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3 years ago

Y=-3/2x+15 ........................

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

8 0

What is the question? This is just a function.

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The area of a rectangular wall of a barn is 54 feet. It’s length is 12 feet longer than twice it’s width. Find the length and wi

matrenka [14]

Answer:the width is 3 feet

Step-by-step explanation:

6 0

3 years ago

find a curve that passes through the point (1,-2 ) and has an arc length on the interval 2 6 given by 1 144 x^-6

taurus [48]

Answer:

$f(x) = \frac{6}{x^2} -8$ or $f(x) = -\frac{6}{x^2} + 4$

Step-by-step explanation:

Given

$(x,y) = (1,-2)$ --- Point

$\int\limits^6_2 {(1 + 144x^{-6})} \, dx$

The arc length of a function on interval [a,b]: $\int\limits^b_a {(1 + f'(x^2))} \, dx$

By comparison:

$f'(x)^2 = 144x^{-6}$

$f'(x)^2 = \frac{144}{x^6}$

Take square root of both sides

$f'(x) =\± \sqrt{\frac{144}{x^6}}$

$f'(x) = \±\frac{12}{x^3}$

Split:

$f'(x) = \frac{12}{x^3}$ or $f'(x) = -\frac{12}{x^3}$

To solve fo f(x), we make use of:

$f(x) = \int {f'(x) } \, dx$

For: $f'(x) = \frac{12}{x^3}$

$f(x) = \int {\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = \frac{12}{2x^2} + c$

$f(x) = \frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = \frac{6}{1^2} + c$

$-2 = \frac{6}{1} + c$

$-2 = 6 + c$

Make c the subject

$c = -2-6$

$c = -8$

$f(x) = \frac{6}{x^2} + c$ becomes

$f(x) = \frac{6}{x^2} -8$

For: $f'(x) = -\frac{12}{x^3}$

$f(x) = \int {-\frac{12}{x^3} } \, dx$

Integrate:

$f(x) = -\frac{12}{2x^2} + c$

$f(x) = -\frac{6}{x^2} + c$

We understand that it passes through $(x,y) = (1,-2)$ .

So, we have:

$-2 = -\frac{6}{1^2} + c$

$-2 = -\frac{6}{1} + c$

$-2 = -6 + c$

Make c the subject

$c = -2+6$

$c = 4$

$f(x) = -\frac{6}{x^2} + c$ becomes

$f(x) = -\frac{6}{x^2} + 4$

3 0

3 years ago

What is the GCF of 104 and 117

solong [7]

Answer: 13

Step-by-step explanation:

104 divided by 13 = 8

117 divided by 13 = 9

13 in this case, is the only number that can be divided in BOTH numbers.

7 0

3 years ago

Never mind. No question. Can’t delete.

Evgen [1.6K]

Oof it be like that

4 0

3 years ago

Read 2 more answers

An ellipse has a co-vertex at (–8, 9) and a foci at (4, 4). If the center of the ellipse is located below the given co-vertex, t

vovikov84 [41]

Answer:

Step-by-step explanation:

“the center of the ellipse is located below the given co-vertex”

Co-vertex and center are vertically aligned, so the ellipse is horizontal.

Equation for horizontal ellipse:

(x-h)²/a² + (y-k)²/b² = 1

with

a² ≥ b²

center (h,k)

vertices (h±a, k)

co-vertices (h, k±b)

foci (h±c,k), c² = a² -b²

One co-vertex is (-8,9), so h = -8.

One focus is (4,4), so k = 4.

Center (h,k) = (-8,4)

c = distance between center and focus = |-8 - 4| = 12

b = |9-k| = 5

a² = c² + b² = 169

(x+8)²/169 + (y-4)²/25 = 1

7 0

3 years ago