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3 years ago

Please hurry ASAP

Mathematics

2 answers:

lukranit [14]3 years ago

5 0

Answer:

It Is The First One

Step-by-step explanation:

ExtremeBDS [4]3 years ago

4 0

Answer:

Step-by-step explanation:

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A shirt has a ticket price of $14.50 courtney used a coupon and got discounted of 20%

cluponka [151]

Answer:

where is the question

Step-by-step explanation:

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3 years ago

Please HELP ME I need help on this final question

I am Lyosha [343]

1 and 2 are right because one has adjacent. 2 vertical angles. Vertical angles mean the same plane and adjacent means two sides of an X.

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1 year 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.

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

So, this integral is divergent.

4 0

3 years ago

Someone annnnswer pls

cupoosta [38]

Answer:

t = 16

Step-by-step explanation:

Hope this helped :]

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3 years ago

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Find the components of the vertical force Bold Upper FFequals=left angle 0 comma negative 10 right angle0,−10 in the directions

quester [9]

Solution :

Let $v_0$ be the unit vector in the direction parallel to the plane and let $F_1$ be the component of F in the direction of $v_0$ and $F_2$ be the component normal to $v_0$ .