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2 years ago

Find the solution set for this equation.

Mathematics

1 answer:

devlian [24]2 years ago

5 0

Answer:
y= 0,-2

Hope this helps!

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Calculate the margin of error

Pepsi [2]

Answer:

The margin of error is 3.465

Step-by-step explanation:

Margin of error = t × sd/√n

n = 92, degree of freedom = n - 1 = 92 - 1 = 91, t-value corresponding to 91 degrees of freedom and 90% confidence level is 1.6618, sd = 20

Margin of error = 1.6618×20/√92 = 33.236/9.592 = 3.465

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3 years ago

Which equation is equivalent to a=2d +5r

denis-greek [22]

Answer:

Step-by-step explanation:

a=2d +5r = 57

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3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Crm%5Cfrac%7Bd%7D%7Bdx%7D%20%20%20%5Cleft%20%28%20%5Cbigg%28%20%5Cint_%7B1%7D%5E%7B%20%7B

Margaret [11]

Applying the product rule gives

$\displaystyle \frac{d}{dx}\int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \int_1^{x^2}\frac{2t}{1+t^2}\,dt \times \frac{d}{dx}\int_1^{\ln(x)}\frac{dt}{(1+t)^2}$

Use the fundamental theorem of calculus to compute the remaining derivatives.

$\displaystyle \frac{4x^3}{1+x^4} \int_1^{\ln(x)}\frac{dt}{(1+t)^2} + \frac{1}{x(1+\ln(x))^2}\int_1^{x^2}\frac{2t}{1+t^2}\,dt$

The remaining integrals are

$\displaystyle \int_1^{\ln(x)}\frac{dt}{(1+t)^2} = -\frac1{1+t}\bigg|_1^{\ln(x)} = \frac12-\frac1{1+\ln(x)}$

$\displaystyle \int_1^{x^2}\frac{2t}{1+t^2}\,dt=\int_1^{x^2}\frac{d(1+t^2)}{1+t^2}=\ln|1+t^2|\bigg|_1^{x^2}=\ln(1+x^4)-\ln(2) = \ln\left(\frac{1+x^4}2\right)$

and so the overall derivative is

$\displaystyle \frac{4x^3}{1+x^4} \left(\frac12-\frac1{1+\ln(x)}\right) + \frac{1}{x(1+\ln(x))^2} \ln\left(\frac{1+x^4}2\right)$

which could be simplified further.

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2 years ago