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Cerrena [4.2K]
2 years ago
15

Find the zeros of the quadratic function: y= 6(7x + 9)(8x - 3).

Mathematics
1 answer:
fiasKO [112]2 years ago
6 0

Answer:

A) (-9/7,0), (3/8,0)

Step-by-step explanation:

Zeros of a quadratic function:

x for which y = 0.

In this question:

y = 6(7x + 9)(8x - 3)

It's 0 if one of the factors is 0. 6 is never 0, now about the other to:

7x + 9 = 0

7x = -9

x = -\frac{9}{7}

So (-9/7, 0) is a zero of the quadratic function.

The other is:

8x - 3 = 0

8x = 3

x = \frac{3}{8}

(3/8,0) is the other zero.

Thus, the correct answer is given by option A.

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Lesechka [4]

Answer:

Alternate exterior

Step-by-step explanation:

They are on opposite sides and both exterior

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3 years ago
Determine the value of x so that the square and the rectangle is shown to have equal perimeters.
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2 years ago
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Hi! Please help, I'd appreciate it!
Romashka-Z-Leto [24]

The volume of sphere is \frac{1372}{3}\pi feet cubed.

Step-by-step explanation:

Given,

Surface area of sphere = 196π square feet

Surface area of sphere = 4πr²

We will find radius of sphere.

4\pi r^2=196\pi

Dividing both sides by 4π

\frac{4\pi r^2}{4\pi}=\frac{196\pi}{4\pi}\\\\r^2=49

Taking square root on both sides

\sqrt{r^2}=\sqrt{49}\\r=7

The radius is 7 feet.

Volume of sphere = \frac{4}{3}\pi r^3

Volume=\frac{4}{3}\pi (7)^3\\\\Volume= \frac{4}{3}\pi *343\\\\Volume = \frac{1372}{3}\pi

The volume of sphere is \frac{1372}{3}\pi feet cubed.

Keywords: volume, surface area

Learn more about surface area at:

  • brainly.com/question/101683
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3 years ago
Use Simpson's Rule with n = 10 to approximate the area of the surface obtained by rotating the curve about the x-axis. Compare y
DiKsa [7]

The area of the surface is given exactly by the integral,

\displaystyle\pi\int_0^5\sqrt{1+(y'(x))^2}\,\mathrm dx

We have

y(x)=\dfrac15x^5\implies y'(x)=x^4

so the area is

\displaystyle\pi\int_0^5\sqrt{1+x^8}\,\mathrm dx

We split up the domain of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [4, 9/2], [9/2, 5]

where the left and right endpoints for the i-th subinterval are, respectively,

\ell_i=\dfrac{5-0}{10}(i-1)=\dfrac{i-1}2

r_i=\dfrac{5-0}{10}i=\dfrac i2

with midpoint

m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4

with 1\le i\le10.

Over each subinterval, we interpolate f(x)=\sqrt{1+x^8} with the quadratic polynomial,

p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}

Then

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx

It turns out that the latter integral reduces significantly to

\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)

which is about 651.918, so that the area is approximately 651.918\pi\approx\boxed{2048}.

Compare this to actual value of the integral, which is closer to 1967.

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AfilCa [17]

Answer:

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