According to these three facts, which statements are true?

Mathematics

1 answer:

natta225 [31]3 years ago

7 0

Answer: circle w has center

Step-by-step explanation:

Because the center and the radius of the circle

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What is the area of a rectangle with a length 3 1/4 of inches and a width of 2 1/2 inches?

qaws [65]

Answer: 8.125 square inches

Explanation/Calculation:
First, convert the fractions to decimals.
3 1/4= 3.25
2 1/2= 2.5

Now multiply 3.25*2.5
You get 8.125

7 0

3 years ago

How many milliliters of 0.30%. sugar solution must you add to 75 mL of 4% sugar solution to get a 0.50% sugar solution? show you

LekaFEV [45]

We have 75 mL of 4% sugar solution.
We have to add a 30% sugar solution to make a 50% sugar solution.
75 * .04 + .30x = .50 * (75 +x)
3 + .30x = 37.5 +.50x
I can't get the equation to solve.
Did you type it correctly? For one thing, you have the percentages typed as .30 % and 4%. Are the decimal places in the correct positions?

Also, no matter how much 30% sugar solution you add, it will NEVER increase to 50%.

6 0

3 years ago

4. (03.06 MC)

bazaltina [42]

Step-by-step explanation:

Volume of a cylinder is:

V = πr²h

V = (3.14) (4 in)² (5 in)

V = 251.2 in³

Volume of a cone is:

V = ⅓ πr²h

V = ⅓ (3.14) (4 in)² (15 in)

V = 251.2 in³

The volumes are equal.

8 0

3 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.