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4 years ago

Help me with math anyone

Mathematics

1 answer:

Vinil7 [7]4 years ago

3 0

A. Is the correct answer because if you divide 60 and 5280 you get 88

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This question is worth 50 points, make sure to provide ALL calculations without using a calculator, though.

Fed [463]

Answer:

Step-by-step explanation:

Use the formula for the volume of a cone:

V = (1/3)(area of base / round opening)(height of cone)

In this case we're interested in the following:

(1/3)(area of base / round opening)(height of cone) ≤ 535 cm³

First, mult. both sides by 3 to elim. the fraction 1/3:

(area of base)(height) ≤ 1605 cm³

Since the height is 8 cm, we now have:

(area of base) / (8 cm) = 1605 cm³, or

(area of base) = (1605 cm³) · (8 cm) = 200.625 cm²

The formula for the area of a circle is A = 3.14·r², where r is the radius.

If the area of the base is 200.625 cm², as found above, this equals 3.14r², and so

r² = (200.625 cm²) / (3.14) = 63.89 cm²

Then the radius must be +√63.89 cm, or 8 cm

and the diameter must be 2r = 2(8 cm) = 16 cm. This would be the largest width of the container.

7 0

3 years ago

What must 111,111,102 be divided by to get 9 as the quotient and 0 as the remainder?

never [62]

ANSWER: 12,345,678

EXPLANATION:
we can work backwards by figuring out what 9(the answer) is multiplied by to get 111,111,102 (the starting number) we can do this by dividing by 9:

111,111,102/9 = 12,345,678

so: 9 * 12,345,678 = 111,111,102
which also means

111,111,102/12,345,678 = 9 (with 0 as a remainder)

5 0

4 years ago

PLEASE HELP ASAP In this task, you will practice finding the area under a nonlinear function by using rectangles. You will use g

mrs_skeptik [129]

Answer:

a) $1280 u^{2}$

b) $1320 u^{2}$

c) $\frac{4000}{3} u^{2}$

Step-by-step explanation:

In order to solve this problem we must start by sketching the graph of the function. This will help us visualize the problem better. (See attached picture)

You can sketch the graph of the function by plotting as many points as you can from x=0 to x=20 or by finding the vertex form of the quadratic equation by completing the square. You can also do so by using a graphing device, you decide which method suits better for you.

So we are interested in finding the area under the curve, so we divide it into 5 rectangles taking a right hand approximation. This is, the right upper corner of each rectangle will touch the graph. (see attached picture).

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=5 so we get:

$\Delta x=\frac{20-0}{5}=\frac{20}{5}=4$

so each rectangle must have a width of 4 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=64

h2=96

h3=96

h4= 64

h5=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(4)(64+96+96+64+0)$

so:

$A= 1280 u^{2}$

B) The same procedure is used to solve part B, just that this time we divide the area in 10 rectangles.

In order to figure the width of each rectangle we can use the following formula:

$\Delta x=\frac{b-a}{n}$

in this case a=0, b=20 and n=10 so we get:

$\Delta x=\frac{20-0}{10}=\frac{20}{10}=2$

so each rectangle must have a width of 2 units.

We can now calculate the hight of each rectangle. So we figure the y-value of each corner of the rectangles. We get the following heights:

h1=36

h2=64

h3=84

h4= 96

h5=100

h6=96

h7=84

h8=64

h9=36

h10=0

so now we can use the following formula to find the area under the graph. Basically what the formula does is add the areas of the rectangles:

$A=\sum^{n}_{i=1} f(x_{i}) \Delta x$

which can be rewritten as:

$A=\Delta x \sum^{n}_{i=1} f(x_{i})$

So we go ahead and solve it:

$A=(2)(36+64+84+96+100+96+84+64+36+0)$

so:

$A= 1320 u^{2}$

In order to find part c, we calculate the area by using limits, the limit will look like this:

$\lim_{n \to \infty} \sum^{n}_{i=1} f(x^{*}_{i}) \Delta x$

so we start by finding the change of x so we get:

$\Delta x =\frac{b-a}{n}$

$\Delta x =\frac{20-0}{n}$

$\Delta x =\frac{20}{n}$

next we find $x^{*}_{i}$

$x^{*}_{i}=a+\Delta x i$

so:

$x^{*}_{i}=0+\frac{20}{n} i=\frac{20}{n} i$

and we find $f(x^{*}_{i})$

$f(x^{*}_{i})=f(\frac{20}{n} i)=-(\frac{20}{n} i)^{2}+20(\frac{20}{n} i)$

cand we do some algebra to simplify it.

$f(x^{*}_{i})=-\frac{400}{n^{2}}i^{2}+\frac{400}{n}i$

we do some factorization:

$f(x^{*}_{i})=-\frac{400}{n}(\frac{i^{2}}{n}-i)$

and plug it into our formula:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{400}{n}(\frac{i^{2}}{n}-i) (\frac{20}{n})$

And simplify:

$\lim_{n \to \infty} \sum^{n}_{i=1}-\frac{8000}{n^{2}}(\frac{i^{2}}{n}-i)$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} \sum^{n}_{i=1}(\frac{i^{2}}{n}-i)$

And now we use summation formulas:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{n(n+1)(2n+1)}{6n}-\frac{n(n+1)}{2})$

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (\frac{2n^{2}+3n+1}{6}-\frac{n^{2}}{2}-\frac{n}{2})$

and simplify:

$\lim_{n \to \infty} -\frac{8000}{n^{2}} (-\frac{n^{2}}{6}+\frac{1}{6})$

$\lim_{n \to \infty} \frac{4000}{3}+\frac{4000}{3n^{2}}$

and solve the limit

$\frac{4000}{3}u^{2}$

4 0

3 years ago

Solve for x: 3(x + 1) = −2(x − 1) + 6.

Firlakuza [10]

The answer would be x=1

8 0

3 years ago

Read 2 more answers

Hello please help i’ll give brainliest:)

nika2105 [10]

Answer:

Its is 85

Step-by-step explanation:

It is 85 because you need to see out of all of those numbers which one is one is in the middle so just mark one number from each side untill you get to the middle number!

7 0

3 years ago