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3 years ago

I just need help with number 9:) if someone wanted to help that would be nice

Mathematics

1 answer:

nikdorinn [45]3 years ago

7 0

Answer: I think the answer is 16.55

Step-by-step explanation: I added 9.80 and 7.25 and got 17.05 and then subtracted 17.05 and 0.5 and got 16.55 I hope this helped!! :)

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57times71= please tell me the answer

Phoenix [80]

57 times 71 = 4047 Hope that helps!

5 0

3 years ago

Read 2 more answers

Cot -90<br> trig function

Licemer1 [7]

Answer:

It doesn't exist,bro.

Step-by-step explanation:

5 0

3 years ago

A test was given to a group of students. The grades and gender are summarized below A B C TotalMale 5 9 2 16Female 7 11 12 30Tot

faust18 [17]

Probability that the student got a 'C' GIVEN they are female = number of females that got a C in the test/number of females

From the information given,

number of females that got a C in the test = 12

number of females = 30

Thus,

Probability that the student got a 'C' GIVEN they are female = 12/30

We would simplify the fraction by dividing the numerator and denominator by 6. Thus,

Probability that the student got a 'C' GIVEN they are female = 2/5

4 0

1 year ago

Ed's class took a field trip to the art museum. They left school at 10:45 A.M. It took them 15 minutes to drive to the museum. T

Brut [27]

Answer:
12:45

10:45 + 15 minutes = 11:00

11:00 + 1 hour and 45 minutes = 12:45

Think of it as:
11:00
+1.45
———
12:45

4 0

3 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:

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

Coming back to x:

We have that:

$x = 3\sin{\theta}$

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

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

8 0

3 years ago