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3 years ago

What is the measure of WXZ

Mathematics

2 answers:

Mrrafil [7]3 years ago

8 0

83 degrees hope this helps

Tatiana [17]3 years ago

5 0

The answer is 83° hope it helps

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Determine the first four terms of the sequence in which the nth term is

Licemer1 [7]

<u>Answer:</u>

The correct answer option is: $\frac{1}{3} ,\frac{1}{4} ,\frac{1}{5} ,\frac{1}{6}$ .

<u>Step-by-step explanation:</u>

We know that the $nth$ term $a_n$ for an arithmetic sequence is given by:

$a_n=\frac{(n+1)!}{(n+2)!}$

where $n$ is the number of the position of the term.

We are supposed to find the first four terms of the sequence so we will substitute the values of $n$ from 1 to 4 in the given formula to get:

1st term:

$a_1=\frac{(1+1)!}{(1+2)!}=\frac{1}{3}$

2nd term:

$a_2=\frac{(2+1)!}{(2+2)!}=\frac{1}{4}$

3rd term:

$a_3=\frac{(3+1)!}{(3+2)!}=\frac{1}{5}$

4th term:

$a_4=\frac{(4+1)!}{(4+2)!}=\frac{1}{6}$

8 0

3 years ago

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

6 0

3 years ago

What is the value of the fourth term in a geometric sequence for which ay =

enot [183]

Answer:

a₄ = 1.25

Step-by-step explanation:

Given that,

First term of GP, a₁ = 10

The common ration of GP, r = 0.5

The nth term of the GP is given by :

$a_4=a_1r^{n-1}\\\\a_4=a_1r^3\\\\a_4=10\times (0.5)^3\\\\a_4=10\times 0.125\\\\a_4=1.25$

So, the fourth term of the GP is 1.25.

5 0

3 years ago

What is another name for like r.

Alina [70]

It would seem that r correlates to line AC. So another name could be Line AC. Remember to put a line above the two letters to let people know it is a line. I hope this helps! =)

3 0

3 years ago

Read 2 more answers

Just 18 please help I'm confused

levacccp [35]

I think you would divide 91 by 35% which is 260 students.

5 0

3 years ago

Read 2 more answers