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Aleksandr [31]
3 years ago
5

Which is a valid prediction about the continuous function f(x)?

Mathematics
2 answers:
uranmaximum [27]3 years ago
6 0

f(x) ≥ 0 over the interval [–1, 1]. is the correct answer

grigory [225]3 years ago
5 0
<span> C f(x) ≥ 0 over the interval [–1, 1]. </span>
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The area of triangle PQR is 273 cm. Given that PQ = 12.8cm and PQR = 107°, find QR.​
storchak [24]

\text{Area of triangle} = \dfrac 12 \times PQ \times QR \times  \sin \angle PQR \\\\\\\implies  273= \dfrac 12 \times 12.8 \times QR \times \sin 107^{\circ}\\\\\implies 273 = 6.4 \sin 107^{\circ}  \times QR\\\\\\\implies QR = \dfrac{273}{6.4 \sin 107^{\circ}} = 44. 606~~ \text{cm}

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The area of a parallelogram is 32 square miles. Find its base and height. While x-7 and x+7.
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\sqrt{x}  \lim_{n \to \infty} a_n  \sqrt[n]{x}
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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Draw a typical approximating
natulia [17]

Answer:

V=25088π vu

Step-by-step explanation:

Because the curves are a function of "y" it is decided to take the axis of rotation as y

, according to the graph 1 the cutoff points of f(y)₁ and f(y)₂ are ±2

f(y)₁ = 7y²-28;  f(y)₂=28-7y²

y=0;   x=28-0 ⇒ x=28

x=0;   0 = 7y²-28 ⇒ 7y²=28 ⇒ y²= 28/7 =4 ⇒ y=√4 =±2

Knowing that the volume of a solid of revolution  V=πR²h, where R²=(r₁-r₂) and h=dy then:

dV=π(7y²-28-(28-7y²))²dy ⇒dV=π(7y²-28-28+7y²)²dy = 4π(7y²-28)²dy

dV=4π(49y⁴-392y²+784)dy integrating on both sides

∫dV=4π∫(49y⁴-392y²+784)dy ⇒ solving  ∫(49y⁴-392y²+784)dy

49∫y⁴dy-392∫y²dy+784∫dy =

V=4π( 49\frac{y^{5} }{5}-392\frac{y^{3}}{3}+784y ) evaluated -2≤y≤2, or 2(0≤y≤2), also

V=8\pi(49\frac{2^{5} }{5}-392\frac{2^{3} }{3}+784.2)  ⇒ V=25088π vu

8 0
3 years ago
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