All categories

3 years ago

Right triangles and trigonometry help!

Mathematics

1 answer:

Maslowich3 years ago

5 0

Step-by-step explanation:

$sin \: 45 \degree = \frac{39}{z} \\ \\ \therefore \: \frac{1}{ \sqrt{2} } = \frac{39}{z} \\ \\ \therefore \: z = 39 \sqrt{2} \\ let \: p \: be \: the \: perpendicular \: \\ \\\therefore \: tan \: 45 \degree = \frac{39}{p} \\ \\ \therefore \:1 = \frac{39}{p} \\ \\ \therefore \:p = 39 \\ \\ sin \: 60 \degree = \frac{39}{x} \\ \\ \therefore \: \frac{ \sqrt{3} }{2} = \frac{39}{x} \\ \\ \therefore \: x = \frac{78}{ \sqrt{3} } \\ \\ \therefore \: x = \frac{78 \sqrt{3} }{ 3 } \\ \\ \therefore \: x = 26 \sqrt{3} \\\\tan\: 60 \degree = \frac{39}{y} \\ \\ \therefore \: \sqrt{3} = \frac{39}{y} \\ \\ \therefore \: y = \frac{39}{ \sqrt{3} } \\ \\ \therefore \: y= \frac{39 \sqrt{3} }{ 3 } \\ \\ \therefore \: y = 13 \sqrt{3} \\$

You might be interested in

Why is it valid to say that both a function and its inverse describe the same relationship

liberstina [14]

A comparison between a function and its inverse would show that the domain and range of the original function swap. The domain of the function becomes the range of the inverse, the range of the function becomes the domain of its inverse.

Looking at ordered pairs of the function and its inverse would look like this:

(2,4) on the original function becomes (4,2) on the inverse.

While the graph of a function and its inverse are noticeably different an important thing to note is that it is merely a reflection across the line y=x.

So even though they appear different you are looking at the same relationship just as y vs. x instead of x vs. y

4 0

3 years ago

Plz help and no spam plz. A large university accepts 60% of the students who apply. Of the students the

mihalych1998 [28]

2,600 students actually enroll.

3 0

3 years ago

Read 2 more answers

I have no idea please help

melamori03 [73]

22. Alternate angles

23. 4: ( $h_{b} = 2 \frac{a}{b}$ , substitute b with 10 and a with 20 and you get 4.)

24. 0.28: 0.12 + 0.60 = 0.72. Probability is always out of 1, so 1 - 0.72 should give you the probability of losing, which is 0.28.

8 0

3 years ago

Write an equation of the circle with center (-3, 7) and radius 2.

jok3333 [9.3K]

Answer:

(x+3)^2 + (y-7)^2 = 2^2

Step-by-step explanation:

Follow the standard form of a circle, (x-h)^2 + (y-k)^2 = r^2, given that (-3,7) is (h,k) and 2 is r.

4 0

2 years ago

Determine whether the equation x^3 - 3x + 8 = 0 has any real root in the interval [0, 1]. Justify your answer.

nikdorinn [45]

Answer:

The equation does not have a real root in the interval $\rm [0,1]$

Step-by-step explanation:

We can make use of the intermediate value theorem.

The theorem states that if $f$ is a continuous function whose domain is the interval [a, b], then it takes on any value between f(a) and f(b) at some point within the interval. There are two corollaries:

If a continuous function has values of opposite sign inside an interval, then it has a root in that interval. This is also known as Bolzano's theorem.
The image of a continuous function over an interval is itself an interval.

Of course, in our case, we will make use of the first one.

First, we need to proof that our function is continues in $\rm [0,1]$ , which it is since every polynomial is a continuous function on the entire line of real numbers. Then, we can apply the first corollary to the interval $\rm [0,1]$ , which means to evaluate the equation in 0 and 1:

$f(x)=x^3-3x+8\\f(0)=8\\f(1)=6$

Since both values have the same sign, positive in this case, we can say that by virtue of the first corollary of the intermediate value theorem the equation does not have a real root in the interval $\rm [0,1]$ . I attached a plot of the equation in the interval $\rm [-2,2]$ where you can clearly observe how the graph does not cross the x-axis in the interval.

6 0

2 years ago