What is the slop formula? A. Y2+Y1 / 2 B. a2 +b2 = c2 C. rise / run

The temperature fell from 0Degrees Fahrenheit to 6 and three-fifthsDegrees Fahrenheit below 0Degrees Fahrenheit in 2 and one-fif

stepan [7]

Answer:

B) -3 (Sorry if its overwhelming)

Explanation:

First, we have the given information of the weather over a 2 and 1/5 span

And we are asked to find the change per hour

We can first start by finding the change for 2 and 1/5 hours:

- Subtract the first amount by the second amount.

= (-6 3/5) - 0

= (-6 3/5) + 0

= -6 3/5

Now that we have the change for 2 and 1/5 hours we must find the change per hour:

- Divide the given dividend by the given divisor

= -6 3/5 ÷ 2 1/5

= -3

Hence, the temperature change per hour over a 2 1/5 hour span is -3° per hour

7 0

3 years ago

Read 2 more answers

What is the order of the set A={-1,20,3,17,5}

laiz [17]

Hey!

Hope this helps...

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I'm assuming your talking about least to greatest, OR greatest to least. If so, that's what I'll do...

Least to Greatest: -1, 3, 5, 17, 20

Greatest to Least: 20, 17, 5, 3, -1

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Unfortunately I did not completely understand the question, for future reference please give AS MUCH detail as possible so that your fellow brainly members can help you ASAP...

And thanks for using Brainly!

3 0

3 years ago

I am confused please help

Otrada [13]

Answer:

3 tiles will not fit together.

Step-by-step explanation:

Measure of an Interior angle of a polygon = $\frac{(n-2)\times 180}{n}$

Here, n = number of sides of the polygon

Therefore, measure of the interior angles of a regular hexagon,

A = $\frac{(6-2)\times 180}{6}$

A = 120°

Similarly, interior angle of the regular pentagon,

B = $\frac{(5-2)\times 180}{5}$

B = 108°

Now m∠A + m∠B + m∠C = 360°

m∠C = 360° - (120° + 108°)

= 132°

To fit the given three tiles perfectly, interior angle (∠D) of the third Octagonal tile should be 132°.

D = $\frac{(8-2)\times 180}{8}$

D = 135°

m∠C ≠ m∠D

Therefore, 3 tiles will not fit together.

8 0

3 years ago

Integrate 1 - x / x(x2 + 1) d x by partial fractions.

solniwko [45]

Answer:

$log x-\frac{log(x^{2}+1) }{2}-tan^{-1} x$

Step-by-step explanation:

step 1:- by using partial fractions

$[tex]\frac{1-x}{x(x^{2}+1) } =\frac{A(x^{2}+1)+(Bx+C)(x }{x(x^{2}+1) }$ ......(1)

step 2:-

solving on both sides

$1-x=A(x^{2} +1)+(Bx+C)x......(2)$

substitute x =0 value in equation (2)

1=A(1)+0

A=1

comparing x^2 co-efficient on both sides (in equation 2)

0 = A+B

0 = 1+B

B=-1

comparing x co-efficient on both sides (in equation 2)

-1 = C

step 3:-

substitute A,B,C values in equation (1)

now

$\\\int\limits^ {} \, \frac{1-x}{x(x^{2}+1) } d x =\int\limits^ {} \frac{1}{x} d x +\int\limits^ {} \frac{-x}{x^{2}+1 } d x -\int\limits \frac{1}{x^{2}+1 } d x$