Andrew has a Frisbee™ with a radius of 10 cm. What is the approximate circumference of the

Mathematics

1 answer:

Ne4ueva [31]3 years ago

3 0

Answer:

The answer is D. 62.8

Step-by-step explanation:

I used this formula.

Circumference = 2πr

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Answer and working out please!

frosja888 [35]

Answer:

17.0710678119 and 2.92893218813

Step-by-step explanation:

Theoretically, the wire should come out to be x+y=80,

and the sum of the areas of the squares should be (x/4)^2+(y/4)^2=300.

x/4 and y/4 being the length of one side of the square, we would then square that to find the area. The total wire is all x+y=80. X being one part and Y being the other.

Theoretically, using systems. i found the answers: 68.2842712474619,11.715728752538098

Again, these numbers are very large, and I they actually do both add up to each amount basically perfectly. If your teacher is asking for a rounded answer, that would've been helpful to know. But again, theoretically, those are the answers.

3 0

4 years ago

WILL MARK BRAINLIEST 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.