Can somebody just give me the answer plz and if you know how to do geometry drop your insta along with the answer pleaseee

Mathematics

2 answers:

SSSSS [86.1K]2 years ago

8 0

Answer:

x = 40

Step-by-step explanation:

3x + (x+20) = 180

bonufazy [111]2 years ago

3 0

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Please help me simplify these expressions

Fantom [35]

Answer:

1. -18x¹¹

2. 3n⁷

Step-by-step explanation:

For these problems, there are two things you need to worry about: negative signs and exponents.

1. Let's look at the signs first. There is only one value with a negative sign, meaning that the negative sign will stay.

When multiplying with exponents, you have to add up the exponents. Don't forget the numerical coefficients.

-3x² · 3x · 2x³ · x⁵ = -18x¹¹

2. There are two negative signs in this probem, meaning that they will cancel out. Multiply the rest like we did in the first problem.

3n² · -n² · -n³ = 3n⁷

7 0

2 years ago

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

AVprozaik [17]

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

$\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\$

So, this integral is convergent.

$\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\$

So, this integral is divergent.

$\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\$