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4 years ago

Which unit of measure would be appropriate for the area of a pictu - that is

Mathematics

1 answer:

sp2606 [1]4 years ago

5 0

Answer:

Can you please look at your question and tell me the whole thing. pls and ty :)

Step-by-step explanation:

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The population of a city on April 15, 1915, was 47,175. During the period between March 1 and July 1, 1915, 1,325 new cases of s

Norma-Jean [14]

Answer:

the incidence rate is 28.47

Step-by-step explanation:

The computation of the monthly incident rate of active cases is shown below:

Incidence rate is

= 1325 ÷ (47175 - 642) × 1000

= (1325 ÷ 46533 ) × 1000

= 28.47

hence, the incidence rate is 28.47

7 0

3 years ago

What is the ful cost of a 59.99 game after adding taxes

dsp73

The correct answer would be 61.32

5 0

4 years ago

Write a word problem that can be solved by finding the numbers that have 4 as a factor

Ksenya-84 [330]

8 10 12 14 16 I think

5 0

4 years ago

Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

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

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$