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3 years ago

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A 185-pound person burns 260 calories during a soccer game. Find the number of calories that a 250-pound person would burn durin

g the same activity. Round to the nearest calorie. 178 calories 422 calories 351 calories 217 calories

1 answer:

Olin [163]3 years ago

5 0

Answer: The answer to this problem is 178 calories.

Step-by-step explanation:

To calculate the answer, simply get the percentage of calories a person would burn during exercise (I am using a calculator BTW).

185 / 260 = 0.711538462

Then, calculate 71.1538462% of 250 pounds.

250 * 0.711538462 = 177.884616

Finally, round up your answer. 177.884616 rounded up is 178.

You might be interested in

Victoria has $200 of her birthday gift money saved at home, and the amount is modeled by the function h(x) = 200. She reads abou

san4es73 [151]

Answer:

h(x) * s(x) = 200(1.05)^(x - 1)

Step-by-step explanation:

Our interest equation is s(x) = (1.05)^(x - 1). This is actually a part of a bigger formula for calculating the amount of money accumulated including interest:

A = P(1 + r)^n, where A is the total, P is the principal amount (initial amount), r is the interest rate, and n is the time

Here, we technically already have the (1 + r)^n part; it's just (1.05)^(x - 1). The principle, though, will actually be the 200 because she starts out at $200.

Thus, to combine these, we simply multiply them together to get:

h(x) * s(x) = 200(1.05)^(x - 1)

4 0

3 years ago

Read 2 more answers

I need help with my math homework. The questions is: Find all solutions of the equation in the interval [0,2π).

Aleksandr-060686 [28]

Answer:

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

Step-by-step explanation:

$\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$

Let's first isolate the trig function.

Add 1 one on both sides:

$\sqrt{3} \tan(x-\frac{\pi}{8})=1$

Divide both sides by $\sqrt{3}$ :

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

Now recall $\tan(u)=\frac{\sin(u)}{\cos(u)}$ .

$\frac{1}{\sqrt{3}}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

or

$\frac{1}{\sqrt{3}}=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

The first ratio I have can be found using $\frac{\pi}{6}$ in the first rotation of the unit circle.

The second ratio I have can be found using $\frac{7\pi}{6}$ you can see this is on the same line as the $\frac{\pi}{6}$ so you could write $\frac{7\pi}{6}$ as $\frac{\pi}{6}+\pi$ .

So this means the following:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$

is true when $x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

where $n$ is integer.

Integers are the set containing {..,-3,-2,-1,0,1,2,3,...}.

So now we have a linear equation to solve:

$x-\frac{\pi}{8}=\frac{\pi}{6}+n \pi$

Add $\frac{\pi}{8}$ on both sides:

$x=\frac{\pi}{6}+\frac{\pi}{8}+n \pi$

Find common denominator between the first two terms on the right.

That is 24.

$x=\frac{4\pi}{24}+\frac{3\pi}{24}+n \pi$

$x=\frac{7\pi}{24}+n \pi$ (So this is for all the solutions.)

Now I just notice that it said find all the solutions in the interval $[0,2\pi)$ .

So if $\sqrt{3} \tan(x-\frac{\pi}{8})-1=0$ and we let $u=x-\frac{\pi}{8}$ , then solving for $x$ gives us:

$u+\frac{\pi}{8}=x$ ( I just added $\frac{\pi}{8}$ on both sides.)

So recall $0\le x$ .

Then $0 \le u+\frac{\pi}{8}$ .

Subtract $\frac{\pi}{8}$ on both sides:

$-\frac{\pi}{8}\le u$

Simplify:

$-\frac{\pi}{8}\le u$

$-\frac{\pi}{8}\le u$

So we want to find solutions to:

$\tan(u)=\frac{1}{\sqrt{3}}$ with the condition:

$-\frac{\pi}{8}\le u$

That's just at $\frac{\pi}{6}$ and $\frac{7\pi}{6}$

So now adding $\frac{\pi}{8}$ to both gives us the solutions to:

$\tan(x-\frac{\pi}{8})=\frac{1}{\sqrt{3}}$ in the interval:

$0\le x$ .

The solutions we are looking for are:

$\frac{\pi}{6}+\frac{\pi}{8}$ and $\frac{7\pi}{6}+\frac{\pi}{8}$

Let's simplifying:

$(\frac{1}{6}+\frac{1}{8})\pi$ and $(\frac{7}{6}+\frac{1}{8})\pi$

$\frac{7}{24}\pi$ and $\frac{31}{24}\pi$

$\frac{7\pi}{24}$ and $\frac{31\pi}{24}$

5 0

3 years ago

9. Khalid drove 66 miles on 3 gallons of gasoline. Which is the number of miles he could drive on 12 gallons of gasoline?

IrinaK [193]

Answer:

264 miles

Step-by-step explanation:

Find the unit rate (mpg):

66 miles

----------------- = 22 mpg

3 gallons

On 12 gallons he could drive (22 mpg)(12 gallons) = 264 miles

8 0

3 years ago

Find angle abc (2x+14) (x+7)

melisa1 [442]

It's a complementary angle, so just solve for x and subtract the CBD from 90,
First, we have to solve for x.
2x + 14 + x + 7 = 90
Combine like terms
3x + 21 = 90
Balance, subtract 21 from each side.
3x = 69
Divide 3 from each side
x = 23.
So know that we know x is 23, we can go about it 2 ways, subtracting the other angle with 90 or just substituting the variable, I'd just substitute the variable for it so just do 23 + 7. That equals 30, angle ABC = 30.

7 0

3 years ago

Read 2 more answers

A water cup is shaped like a cone. It has a diameter of 3 inches and a slant height of 5 inches. About how many square inches of

daser333 [38]

Options were missing in the question;

A water cup is shaped like a cone. It has a diameter of 3 inches and a slant height of 5 inches. About how many square inches of paper do you need to make a cup? Use 3.14 to approximate pi.

24 square inches

31 square inches

47 square inches

52 square inches

Answer:

24 square inches

Step-by-step explanation:

Given:

Diameter = 3 in

So we can say that;

Radius is half of diameter.

radius (r) = $\frac{3}{2}=1.5\ in$

Slant height (l) = 5 in

We need to find how many square inches of paper do you need to make a cup.

Solution:

Now according to question we required to find the amount of paper used to make cup.

Also Water cup is shaped liked cone.

Hence we can say that the base of cone is not covered since it will remain open.

So we will use Lateral surface area of the cone to find the paper requirement.

Now we know that;

Lateral surface area of cone is equal to π times radius times slant height.

framing in equation form we get;

$A_L = \pi rl=3.14\times1.5\times5 = 23.55 in^2 \approx 24\ in^2$

Hence We can say that 24 square inches of paper will be needed to make the cup.

5 0

3 years ago