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3 years ago

8

PLEASE HELP!

1 answer:

Marrrta [24]3 years ago

3 0

Answer:

15.6 feet

Step-by-step explanation:

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∠3 and ∠6 can be classified as:

anastassius [24]

Answer:

Alternate Interior Angles

Step-by-step explanation:

Since they are inside the parallel lines, Alternate Exterior Angles and any other similar theorems can be ruled out.

Since they are on opposite sides of each other, Corresponding Angles and any other similar theorems can be ruled out.

Since they are far apart from each other, Supplementary Angles, Adjacent Angles, Vertical Angles, and any other similar definitions can be ruled out.

Therefore, we are left with Alternate Interior Angles.

6 0

3 years ago

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An example of a rectangular Prism can be: A globe or ball O Fish tank o Bowl Can

garik1379 [7]

Answer:

Fish Tank

Step-by-step explanation:

This of a rectangular prism as a cardboard box. It has rectangles on the top and the bottom. The fish tank is somewhat similar to this so it will be a rectangular prism.

3 0

3 years ago

Factorize the algebraic expression:<br>10y²+13y-3

laila [671]

<h2>⟰ <u>A</u><u>N</u><u>S</u><u>W</u><u>E</u><u>R</u> ⟰</h2>

<h2>U R ANS IS IN THE PIC !!</h2>

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4 0

3 years ago

Pls help will give brainlist if right

sergiy2304 [10]

Answer:

1) supplementary

2) ∠ABD = 180° - 123°

Step-by-step explanation:

3 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

2 years ago