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3 years ago

Find the size of angle xyz give your answer to 3 significant figures

Mathematics

1 answer:

KengaRu [80]3 years ago

8 0

Answer:

68.2°

Step-by-step explanation:

This is a right triangle, so we can use SIN,COS,TAN to solve for the angle.

We want ∠YXZ

THe side opposite to the angle is given as 20

The side adjacent to the angle is given as 8

Which trig ratio relates "opposite" and "adjacent"??

It is Tan!!

Thus, we can write:

$Tan(\angle YXZ)=\frac{20}{8}\\Tan(\angle YXZ)=\frac{5}{2}\\Tan(\angle YXZ)=2.5\\\angle YXZ = ArcTan(2.5)\\\angle YXZ = 68.2$

THis is the angle

68.2°

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17 in long 1.3 ft wide and 8in high what is the volume

OverLord2011 [107]

Answer:

135.2

Step-by-step explanation:

Volume is LWH and that is 17 * 1.3 * 8=135.2

5 0

2 years ago

Read 2 more answers

Assume a normal distribution and that the average phone call in a certain town lasted 4 min, with a standard deviation of 1 min.

olya-2409 [2.1K]

SOLUTION

The mean is 4min

standard deviation is 1min

the z score is

$z=\frac{x-\bar{x}}{\sigma}$

where

$\begin{gathered} x=3 \\ \bar{x}=4 \\ \sigma=1 \end{gathered}$

then we have

$\begin{gathered} z=\frac{3-4}{1}=\frac{-1}{1}=-1 \\ z=-1 \end{gathered}$

The probability the call lasted less than 3 min will be

Therefore, the probability that (z < -1 ) is

[tex]\begin{gathered} Pr(z<-1)=Pr(0Hence, the percentage of the calls that lasted less than 3 min is 16%

4 0

11 months ago

There are four times as many girls in a chemistry class as there are boys. if there are 45 students in the class, how many boys

puteri [66]

In order to determine the number of boys in the class, first, we set-up an algebraic equation. Let x = number of boys in the class. The equation is then: x + 4x = 45. This is because the number of girls is 4 times greater than the number of boys. After solving the algebraic equation, it is then determined that there are 9 boys in the Chemistry class.

4 0

2 years ago

Solve tan X = 2cot X

vichka [17]

If x = 1 the the answer would be 2.01715045524.

4 0

3 years ago

General solutions of sin(x-90)+cos(x+270)=-1 {both 90 and 270 are in degrees}

mixer [17]

Answer:

$\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

Step-by-step explanation:

Given:

$\sin (x-90^{\circ})+\cos(x+270^{\circ})=-1$

First, note that

$\sin (x-90^{\circ})=-\cos x\\ \\\cos(x+270^{\circ})=\sin x$

So, the equation is

$-\cos x+\sin x= -1$

Multiply this equation by $\frac{\sqrt{2}}{2}:$

$-\dfrac{\sqrt{2}}{2}\cos x+\dfrac{\sqrt{2}}{2}\sin x= -\dfrac{\sqrt{2}}{2}\\ \\\dfrac{\sqrt{2}}{2}\cos x-\dfrac{\sqrt{2}}{2}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos 45^{\circ}\cos x-\sin 45^{\circ}\sin x=\dfrac{\sqrt{2}}{2}\\ \\\cos (x+45^{\circ})=\dfrac{\sqrt{2}}{2}$

The general solution is

$x+45^{\circ}=\pm \arccos \left(\dfrac{\sqrt{2}}{2}\right)+2\pi k,\ \ k\in Z\\ \\x+\dfrac{\pi }{4}=\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{4}\pm \dfrac{\pi }{4}+2\pi k,\ \ k\in Z\\ \\\left[\begin{array}{l}x=2\pi k,\ \ k\in Z\\ \\x=-\dfrac{\pi }{2}+2\pi k,\ k\in Z\end{array}\right.$

4 0

3 years ago