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3 years ago

What distance was used to define the astronomical unit (AU)

1 answer:

5 0

<h2>approximately 93 million miles Definition of astronomical unit. For general reference, we can say that one astronomical unit (AU) represents the mean distance between the Earth and our sun. An AU is approximately 93 million miles (150 million km). It's approximately 8 light-minutes.</h2>

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Line p has an equation of y=-8x+6. Line q, which is perpendicular to line p, includes the point (2,–2). What is the equation of

Arturiano [62]

Answer:

$y = \frac{1}{8} x - 2 \frac{1}{4}$

Step-by-step explanation:

Slope-intercept form

y= mx +c, where m is the slope and c is the y-intercept

Line p: y= -8x +6

slope= -8

The product of the slopes of perpendicular lines is -1. Let the slope of line q be m.

m(-8)= -1

m= -1 ÷(-8)

m= ⅛

Substitute m= ⅛ into the equation:

y= ⅛x +c

To find the value of c, substitute a pair of coordinates that the line passes through into the equation.

When x= 2, y= -2,

-2= ⅛(2) +c

$- 2 = \frac{1}{4} + c$

$c = - 2 - \frac{1}{4}$

$c = - 2 \frac{1}{4}$

Thus, the equation of line q is $y = \frac{1}{8} x - 2 \frac{1}{4}$ .

4 0

2 years ago

Use the fundamental theorem of calculus to find the area of the region between the graph of the function x^5 + 8x^4 + 2x^2 + 5x

BaLLatris [955]

Answer:

The area of the region is 25,351 $units^2$ .

Step-by-step explanation:

The Fundamental Theorem of Calculus: if $f$ is a continuous function on $[a,b]$ , then

$\int_{a}^{b} f(x)dx = F(b) - F(a) = F(x) | {_a^b}$

where $F$ is an antiderivative of $f$ .

A function $F$ is an antiderivative of the function $f$ if

$F^{'}(x)=f(x)$

The theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

To find the area of the region between the graph of the function $x^5 + 8x^4 + 2x^2 + 5x + 15$ and the x-axis on the interval [-6, 6] you must:

Apply the Fundamental Theorem of Calculus

$\int _{-6}^6(x^5+8x^4+2x^2+5x+15)dx$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\\\\int _{-6}^6x^5dx+\int _{-6}^68x^4dx+\int _{-6}^62x^2dx+\int _{-6}^65xdx+\int _{-6}^615dx$

$\int _{-6}^6x^5dx=0\\\\\int _{-6}^68x^4dx=\frac{124416}{5}\\\\\int _{-6}^62x^2dx=288\\\\\int _{-6}^65xdx=0\\\\\int _{-6}^615dx=180\\\\0+\frac{124416}{5}+288+0+18\\\\\frac{126756}{5}\approx 25351.2$

3 0

3 years ago

PLEASE HELP ME ASAP. I REALLY NEED THIS DONE

Juli2301 [7.4K]

Answer:

x = 65

Step-by-step explanation:

using the Altitude- on- Hypotenuse theorem

(leg of big Δ )² = (part of hypotenuse below it ) × (whole hypotenuse)

x² = 25 × 169 = 4225 ( take square root of both sides )

x = $\sqrt{4225}$ = 65

6 0

2 years ago

What is 1 half subtract 1 tenth

Anna [14]

To solve for $\frac{1}{2}- \frac{1}{10}$ You first need to find a common denominator.

To do so, you need to make both denominators 10 by multiplying the top and bottom of $\frac{1}{2}$ by 5

$\frac{5}{10}- \frac{1}{10}$ = $\frac{4}{10}$ Reduce by dividing both the top and bottom by 2

Your answer is $\frac{2}{5}$

8 0

3 years ago

Read 2 more answers

F(x) = (1/4)^x. for x = -3

Trava [24]

Plug in -3 for x into the equation.

(1/4)^-3

And solve

(1/4)^-3 = 64

7 0

2 years ago

Read 2 more answers