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4 years ago

HELP SOMEONE PLEASE HELP

Mathematics

1 answer:

romanna [79]4 years ago

8 0

Answer:

Step-by-step explanation:

Evaluate the numerator/ denominator using order of operations

numerator = 16 + 2(4 × 2)

using the order parenthesis, multiplication followed by addition

= 16 + 2(8) ← evaluating parenthesis

= 16 + 2 × 8

= 16 + 16 ← performing multiplication

= 32 ← addition

denominator = 8 + 2³

using the order exponents then addition

= 8 + 8 ← evaluating exponent

= 16 ← addition

Hence

$\frac{numerator}{denominator}$ = $\frac{32}{16}$ = 2 → B

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Stuart planted a shrub that was 64 centimeters tall. He measured the shrub regularly and found that it grew 1.8 centimeters each

andreev551 [17]

Answer:

C. 7 weeks.

Step-by-step explanation:

Since the shrub was 64 cm tall at first, take away 64 from 76.6. Now we are left with 12.6. Let's divide 12.6 by 1.8 (that's how many cm it grows per week). When we do this, we are left with 7. This means it took 7 weeks for the shrub to reach 76.6 cm. I hope this helps!

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3 years ago

If you park 3 levels below the lobby floor of a hotel and take the elevator to the 15th floor, how many floors apart are you fro

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So the lobby would be the 1st floor, and the level you parked your car on would be the -3rd level.

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4 years ago

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nirvana33 [79]

Answer:

False.

Step-by-step explanation:

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4 years ago

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As a puppy, Cricket weighed 3.75 pounds. At her 1-year check-up with the vet, Cricket

k0ka [10]

Answer:

8.75

Step-by-step explanation:

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3 years ago

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An equilateral triangle is inscribed in a circle of radius 6r. Express the area A within the circle but outside the triangle as

Paul [167]

Answer:

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Step-by-step explanation:

We have been given that an equilateral triangle is inscribed in a circle of radius 6r. We are asked to express the area A within the circle but outside the triangle as a function of the length 5x of the side of the triangle.

We know that the relation between radius (R) of circumscribing circle to the side (a) of inscribed equilateral triangle is $\frac{a}{\sqrt{3}}=R$ .

Upon substituting our given values, we will get:

$\frac{5x}{\sqrt{3}}=6r$

Let us solve for r.

$r=\frac{5x}{6\sqrt{3}}$

$\text{Area of circle}=\pi(6r)^2=\pi(6\cdot \frac{5x}{6\sqrt{3}})^2=\pi(\frac{5x}{\sqrt{3}})^2=\frac{25\pi x^2}{3}$

We know that area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4}s^2$ , where s represents side length of triangle.

$\text{Area of equilateral triangle}=\frac{\sqrt{3}}{4}s^2=\frac{\sqrt{3}}{4}(5x)^2=\frac{25\sqrt{3}}{4}x^2$

The area within circle and outside the triangle would be difference of area of circle and triangle as:

$A(x)=\frac{25\pi x^2}{3}-\frac{25\sqrt{3}x^2}{4}$

We can make a common denominator as:

$A(x)=\frac{4\cdot 25\pi x^2}{12}-\frac{3\cdot 25\sqrt{3}x^2}{12}$

$A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$

Therefore, our required expression would be $A(x)=\frac{100\pi x^2-75\sqrt{3}x^2}{12}$ .

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3 years ago