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3 years ago

Help ASAP please like rn

Mathematics

2 answers:

GenaCL600 [577]3 years ago

6 0

Answer:

8/35 is correct

Step-by-step explanation:

Just multiply the numerator by numerator and denominator by denominator, then simplify

Sedbober [7]3 years ago

5 0

Answer:

14/25

hope you answer is right, but I am sorry if my answer is wrong, but have a great weekends!

- friend

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Which of the following numbers is irrational

Troyanec [42]

Answer:

Step-by-step explanation:

Irrational means cannot be put into a fraction. Or random repeating decimals is how I remember it.

√48 can be simplified to equal 4√3. √3 cannot be put into fraction form.

6 0

3 years ago

Describe the number of faces and thier shapes

Marysya12 [62]

Do you have pictures

3 0

3 years ago

Calculus Problem

Roman55 [17]

The two parabolas intersect for

$8-x^2 = x^2 \implies 2x^2 = 8 \implies x^2 = 4 \implies x=\pm2$

and so the base of each solid is the set

$B = \left\{(x,y) \,:\, -2\le x\le2 \text{ and } x^2 \le y \le 8-x^2\right\}$

The side length of each cross section that coincides with B is equal to the vertical distance between the two parabolas, $|x^2-(8-x^2)| = 2|x^2-4|$ . But since -2 ≤ x ≤ 2, this reduces to $2(x^2-4)$ .

a. Square cross sections will contribute a volume of

$\left(2(x^2-4)\right)^2 \, \Delta x = 4(x^2-4)^2 \, \Delta x$

where ∆x is the thickness of the section. Then the volume would be

$\displaystyle \int_{-2}^2 4(x^2-4)^2 \, dx = 8 \int_0^2 (x^2-4)^2 \, dx \\\\ = 8 \int_0^2 (x^4-8x^2+16) \, dx \\\\ = 8 \left(\frac{2^5}5 - \frac{8\times2^3}3 + 16\times2\right) = \boxed{\frac{2048}{15}}$

where we take advantage of symmetry in the first line.

b. For a semicircle, the side length we found earlier corresponds to diameter. Each semicircular cross section will contribute a volume of

$\dfrac\pi8 \left(2(x^2-4)\right)^2 \, \Delta x = \dfrac\pi2 (x^2-4)^2 \, \Delta x$

We end up with the same integral as before except for the leading constant:

$\displaystyle \int_{-2}^2 \frac\pi2 (x^2-4)^2 \, dx = \pi \int_0^2 (x^2-4)^2 \, dx$

Using the result of part (a), the volume is

$\displaystyle \frac\pi8 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{256\pi}{15}}}$

c. An equilateral triangle with side length s has area √3/4 s², hence the volume of a given section is

$\dfrac{\sqrt3}4 \left(2(x^2-4)\right)^2 \, \Delta x = \sqrt3 (x^2-4)^2 \, \Delta x$

and using the result of part (a) again, the volume is

$\displaystyle \int_{-2}^2 \sqrt 3(x^2-4)^2 \, dx = \frac{\sqrt3}4 \times 8 \int_0^2 (x^2-4)^2 \, dx = \boxed{\frac{512}{5\sqrt3}}$

7 0

2 years ago

Emily and her children went into a grocery store and she bought $20.80 worth of bananas and peaches. Each banana costs $0.80 and

MakcuM [25]

Answer:

The number of bananas that Emily bought was 6 and the number of peaches that Emily bought was 8

Step-by-step explanation:

<u><em>The complete question is</em></u>

Emily and her children went into a grocery store and she bought $20.80 worth of bananas and peaches. Each banana costs $0.80 and each peach costs $2. She bought a total of 14 peaches and bananas altogether. Determine the number of peaches and the number of bananas that Emily bought

Let

x ----> the number of bananas that Emily bought

y ----> the number of peaches that Emily bought

we know that

She bought a total of 14 bananas and peaches altogether

$x+y=14$ -----> equation A

She bought $20.80 worth of bananas and peaches

$0.80x+2y=20.80$ -----> equation B

Solve the system by graphing

Remember that the solution is the intersection point both graphs

using a graphing tool

The solution is the point (6,8)

see the attached figure

therefore

The number of bananas that Emily bought was 6 and the number of peaches that Emily bought was 8

3 0

3 years ago

the sum of two consecutive even integers is greater than or equal to 34 what are the smallest possible integers?

DedPeter [7]

n, n + 2 - two consecutive even integers

the sum of two consecutive even integers is greater than or equal to 34

n + (n + 2) ≥ 34

n + n + 2 ≥ 34

2n + 2 ≥ 34 <em>subtract 2 from both sides</em>

2n ≥ 32 <em>divide both sides by 2</em>

n ≥ 16

<h3>Answer: The smallest possible integers is equal 16.</h3>

3 0

3 years ago