2 years ago

Help with maths question please

Mathematics

1 answer:

mars1129 [50]2 years ago

6 0

<h3>✽ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ✽</h3>

➷ Substitute -3 into g(x)

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

Solve:

g(x) = -2

Substitute this value into f(x)

f(-2) = 3(-2)^2

Solve:

f(-2) = 12

The answer is B. 12

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

You might be interested in

Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions?

REY [17]

<span>Which value of b will cause the quadratic equation x2 + bx + 5 = 0 to have two real number solutions?

–5</span>

5 0

2 years ago

Read 2 more answers

Factor 18bc + 9ab + 3cb completely.

Natasha_Volkova [10]

Answer:

$3b\left(7c+3a\right)$

Step-by-step explanation:

$18bc+9ab+3cb$

$=6\cdot \:3bc+3\cdot \:3ba+3bc$

$=3b\left(6c+3a+c\right)$

$=3b\left(7c+3a\right)$

6 0

2 years ago

a flower garden is in the shape of a right triangle. The longest side of the triangle measures 13 meters. One of the shorter sid

netineya [11]

The answer is 10.4 Hope this helps

7 0

2 years ago

Read 2 more answers

6. How could you change the simulation to generate data for three pets? Result Frequency Heads, Heads 9 Heads, Tails 14 Tails, H

ArbitrLikvidat [17]

9 plus 14 plus 18 plus 9 is 50

8 0

2 years ago

Laplace Transform Let f be a function defined for t ≥ 0. Then the integral ℒ{f(t)} = [infinity] e−stf(t) dt 0 is said to be the

Anastaziya [24]

Answer:

The laplace transform is $\frac{2}{(s+6)^3}$

Step-by-step explanation:

We will solve this problem by applying the laplace transform properties (their proofs are beyond the scope of this explanation).

Consider first the function f(t) = 1. By definition of the laplace transform, we have

$F(s) = \int_{0}^{\infty}f(t)e^{-st}dt$

when f(t) = 1 we get

$F(s) = \int_{0}^{\infty}e^{-st}dt = \left.\frac{-e^{-st}}{s}\right|_{0}^{\infty} = \frac{1}{s}$

We will apply the following properties: Define L(f) as applying the laplace transform

$L(e^{at}f(t)) = F(s-a)$ (this means, multiplying by an exponential corresponds to a shift in the s parameter of the transform of f)

$L(t^nf(t)) = (-1)^n\frac{d^n F}{ds^n}$ (this is, multypling by $t^n$ is equivalent to taking the n-th derivative of the transform.

We are given the function $g(t) = t^2e^{-6t}\cdot 1$ . Since the transform of the constant function 1 is 1/s, by applying the first property we get

$L(e^{-6t}\cdot 1 ) = \frac{1}{s+6}$

By applying the second property we get

$L(g(t)) = L(t^2 e^{-6t} \cdot 1) = (-1)^2\frac{d^2}{ds^2}(\frac{1}{s+6}) = \frac{2}{(s+6)^3}$

3 0

2 years ago