Determine g(-1) for g(t)=(t^2-1)(t+4) A)-2 B)-6 C)3 D)0

2 answers:

7 0

It is C because Since {{g(x)}={x^3+2}}g(x)=x3+2g, left parenthesis, x, right parenthesis, equals, x, start superscript, 3, end superscript, plus, 2, we can substitute {x^3+2}x3+2x, start superscript, 3, end superscript, plus, 2 in for {g(x)}g(x)g, left parenthesis, x, right parenthesis.\begin{aligned}{f(g(x))}&=3(g(x))-1 \\\\ &=3({x^3+2})-1 \\\\ &=3x^3+6-1\\\\ &=3x^3+5 \end{aligned}f(g(x))=3(g(x))−1=3(x3+2)−1=3x3+6−1=3x3+5

Zanzabum2 years ago

3 0

D option that is 0 will be answer

You might be interested in

Solve for y: 5x-3y<75

Arada [10]

y<-5/3x+25 Is your answer. To get it you subtract the 5x to both sides and then you divide the 3 on both sides

answer : y < -5/3x +25

4 0

3 years ago

Bailey tossed a coin 10 times. the results were 7 heads and 3 tails. What is the best comparison between the theoretical and exp

erastova [34]

Answer:

3/10

Step-by-step explanation:

i answered it already

8 0

3 years ago

What's the slope-intercept form that passes through the points (0, -1) and (1, 5)?

Rzqust [24]

Answer:

Step-by-step explanation:

$\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}$

(0, -1) ⇒ x₁ = 0, y₁ = -1

(1, 5) ⇒ x₂ = 1, y₂ = 5

So the slope:

$\bold{m=\dfrac{5+1}{1-0}=\dfrac{6}{1}=6}$

The slope-intercept form of the equation of line is y = mx + b, where m is the slope and b is the y-intercept of the line.

(0, -1) ⇒ x₀ = 0, y₀ = -1 ⇒ b = -1

Therefore:

y = 6x - 1 ← the slope-intercept form of the equation

5 0

3 years ago

I need help, Now please

Sav [38]

Answer
D

Explanation:

7 0

2 years ago

#1. Which is equal to z3 8? A. (z - 2)(z^2 - 2z - 4). B. (z + 2)(z^2 - 2z + 4) C.(z - 2)(z^2+ 2z + 4). D. (z + 2)(z^2+ 2z - 4) #

tankabanditka [31]

#1. B
(z * z^2 + z * 2z + z * 4) – (-2 *z^2 – (-2) 2z – (-2) 4)
Z^3 + 2z^2 + 4z – 2z^2 -4z – 8
Z^3 + 2z^2 – 2z^2 + 4z – 4z – 8
Z^3 - 8

#2 and #3. D
(x + y)(x + 2)
x^2 + 2x + yx + 2y

#4. D.
(x - 7)(x + 7)(x- 2)
x^2 + 7x – 7x -49
x^2 + x – 49
x^2 -49
(x^2 – 49 ) (x – 2)
x^3 – 2x^2 – 49x + 98

#5. C
(y - 4) = 0
y = 4
(x + 3)= 0
x = -3

#6. A and B

3 0

3 years ago