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3 years ago

1. x^2-5x-14=0 2. 5x^2+12x=9 3. 6x^2=5x+6 Please help :(

Mathematics

1 answer:

yuradex [85]3 years ago

3 0

Answer:

1. x = 7 or x = -2

2. x = 3/5 or x = -3

3. x = 3/2 or x = -2/3

Step-by-step explanation:

1. Solve for x over the real numbers:

x^2 - 5 x - 14 = 0

The left hand side factors into a product with two terms:

(x - 7) (x + 2) = 0

Split into two equations:

x - 7 = 0 or x + 2 = 0

Add 7 to both sides:

x = 7 or x + 2 = 0

Subtract 2 from both sides:

Answer: x = 7 or x = -2

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2> Solve for x over the real numbers:

5 x^2 + 12 x = 9

Divide both sides by 5:

x^2 + (12 x)/5 = 9/5

Add 36/25 to both sides:

x^2 + (12 x)/5 + 36/25 = 81/25

Write the left hand side as a square:

(x + 6/5)^2 = 81/25

Take the square root of both sides:

x + 6/5 = 9/5 or x + 6/5 = -9/5

Subtract 6/5 from both sides:

x = 3/5 or x + 6/5 = -9/5

Subtract 6/5 from both sides:

Answer: x = 3/5 or x = -3

______________________________________________

3. Solve for x over the real numbers:

6 x^2 = 5 x + 6

Subtract 5 x + 6 from both sides:

6 x^2 - 5 x - 6 = 0

The left hand side factors into a product with two terms:

(2 x - 3) (3 x + 2) = 0

Split into two equations:

2 x - 3 = 0 or 3 x + 2 = 0

Add 3 to both sides:

2 x = 3 or 3 x + 2 = 0

Divide both sides by 2:

x = 3/2 or 3 x + 2 = 0

Subtract 2 from both sides:

x = 3/2 or 3 x = -2

Divide both sides by 3:

Answer: x = 3/2 or x = -2/3

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Answer:

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3 years ago

Volumes of the solids with disk or shell method?

lina2011 [118]

The answer to the questions of volumes are given as follows

a) $v=128 \pi$

b) $v=\frac{128}{3} \pi$

c) $v=\frac{1024}{5} \pi$

d) $V=\frac{13568}{15} \pi$

Generally, the questions are mathematically solved below

$y=\sqrt{x}, y=4, x=0$

a) x-axis

if $y=4, x=16, x=0$

Using disk method

$} v=\pi \int_{0}^{16}(\sqrt{x})^{2} d x \\$

$v=\pi\left(\frac{x^{2}}{2}\right)_{0}^{16} \\$

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b) line y=4

if x=0, y=0 ;

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Using shell method

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$v=2 \pi \int_{0}^{4}\left(4 y^{2}-y^{3}\right) d y$

$v=2 \pi\left[\frac{4 y^{3}}{3}-\frac{y^{4}}{4}\right]_{0}^{4} \\$

$v=\frac{2 \pi}{12}[1024-768] \\$

$v=\frac{512 \pi}{12} \\$

$v=\frac{128}{3} \pi$

c) y-axis

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x=y^2

Using disk method

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$v=\pi \int_{0}^{4} y^{4} d y$$

$v=\pi\left(\frac{y 5}{5}\right)_{0}^{4} \\$

$v=\frac{1024}{5} \pi$

d) line x=-1

y=√x, y=4, x=0

0 ≤ x ≤ 6

Using shell method

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$V=\int_{0}^{16} 2 \pi(1+x) \sqrt{x} d x$$

$V=2 \pi\int_{0}^{16}(x^{1 / 2}+x^{3 / 2}\right) )d x\right. \\$

$V=2 \pi\left[\frac{x^{3 / 2}}{3 / 2}+\frac{x^{5 / 2}}{5 / 2}\right]_{0}^{16} \\$

$V=2 \pi\left[2 / 3 \cdot\left(4^{2}\right)^{3 / 2}+2 / 5\left(4^{2}\right)^{5 / 2}\right] \\$

$V=4 \pi / 3 \cdot 4^{3}+4 \pi / 5 \cdot 4^{5} \\$

$V=4 \pi\left(\frac{1}{3}+\frac{4^{2}}{5}\right)$

$V=\frac{256}{15}(5+48) \pi \\$

$V=\frac{256 \times 53}{15} \pi \\$

$V=\frac{13568}{15} \pi$

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Answer:

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Again, let me know if you don't see how they are supplementary

Step-by-step explanation:

This looked super overwhelming when I first looked at it. Thankfully it's actually pretty simple.

We want p and q to be parallel and want to deal with y. Well both of those are on n as the transversal, so we can pretend m doesn't exist. The two angles labeled are supplementary to each other, if you cannot tell that let me know, but adding the two together make 180.

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