Ask question

Ask question

All categories

3 years ago

8

The art club is raising money for a new equipment. They need $842.36 to purchase new supplies and have received a $300.00 donati

on. The remaining amount will be equally divided among 4 different fundraisers throughout the summer. How much money will each fundraiser need to raise in order to purchase the new net and equipment?

1 answer:

Bumek [7]3 years ago

7 0

Answer: Each fundraiser need to raise $ 135.59 in order to purchase the new net and equipment.

Step-by-step explanation:

Money required = $ 842.36

Donation amount = $ 300.00

Amount left to arrange = (Money required - Donation amount )

= $ (842.36- 300.00)

= $ 542.36

Number of fundraisers = 4

Money required to arrange by each fundraiser = (Amount left to arrange) ÷4

= $ (542.36÷4)

= $ 135.59

Hence, each fundraiser need to raise $ 135.59 in order to purchase the new net and equipment.

You might be interested in

A school is having a canned food drive. Each class is challenged to collect 220 cans. If there are x classes in the school, whic

lara31 [8.8K]

Answer:

B)

Step-by-step explanation:

no. of Classes = x

cans = 220

220×x = 220x

no.of classes × cans = total number of cans

5 0

2 years ago

Read 2 more answers

A square is inscribed in a circle with an area of 10 pi square inches. what is the area of the square

Naddik [55]

We need to determine the radius and diameter of the circle. If the area of the circle is 10 pi in^2, then, according to the formula for the area of a circle,

A = 10 pi in^2 = pi*r^2. Thus, 10 in^2 = r^2, and r = radius of circle = sqrt(10) in.

Thus, the diam. of the circle is 2sqrt(10) in. This diam. has the same length as does the hypotenuse of one of the triangles making up the square.

Thus, [ 2*sqrt(10) ]^2 = x^2 + x^2, where x represents the length of one side of the square. So, 4(10) in^2 = 2x^2. Then:
40 in^2 = 2x^2, or 20 in^2 = x^2, and so the length x of one side of the square is sqrt(20). The area of the square is the square of this result:

Area of the square = x^2 = [ sqrt(20) ]^2 = 20 in^2 (answer). Compare that to the 10 pi sq in area of the circle (31.42 in^2).

8 0

3 years ago

3. A penny is made by combining 0.1 g of a copper alloy that costs $5.50 per kilogram (1000 grams) with

lbvjy [14]

.0001*5.5 + 1.5*x = 2
x=1.33 kg

6 0

2 years ago

X ^-4 + x ^ 3 - 4x ^ 2 = 2x - 4

Lady bird [3.3K]

Answer:

Ok, If you are trying to solve for x then your answer in fraction form is x=4/11. If you want it in decimal form the answer is 0.36(Repeats forever). And these answers are also implying that when you typed a "^" it meant multiplcation

Step-by-step explanation:

I have used an Online Algebra Calculator to help guide me to my answer and this answer is 100% correct so,

PLEASE MARK BRAINLIEST

4 0

2 years ago

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