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2 years ago

11

Can anyone help solve algebra 2? (4x)^(4/3)

1 answer:

Fantom [35]2 years ago

5 0

See picture for solution steps and answer.

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f the line passing through the points (1, a) and (5, −2) is parallel to the line passing through the points (2, 8) and (−5, a +

aleksandrvk [35]

Answer:

$a=-6$

Step-by-step explanation:

Line 1: (1, a) and (5, −2)

slope of the line is

$slope = \frac{y_2-y_1}{x_2-x_1} =\frac{-2-a}{5-1} =\frac{-2-a}{4}$

Line 2: (2, 8) and (−5, a + 7)

slope of the line is

$slope = \frac{y_2-y_1}{x_2-x_1} =\frac{a+7-8}{-5-2} =\frac{a-1}{-7}$

when the lines are parallel then slopes are equal

$\frac{a-1}{-7} =\frac{-2-a}{4}$

cross multiply it

$4(a-1)=-7(-2-a)$

$4a-4=14+7a$

subtract 4a from both sides

$-4=14+3a$

$-18=+3a$

divide both sides by 3

$a=-6$

7 0

3 years ago

Given the function f(x) = 4(2)x, Section A is from x = 1 to x = 2 and Section B is from x = 3 to x = 4.

Ksju [112]

F(x) = 4(2)x
fx = 8x
f = 8

5 0

2 years ago

Need help please answer due tonight

irakobra [83]

Answer: I think 9

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Graph the linear equation. Find three points that solve the equation. 3x-4y=-18

myrzilka [38]

Answer:

Using a graphing calc.

Step-by-step explanation:

5 0

3 years ago

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the ra

podryga [215]

Answer:

$-\frac{1}{2\pi}$ cm³/min

Step-by-step explanation:

Let h = height of water and r = radius of its surface

$V = \frac{ 1}{ 3}\pi r^{2}h$

At all times,

$\frac{r }{h } = \frac{4 }{16 } = \frac{ 1}{4 }$

So, $r = \frac{h }{4 }$

$V = \frac{ 1}{ 3}\pi (\frac{h }{ 4})^{2}h$

$V = \frac{ 1}{ 48}\pi h^{3}$

$\frac{\text{d}V}{\text{d}h} = \frac{ 1}{ 16}\pi h^{2}$

We want $\frac{\text{d}h}{\text{d}t}$

$\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \frac{\text{d}h}{\text{d}t}$

$\frac{\text{d}V}{\text{d}t}$ = -2 cm³/min

$-2 = \frac{ 1}{ 16}\pi h^{2} \frac{\text{d}h}{\text{d}t}$

$\frac{\text{d}h}{\text{d}t} = -\frac{2}{\frac{ 1}{ 16}\pi h^{2} }$

$\frac{\text{d}h}{\text{d}t} = -\frac{32}{\pi h^{2}}$

When h = 8 cm

$\frac{\text{d}h}{\text{d}t} = -\frac{32}{\pi\times8^{2}}$

$\frac{\text{d}h}{\text{d}t} = -\frac{32}{64\pi}$

$\frac{\text{d}h}{\text{d}t} = -\frac{1}{2\pi}$ cm³/min

The depth of the water is decreasing at a rate of $\frac{1}{2\pi}$ cm³/min

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7 0

2 years ago