What is the y-Intercept for the equation 4x -3y= -18?

100 POINTS AND BRAINLIEST TO WHOEVER SOLVES IT

Butoxors [25]

Volume of a tetrahedron with edge s

$\\ \rm\Rrightarrow V=\dfrac{s^3}{6\sqrt{2}}$

According to question

s=4in

$\\ \rm\Rrightarrow V=\dfrac{4^3}{6\sqrt{2}}$

$\\ \rm\Rrightarrow V=\dfrac{64}{6\sqrt{2}}$

$\\ \rm\Rrightarrow V=\dfrac{32}{3\sqrt{2}}$

$\\ \rm\Rrightarrow V=\dfrac{32}{4.2426406871192}$

$\\ \rm\Rrightarrow V\approx 7.5425in^3$

6 0

2 years ago

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5. Which of the following linear functions has a graph which passes through points (−5,−2) and (−3,0)?

sergeinik [125]

Answer:

$f(x) = x + 3$

Step-by-step explanation:

Given

Points (−5,−2) and (−3,0)

Required

Find a linear function that passes through the given points

The question implies that we solve for the equation for the line;

First, the slope of the line must be calculated;

This is calculated as thus:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Where $(x_1,y_1) = (-5,-2)\ and\ (x_2,y_2) = (-3,0)$

So, $m = \frac{y_2 - y_1}{x_2 - x_1}$ becomes

$m = \frac{0 - (-2)}{-3 - (-5)}$

$m = \frac{0 + 2}{-3 + 5}$

$m = \frac{2}{2}$

$m = 1$

The equation of the line can then be calculated using any of the given points;

Using

$m = \frac{y - y_1}{x - x_1}$

$Where\ (x_1,y_1) = (-5,-2)\ and\ m =1$

We have

$1 = \frac{y-(-2)}{x-(-5)}$

$1 = \frac{y+2}{x+5}$

Multiply both sides by x + 5

$(x+5)*1 = \frac{y+2}{x+5} * (x+5)$

$x + 5 = y + 2$

Subtract 2 from both sides

$x + 5 - 2 = y + 2 - 2$

$x + 3 = y$

$y = x + 3$

Replace y with f(x)

$f(x) = x + 3$

Hence, from the list of given options; Option B is correct

8 0

3 years ago

The answer <br> Don't know how to solve

taurus [48]

-5n - 8(1+7n) = -8

-5n - 8 - 56n = -8

-61n = 0

n = 0

5 0

3 years ago

What is the antiderivative of sin^2(x)cos^2(x)?

marin [14]

Answer:

$\frac{x}{8}-\frac{\sin(4x)}{32}+C$

Step-by-step explanation:

[Most of the work here comes from manipulating the trig to make the term (integrand) integrable.]

Recall that we can express the squared trig functions in terms of cos(2x). That is,

$\cos(2x)=2\cos^2x-1 \\ \cos(2x)=1 - 2\sin^2x.$

And so inverting these,

$\cos^2x=\frac{1}{2} (1+\cos2x) \\ \sin^2x=\frac{1}{2} (1-\cos2x)$ .

Multiply them together to obtain an equivalent expression for sin^2(x)cos^2(x) in terms of cos(2x).

$\sin^2x \cdot \cos^2x =\frac{1}{2} (1-\cos2x) \cdot \frac{1}{2} (1+\cos2x) = \frac{1}{4}(1-\cos^2(2x)).$

Notice we have cos^2(2x) in the integrand now. We've made it worse! Let's try plugging back in to the first identity for cos^2(2x).

$\cos(2x)=2\cos^2x-1 \Rightarrow \cos(4x)=2\cos^2(2x)-1 \Rightarrow \cos^2(2x) = \frac{1}{2}(1+\cos(4x))$

So then,

$\sin^2x \cdot \cos^2x = \frac{1}{4}(1-\cos^2(2x)) = \frac{1}{4}(1-\frac{1}{2}(1+\cos(4x))) = \frac{1}{4}(1-\frac{1}{2}-\frac{1}{2}\cos(4x))=\frac{1}{8}(1-\cos(4x)).$

This is now integrable (phew),

$\int \sin^2x\cos^2x \ dx = \int \frac{1}{8}(1-\cos(4x)) \ dx = \frac{1}{8} \int (1-\cos(4x)) \ dx = \frac{1}{8}(x-\frac{1}{4}\sin(4x))+C.$

7 0

3 years ago

Spencer surveyed five of his friends to find out how many pets they have. His results are shown in the table below. What is the

Naily [24]

Answer:

6 pets

Step-by-step explanation:

5 0

2 years ago