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3 years ago

Find the slope of the line passing through the points (-4,-3) and (8,-9)

1 answer:

4 0

The answer is -1/2

Rise/run
X=run
Y=rise

y2-y1/x2-x1
(-9)-(-3)/(8)-(-4)
-6/12
-1/2

Brainliest my answer if it helped you out?

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<img src="https://tex.z-dn.net/?f=%20%5Crm%20%5Cint_%7B0%7D%5E%20%5Cinfty%20%20%5Cfrac%7B%20%5Csqrt%5B%20%20%5Cscriptsize%5Cphi%

Rasek [7]

With ϕ ≈ 1.61803 the golden ratio, we have 1/ϕ = ϕ - 1, so that

$I = \displaystyle \int_0^\infty \frac{\sqrt[\phi]{x} \tan^{-1}(x)}{(1+x^\phi)^2} \, dx = \int_0^\infty \frac{x^{\phi-1} \tan^{-1}(x)}{x (1+x^\phi)^2} \, dx$

Replace $x \to x^{\frac1\phi} = x^{\phi-1}$ :

$I = \displaystyle \frac1\phi \int_0^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

Split the integral at x = 1. For the integral over [1, ∞), substitute $x \to \frac1x$ :

$\displaystyle \int_1^\infty \frac{\tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx = \int_0^1 \frac{\tan^{-1}(x^{1-\phi})}{\left(1+\frac1x\right)^2} \frac{dx}{x^2} = \int_0^1 \frac{\pi2 - \tan^{-1}(x^{\phi-1})}{(1+x)^2} \, dx$

The integrals involving tan⁻¹ disappear, and we're left with

$I = \displaystyle \frac\pi{2\phi} \int_0^1 \frac{dx}{(1+x)^2} = \boxed{\frac\pi{4\phi}}$

8 0

2 years ago

What is used to repair big brass band instruments

alina1380 [7]

<span> the answer tuba glue!

</span>

5 0

3 years ago

If h(x)=x-7 and g(x)=x ^2, which expression is equivalent to (g•h)(5)

pashok25 [27]

x^3 ‐ 7x^2

Explanation:

x^2(x-7) = x^3 ‐ 7x^2

7 0

3 years ago

The table shows a basketball player score in 5 Games how many points must the basketball players score in the next game to achie

bogdanovich [222]

Answer:

You would have to score 17 points.

Step-by-step explanation:

13x6=78 is the points you have to reach to get 13 points per game.

Add up all numbers, which is 61.

78-61=17

That is your answer.

3 0

3 years ago

Help help area and perimeter

natita [175]

Answer:

Perimeter= 49

area= 360

Step-by-step explanation:

perimeter means you need to add 40+9 and area means you must multiple the numbers.

4 0

2 years ago