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3 years ago

What is the range of the function f(x) = 4x + 9, given the domain D = {-4, -2, 0, 2}? please help

1 answer:

8 0

Since this is a linear function, filling in the minimum and maximum of the domain is sufficient.

f(-4) = -16 + 9 = -7
f(2)= 8 + 9 = 17

So the range of the function (given the domain) :
R = {-7, 17}

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Which expression shows the prime factorization of 36?

Svetradugi [14.3K]

36....we need to break this down...12 * 3 = 36
12 * 3....we can break this down because 12 is not prime...4 * 3 = 12
4 * 3 * 3...we can break the 4 down because it is not prime
2 * 2 * 3 * 3 <==== prime factorization

6 0

3 years ago

2 less then 3 times a number is 9

Masteriza [31]

$\huge\boxed{\text{As an expression:}\ 3x-2=9}$

$\huge\boxed{\text{Solved for x:}\ x=\frac{11}{3}}$

Let's start with " $3$ times a number". We will use $x$ to represent the number. $3x$

" $2$ less than" is next. $-2$

Finally, "is $9$ ". $=9$

Put it all together to get an expression. $\boxed{3x-2=9}$

Now, let's solve for the unknown number.

Add $2$ on both sides of the equation. $3x=11$

Divide both sides of the equation by $3$ to isolate $x$ . $x=\boxed{\frac{11}{3}}$

5 0

3 years ago

Read 2 more answers

Order the numbers from least to greatest.<br> 54%, 2/5, -0.45<br><br> SHOW YOUR WORK!

Tom [10]

Answer:

-0.45, 2/5, 54%

Step-by-step explanation:

54%=54/100

=87/50

2/5=20/50

-0.45 must be the smallest as it is negative

7 0

3 years ago

(5) Find the Laplace transform of the following time functions: (a) f(t) = 20.5 + 10t + t 2 + δ(t), where δ(t) is the unit impul

Aloiza [94]

Answer

(a) $F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

Step-by-step explanation:

(a) $f(t) = 20.5 + 10t + t^2 +$ δ(t)

where δ(t) = unit impulse function

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 f(s)e^{-st} \, dt$

where a = ∞

=> $F(s) = \int\limits^a_0 {(20.5 + 10t + t^2 + d(t))e^{-st} \, dt$

where d(t) = δ(t)

=> $F(s) = \int\limits^a_0 {(20.5e^{-st} + 10te^{-st} + t^2e^{-st} + d(t)e^{-st}) \, dt$

Integrating, we have:

=> $F(s) = (20.5\frac{e^{-st}}{s} - 10\frac{(t + 1)e^{-st}}{s^2} - \frac{(st(st + 2) + 2)e^{-st}}{s^3} )\left \{ {{a} \atop {0}} \right.$

Inputting the boundary conditions t = a = ∞, t = 0:

$F(s) = \frac{20.5}{s} - \frac{10}{s^2} - \frac{2}{s^3}$

(b) $f(t) = e^{-t} + 4e^{-4t} + te^{-3t}$

The Laplace transform of function f(t) is given as:

$F(s) = \int\limits^a_0 (e^{-t} + 4e^{-4t} + te^{-3t} )e^{-st} \, dt$

$F(s) = \int\limits^a_0 (e^{-t}e^{-st} + 4e^{-4t}e^{-st} + te^{-3t}e^{-st} ) \, dt$

$F(s) = \int\limits^a_0 (e^{-t(1 + s)} + 4e^{-t(4 + s)} + te^{-t(3 + s)} ) \, dt$

Integrating, we have:

$F(s) = [\frac{-e^{-(s + 1)t}} {s + 1} - \frac{4e^{-(s + 4)}}{s + 4} - \frac{(3(s + 1)t + 1)e^{-3(s + 1)t})}{9(s + 1)^2}] \left \{ {{a} \atop {0}} \right.$

Inputting the boundary condition, t = a = ∞, t = 0:

$F(s) = \frac{-1}{s + 1} - \frac{4}{s + 4} - \frac{4}{9(s + 1)^2}$

3 0

3 years ago

larry ran 220 meters farther that leigh ran. leigh ran 2.3 kilometers. how many kilometers did larry run?

Gala2k [10]

Larry ran 2.52 killometers. Since a meter is 0.001 killometers, you have to do 0.001 x 220 which is 0.22 then do 2.3 + 0.22 and get 2.52.

6 0

4 years ago