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tino4ka555 [31]

3 years ago

14

Plies sambedy help problem 5.

2 answers:

DaniilM [7]3 years ago

4 0

No, because it does not land on day 10

NikAS [45]3 years ago

3 0

The answer is no because 10 days from may 5 is the 15 and then 10 days after that is the 25

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Construct a linear equation for the linear data presented in the table

yaroslaw [1]

Answer:

y = -7x

Explanation:

Take two points from table: (-4, 28), (-3, 21)

Find slope:

$\sf slope: \dfrac{y_2 - y_1}{x_2- x_1} \ \ where \ (x_1 , \ y_1), ( x_2 , \ y_2) \ are \ points$

$\rightarrow \sf slope\ (m): \dfrac{21-28}{-3-(-4)} = -7$

Now find equation:

$\sf y - y_1 = m (x - x_1)$

$\rightarrow \sf y - 21 = -7(x - (-3))$

$\rightarrow \sf y = -7(x +3) + 21$

$\rightarrow \sf y = -7x -21 + 21$

$\rightarrow \sf y = -7x$

5 0

2 years ago

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Which statements are true about the place values in the number 88.888? Choose all answers that are correct. A. The 8 in the hund

Vadim26 [7]

<span>Answers:
______________________________________________________

</span><span>[B]: The 8 in the tenths place is 10 times the value of the 8 in the hundredths place.
</span>______________________________________________________
<span>[C]: The 8 in the hundredths place is 10 times the value of the 8 in the tenths place.
______________________________________________________</span>

6 0

3 years ago

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What is the simplified form of the expression?

RSB [31]

The 404 wasnt found... private message me and ill help you out<span />

8 0

3 years ago

Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.

Maru [420]

Answer:

$9\text{ln}|x|+2\sqrt{x}+x+C$

Step-by-step explanation:

We have been an integral $\int \frac{9+\sqrt{x}+x}{x}dx$ . We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

$\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx$

$\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx$

Now, we will apply the sum rule of integrals as:

$\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx$

$9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx$

Using common integral $\int \frac{1}{x}dx=\text{ln}|x|$ , we will get:

$9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx$

Now, we will use power rule of integrals as:

$9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx$

$9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx$

$9\text{ln}|x|+2\sqrt{x}+\int 1dx$

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

$9\text{ln}|x|+2\sqrt{x}+x+C$

Therefore, our required integral would be $9\text{ln}|x|+2\sqrt{x}+x+C$ .

4 0

3 years ago

Someone please help me

pickupchik [31]

The answer for this question is -3

5 0

3 years ago