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marishachu [46]
3 years ago
15

When labeling a pre-image and it's image, we use a small tick mark. It's called a(n):

Mathematics
2 answers:
Dmitriy789 [7]3 years ago
7 0

Answer:

It’s a prime

Step-by-step explanation:

IgorLugansk [536]3 years ago
4 0

Answer:

Prime

Step-by-step explanation:

If we have a point on a pre-image, we can have it labelled as A(x,y)

Now, we can write this as the image

To write this as the image, we simply have to add a mark on the A

This mark is at the upper right

We have it as A’ (x,y)

The part added is called a prime

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Answer:

Step-by-step explanation:

22 dollars

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3 years ago
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Find the value of x (sines and cosines)
Sav [38]
You would begin by choosing which trigonometric function to use. 
In this case, you would use sine.

Sine is opposite over hypotenuse. 
Opposite of the 35 degree angle is 18, and you want to find the hypotenuse which is x.

sin(35) = \frac{18}{x} (multiply x to both sides)
x sin(35) = 18 (divide sin(35) to both sides)
x = 31.38204...

So, therefore x = 31.38. I don't know what decimal point you're supposed to round to, so I'm gonna guess the hundredths place. :)
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4 years ago
Match each of the trigonometric expressions below with the equivalent non-trigonometric function from the following list. Enter
Levart [38]

Answer:

Match each of the trigonometric expressions below with theequivalent non-trigonometric function from the following list.Enter the appropiate letter(A,B, C, D or E)in each blank

A . tan(arcsin(x/8))

B . cos (arsin (x/8))

C. (1/2)sin (2arcsin (x/8))

D . sin ( arctan (x/8))

E. cos (arctan (x/8))

These are the spaces to fill out :

.. ..........x/64 (sqrt(64-x^2))

.............x/sqrt(64+x^2)

.............sqrt(64-x^2)/8

..............x/sqrt(64-x^2)

..............8/sqrt(64+x^2)

A. ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

Step-by-step explanation:

To solve this we have to find the missing sides to each of the triange discribed in prenthesis thus

A we have the sides of the triangle given by x, 8 and  \sqrt{8^{2} - x^{2} }or  \sqrt{64 - x^{2} }

thus tan(arcsin(x/8))  = \frac{x}{\sqrt{64 - x^{2} }}  =

Therefore  ........tan(arcsin(x/8))  =......x/sqrt(64-x^2)

B

Here we have cos = adjacent/hypotenuse where adjacent side is \sqrt{64 - x^{2} } and hypothenuse = 8 we have \sqrt{64 - x^{2} }/8

B .      cos (arsin (x/8))  ....sqrt(64-x^2)/8

4 0
3 years ago
What is an equation of the line that passes through the points (3,4) and (5,8)
algol [13]

Answer:

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Step-by-step explanation:

6 0
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In a sample of 230 teenagers, 202 would like to have smaller class sizes at their school. Find the sample
marishachu [46]

Answer:

The sample proportion is 0.88.

The margin of error is of 0.04.

The interval likely to contain the true population proportion is (0.84,0.92).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the z-score that has a p-value of 1 - \frac{\alpha}{2}.

Margin of error:

The margin of error is of:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

Sample of 230 teenagers, 202 would like to have smaller class sizes at their school.

This means that n = 230, and that the sample proportion is:

\pi = \frac{202}{230} = 0.88

The sample proportion is 0.88.

Standard 95% confidence level

So \alpha = 0.05, z is the value of Z that has a p-value of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

Margin of error:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

M = 1.96\sqrt{\frac{0.88*0.12}{230}}

M = 0.04

The margin of error is of 0.04.

Confidence interval:

Sample proportion plus/minus margin of error. So

0.88 - 0.04 = 0.84

0.88 + 0.04 = 0.92

The interval likely to contain the true population proportion is (0.84,0.92).

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