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hoa [83]
3 years ago
11

Awnser this please i will give you 16

Mathematics
1 answer:
Aleks04 [339]3 years ago
4 0

Answer:

Its EFBGCAHDI

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Y=x^2+3x-7 y-5x+8=0 how many solutions are there
motikmotik
X^2+3×-7=5x-8
X^2-2x+1=0
(X-1)(x-1)=0
X=1 x=1


Y=1^2+3*1-7
Y=1+3-7
Y=4-7
Y=-3


×=1 y=-3
Solution = (1,-3)
Answer= one solution
7 0
3 years ago
<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
Helppppp thanksssssssss
klemol [59]
The 3rd or 1st one I'm not sure which one but u have an idea of what it might be
4 0
3 years ago
At what value of x does the graph of the following function f(x) have a vertical asymptote? f(x)=1/x+3
ratelena [41]

Answer:

d

Step-by-step explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined.

Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non zero for this value then it is a vertical asymptote.

x + 3 = 0 ⇒ x = - 3 ← equation of vertical asymptote

5 0
3 years ago
Read 2 more answers
Morgan uses 2 oz of dog shampoo to bathe her dog each week. After 4 wk, 34 oz of shampoo remains.
Flauer [41]
The amount of shampoo required by Morgan each week to bathe her dog = 2 oz
So
The amount of shampoo required by Morgan in 7 days to bathe her dog = 2 oz
The amount of shampoo remaining after 4 weeks = 34 oz
So the amount of shampoo remaining after (4 * 7) days = 34 oz
The amount of shampoo remaining after 28 days = 34 oz
The amount of shampoo that Morgan uses in 28 days = (2/7) * 28 oz
                                                                                      = 2 * 4 oz
                                                                                      = 8 oz
Then
8 oz of shampoo is required by Morgan in = 28 days
Then
34 oz of shampoo will be used in = (28/8) * 34 days
                                                      = 7 * 17 days
                                                      = 119 days
So
The total number of
days before the bottle becomes empty = 119 + 8 oz
                                                               = 127 days
6 0
3 years ago
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