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Ainat [17]
3 years ago
6

What duos e=mc² mean

Mathematics
2 answers:
babymother [125]3 years ago
6 0

Answer:

energy equals mass times the speed of light squared

Step-by-step explanation:

polet [3.4K]3 years ago
3 0
Energy equals mass times the speed of light squared

explanation : Albert Einstein was the one who made that!
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In the United States, how many children under age 11 studied a language other than English in school
patriot [66]

Answer:

<u>less</u><u> </u><u>than</u><u> </u><u>5</u><u>%</u>

Step-by-step explanation:

In the USA <u>less</u><u> </u><u>than</u><u> </u><u>5</u><u>%</u><u> </u> of children under the age of 11 study a language other than English at school

Hopefully this helped- have a good day)

6 0
2 years ago
PLEASEE HELP ME IM VERY CONFUSED ON THIS!!!
Anastasy [175]
SIDE LENGTH OF TRIANGLE: 2.14 inches
SIDE LENGTH OF HEXAGON: 6 inches

To solve this problem, we know that the shapes have equal sides as it states “equilateral triangle”. A triangle has 3 sides and a hexagon has 6 sides. We are told the perimeters are the same so you can set their perimeters equal to each other to solve for x. You would get this : 3(1.4x + 2) = 6(0.5x +2)
With basic algebra you would get x= 5
Then you substitute that value into the length sides of the triangle and hexagon. For the triangle you would approx get 2.14 inches and for the hexagon 6 inches

6 0
2 years ago
Read 2 more answers
A=1/2h(a+b), solve for a
telo118 [61]

Step-by-step explanation:

2A=h(a+b)

2A=ha+hb

2A-hb=ha

2A-hb/h=a

4 0
4 years ago
Read 2 more answers
Eric’s average income for the first 4 months of the year is $1,450.25, what must be his average income for the remaining 8 month
netineya [11]
Too harddd sorryyybxnskzmskndjwjnekkks
6 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
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