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4 years ago

Find the range of each function, given the domain

Mathematics

1 answer:

Olenka [21]4 years ago

7 0

In #12, substitute into the given function the x values {-3, 0, 6} to obtain the range: the function takes on the following values: {0, -1, -3}. Thus, the range for #13 is {0, -1, -3}.

Do the others in precisely the same way.

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Two siblings are racing. The older sibling can run at a constant speed of 11 feet per second. The younger sibling can run at a c

Gwar [14]

Answer:

it would take approximately 8 seconds

Step-by-step explanation:

7x8 + 32 =88 but will take 8 seconds to get 88 with going 11 feet per sec 8×11=88

8 0

3 years ago

Find the truth value of (p ^ q) v r. (Make a truth table.)

alexgriva [62]

Answer:

p: (-4)2 > 0

q: An isosceles triangle has two congruent sides.

r: Two angles, whose measure have a sum of 90, are supplements.

Step-by-step explanation:

p: (-4)2 > 0

q: An isosceles triangle has two congruent sides.

r: Two angles, whose measure have a sum of 90, are supplements.

4 0

3 years ago

Between which two consecutive integer<br> Is 3√11

VladimirAG [237]

I believe it would be between 3 and 4

~Hope This Helped~ :D

4 0

4 years ago

I have a rectangular garden. I usually grow cucumbers in 2/3 of my garden, but I want to take 3/4 of the cucumber section to gro

KonstantinChe [14]

Answer:

$\frac{1}{2}$ of the garden will be radishes.

Step-by-step explanation:

Given a rectangular garden which grows cucumbers in $\frac{2}{3}$ of the garden.

$\frac{3}{4}$ of the cucumber section to grow radishes.

To find:

How much of the whole garden will be radishes?

Solution:

Let the total area of the garden = $x$ sq units

Then as per our assumption, total area to grow cucumbers:

$\frac{2x}{3}$ sq units

Cucumber area to grow radishes = $\frac{3}{4}$ of cucumber area

$= \dfrac{3}{4}\times \dfrac{2x}{3}\\\Rightarrow \dfrac{x}{2}\ sq\ units$

Fraction of whole garden to grow radishes = Area to grow radishes divided by whole area of garden

$\Rightarrow \dfrac{\frac{x}{2}}{x}\\\\\Rightarrow \bold{\dfrac{1}{2}}$

Therefore, $\frac{1}{2}$ of the garden will be radishes.

4 0

3 years ago

Inveres laplace transform (3s-14)/s^2-4s+8

Olenka [21]

Complete the square in the denominator.

$s^2 - 4s + 8 = (s^2 - 4s + 4) + 4 = (s-2)^2 + 4$

Rewrite the given transform as

$\dfrac{3s-14}{s^2-4s+8} = \dfrac{3(s-2) - 8}{(s-2)^2+4} = 3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}$

Now take the inverse transform:

$L^{-1}_t\left\{3\times\dfrac{s-2}{(s-2)^2+2^2} - 4\times\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3L^{-1}_t\left\{\dfrac{s-2}{(s-2)^2+2^2}\right\} - 4L^{-1}_t\left\{\dfrac{2}{(s-2)^2+2^2}\right\} \\\\ 3e^{2t} L^{-1}_t\left\{\dfrac s{s^2+2^2}\right\} - 4e^{2t} L^{-1}_t\left\{\dfrac{2}{s^2+2^2}\right\} \\\\ \boxed{3e^{2t} \cos(2t) - 4e^{2t} \sin(2t)}$

5 0

3 years ago