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Katen [24]
3 years ago
11

For which function is 5 NOT an element of the range?

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0
B. y = -2 + 4
y = 2 not 5
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Are 17 heartbeats in 15 seconds and 68 heartbeats in 60 seconds Equivalent
Andre45 [30]

Answer:

1.133.

Step-by-step explanation:

Yes because if you divide 17 by 15 you get about 1.133 and if you divide 68 by 60 you also get about 1.133.

4 0
3 years ago
Read 2 more answers
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
Estimate the sum of numbers 9488 and 857 to the nearest 10 <br><br>Ans: ?
UNO [17]
For the first one it’s 9490 & for the second it’s 860
5 0
3 years ago
Use the GCF and the Distributive Property to express the sum as a product.
serious [3.7K]

Answer: 3 x 13 ?

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
4. The area of a rhombus with one diagonal is 8.72 cm long is the same as the area of a square of side 15.6 cm. Find the length
Lubov Fominskaja [6]

Answer:

55.82 cm

Step-by-step explanation:

d1= 8.72 cm

a= 15.6 cm

A rhombus= 1/2*d1*d2 = A square

A square= 15.6²= 243.36 cm²

d2= 2A/d1= 2*243.36/8.72 ≈55.82 cm

4 0
4 years ago
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