can be expressed in the form $y/x=k$ or $y=kx$

in this problem we have

the point $(5,12)$ is on the line of direct variation

Find the constant of proportionality k

$y/x=k$ -------> substitute ------> $k=12/5$

the equation is

$y=\frac{12}{5}x$

Remember that

If a point is on the line of direct variation

then

the point must satisfy the equation of direct variation

we're proceeding to verify each point

case A) point $(0,0)$

$x=0\ y=0$

Substitute the value of x and y in the direct variation equation

$0=\frac{12}{5}*0$

$0=0$ -------> is true

therefore

the point $(0,0)$ is on the line of direct variation

case B) point $(2.5,6)$

$x=2.5\ y=6$

Substitute the value of x and y in the direct variation equation

$6=\frac{12}{5}*2.5$

$6=6$ -------> is true

therefore

the point $(2.5,6)$ is on the line of direct variation

case C) point $(3,10)$

$x=3\ y=10$

Substitute the value of x and y in the direct variation equation

$10=\frac{12}{5}*3$

$10=7.2$ -------> is not true

therefore

the point $(3,10)$ is not on the line of direct variation

case D) point $(7.5,18)$

$x=7.5\ y=18$

Substitute the value of x and y in the direct variation equation

$18=\frac{12}{5}*7.5$

$18=18$ -------> is true

therefore

the point $(7.5,18)$ is on the line of direct variation

case E) point $(12.5,24)$

$x=12.5\ y=24$

Substitute the value of x and y in the direct variation equation

$24=\frac{12}{5}*12.5$

$18=30$ -------> is not true

therefore

the point $(12.5,24)$ is not on the line of direct variation

case F) point $(15,36)$

$x=15\ y=36$

Substitute the value of x and y in the direct variation equation

$36=\frac{12}{5}*15$

$36=36$ -------> is true

therefore

the point $(15,36)$ is on the line of direct variation

therefore

the answer is

$(0,0)$

$(2.5,6)$

$(7.5,18)$

$(15,36)$

5 0

2 years ago

Read 2 more answers

A video game that usually costs $30.65 is marked down 60%. Kelvin determined that the new price of the game would be $18.39. Loo

sladkih [1.3K]

The error.....Kelvin just figured the 60% markdown cost...not the final price

the final price would be : 30.65 - 0.60(30.65) = 30.65 - 18.39 = 12.26
so 12.26 is the final price

or another way to do this is : if it is marked down 60%, then u r actually paying 40%......0.40(30.65) = 12.26

6 0

2 years ago

Read 2 more answers

Which of the following is equivalent to the expression (3ab)(-5ab)?

Scilla [17]

$-15a^{2}b^{2}$

6 0

2 years ago

What is the value of n in the equation –negative StartFraction one-half EndFraction left-parenthesis 2 n plus 4 right-parenthesi

charle [14.2K]

Answer:

one

Step-by-step explanation:

10 0

2 years ago

Read 2 more answers

Define f(0,0) in a way that extends f to be continuous at the origin. f(x, y) = ln ( 19x^2 - x^2y^2 + 19 y^2/ x^2 + y^2) Let f (

kirill115 [55]

Answer:

f(0,0)=ln19

Step-by-step explanation:

$f(x,y)=ln(\frac{19x^2-x^2y^2+19y^2}{x^2+y^2})$ is given as continuous function, so there exist $lim_{(x,y)\rightarrow(0,0)}f(x,y)$ and it is equal to f(0,0).

Put x=rcosA annd y=rsinA

$f(r,A)=ln(\frac{19r^2cos^2A-r^2cos^2A*r^2sin^2A+19r^2sin^2A}{r^cos^2A+r^2sin^2A})=ln(\frac{19r^2(cos^2A+sin^2A)-r^4cos^2Asin^a}{r^2(cos^2A+sin^2A)})$

we know that $cos^2A+sin^2A=1$ , so we have that

$f(r,A))=ln(\frac{19r^2-r^4cos^2Asin^a}{r^2})=ln(19-r^2cos^2Asin^2A)$

$lim_{(x,y)\rightarrow(0,0)}f(x,y)=lim_{r\rightarrow0}f(r,A)=ln19$

So f(0,0)=ln19.

8 0

2 years ago