All categories

2 years ago

Vhich real-world scenario can be described using the algebraic expression c/4?

Mathematics

1 answer:

Law Incorporation [45]2 years ago

3 0

Answer:

Baking c cookies and dividing them evenly among 4 friends.

Brainiest???

Hope this helps

You might be interested in

Colossus Added to six flags st. Louis in 1986, the Colossus is a giant Ferris wheel. Its diameter is 165 feet, it rotates at a r

NemiM [27]

Given:

D=165 feet and the frequency of the motion is 1.6 revolutions per minute.

Solution:

The radius is half of the diameter.

The radius of the wheel is 82.5 feet.

$T=\frac{1}{1.6} \text{ minutes}$

As we know: $\omega=\frac{2\pi}{T}$

Substitute the value of T in the above formula.

$\omega=\frac{2\pi}{\frac{1}{1.6}}\\\omega=3.2\pi$

If the center of the wheel is at the origin then for $t=0$ the rest position is $-a$ .

This can be written as:

$h(t)=-a\cos(\omega t)\\h(t)=-82.5cos(32.\pi t)$

The actual height of the rider from the ground is:

$h(t)=\text{ Initial height from bottom}+\text{ radius}-82.5\cos(3.2\pi t)\\h(t)=15+82.5-82.5\cos(3.2\pi t)\\h(t)=97.5-82.5\cos(3.2\pi t)$

The required equation is $h(t)=97.5-82.5\cos(3.2\pi t)$ .

3 0

3 years ago

WILL MARK BRAINLIEST <br><br> what is the domain of (f/g) (x)?

tatyana61 [14]

Given that $f(x) = \sqrt{7-x}$ and $g(x) = \sqrt{x + 2}$ , we can say the following:

$\Bigg(\dfrac{f}{g}\Bigg)(x) = \dfrac{f (x)}{g(x)} = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Now, remember what happens if we have a negative square root: it becomes an imaginary number. We don't want this, so we want to make sure whatever is under a square root is greater than 0 (given we are talking about real numbers only).

Thus, let's set what is under both square roots to be greater than 0:

$\sqrt{7 - x} \Rightarrow 7 - x \geq 0 \Rightarrow x \leq 7$

$\sqrt{x + 2} \Rightarrow x + 2 \geq 0 \Rightarrow x \geq -2$

Since both of the square roots are in the same function, we want to take the union of the domains of the individual square roots to find the domain of the overall function.

$x \leq 7 \,\,\cup x \geq -2 = \boxed{-2 \leq x \leq 7}$

Now, let's look back at the function entirely, which is:

$\Bigg( \dfrac{f}{g} \Bigg)(x) = \dfrac{\sqrt{7 - x}}{\sqrt{x+2}}$

Since $\sqrt{x + 2}$ is on the bottom of the fraction, we must say that $\sqrt{x + 2} \neq 0$ , since the denominator can't equal 0. Thus, we must exclude $\sqrt{x + 2} = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2$ from the domain.

Thus, our answer is Choice C, or $\boxed{ \{ x | -2 < x \leq 7 \}}$ .

<em>If you are wondering why the choices begin with the $x |$ symbol, it is because this is a way of representing that $x$ lies within a particular set.</em>