What is the volume of the figure below?

Line j has a slope of − 2. Line d is perpendicular to line j and passes through the point( − 8 , 3 ). Write the equation for

Zina [86]

Answer:

The equation is: y = 1/2x + 7

Step-by-step explanation:

The slope of j is -2. The slope of a line perpendicular to line j has a negative inverse: 1/2. The point is (-8, 3).

Use the point slope form of the equation:

y - y1 = m(x - x1)

Substitute:

y - 3 = 1/2(x - (-8))

y - 3 = 1/2x + 4

y = 1/2x + 4 + 3

y = 1/2x + 7

Proof:

Solve for f(x) when x = -8.

f(x) = 1/2x + 7

f(-8) = 1/2(-8) + 7

= -8/2 + 7

= -4 + 7

= 3, giving the point (-8, 3)

4 0

3 years ago

Find the sum of the measures of the interior angles in the figure shown.

ZanzabumX [31]

Step-by-step explanation:

A: 360/5=72

B:540/5=180

C:720/5=144

D:900/5=180

E:1260/5=252

8 0

4 years ago

Read 2 more answers

After a long study, tree scientists conclude that a eucalyptus tree will grow at the rate of 0.5 6/ (t+4)3 feet per year, where

kipiarov [429]

Answer:

Step-by-step explanation:

Given the rate at which an eucalyptus tree will grow modelled by the equation 0.5+6/(t+4)³ feet per year, where t is the time (in years).

The amount of growth can be gotten by integrating the given rate equation as shown;

$\int\limits {0.5 + \frac{6}{(t+4)^{3} } } \, dt \\= \int\limits {0.5} \, dt + \int\limits\frac{6}{(t+4)^{3} } } \, dx } \, \\= 0.5t +\int\limits {6u^{-3} } \, du \ where \ u = t+4 \ and\ du = dt\\= 0.5t + 6*\frac{u^{-2} }{-2} + C\\= 0.5t-3u^{-2} +C\\= 0.5t-3(t+4)^{-2} + C$

a) The number of feet that the tree will grow in the second year can be gotten by taking the limit of the integral from t =1 to t = 2

$\int\limits^2_1 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^2_1\\= [0.5(2)-3(2+4)^{-2}] - [0.5(1)-3(1+4)^{-2}]\\= [1-3(6)^{-2}] - [0.5-3(5)^{-2}]\\ = [1-\frac{1}{12}] - [0.5-\frac{3}{25} ]\\= \frac{11}{12}-\frac{1}{2}+\frac{3}{25}\\ = 0.9167 - 0.5 + 0.12\\= 0.5367feet$

b) The number of feet that the tree will grow in the third year can be gotten by taking the limit of the integral from t =2 to t = 3

$\int\limits^3_2 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^3_2\\= [0.5(3)-3(3+4)^{-2}] - [0.5(2)-3(2+4)^{-2}]\\= [1.5-3(7)^{-2}] - [1-3(6)^{-2}]\\ = [1.5-\frac{3}{49}] - [1-\frac{1}{12} ]\\ = 1.439 - 0.9167\\= 0.5223feet$

c) The total number of feet grown during the second year can be gotten by substituting the value of limit from t = 0 to t = 2 into the equation as shown

$\int\limits^2_0 {0.5 + \frac{6}{(t+4)^{3} } } \, dt = [0.5t-3(t+4)^{-2}]^2_0\\= [0.5(2)-3(2+4)^{-2}] - [0.5(0)-3(0+4)^{-2}]\\= [1-3(6)^{-2}] - [0-3(4)^{-2}]\\ = [1-\frac{1}{12}] - [-\frac{3}{16} ]\\= \frac{11}{12}+\frac{3}{16}\\ = 0.9167 - 0.1875\\= 0.7292feet$