Help plzzzzz i needs halp!

Can any one help me with this

jonny [76]

Answer: C

Step-by-step explanation:

Since this is an isosceles triangle as indicated by the markers on QP and PR, we know that QS and SR are equivalent.

To find the value of n, we set QS and SR equal to each other.

6n+3=4n+11 [combine like terms]

2n=8 [divide both sides by 2]

n=4

Now that we know n=4, we know that A is incorrect. What we can do is use the value of n to solve for QS, SR, and QR.

QS

6(4)+3=13

Since the length of QS is 13, we know B is incorrect.

SR

4(4)+11=27

Since SR is 27, C is a correct answer.

QR

13+27=40

Since QR is 40, the only correct answer is C.

3 0

3 years ago

Is my answer correct? If not could you plz show the explanation of how to solve some might think it’s easy !!!

Monica [59]

Answer:

1

Step-by-step explanation:

( 50 / 5² )² ÷ 4 (by PEDMAS, do the terms inside the parentheses first)

= [ (5 · 10) / (5 · 5) ]² ÷ 4 (the 5's cancel out in the numerator and denominator)

= [ (10) / (5) ]² ÷ 4

= [ 2 ] ² ÷ 4

= 4 ÷ 4

= 1

Hope this helps

7 0

4 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

1) Provide the sine, cosine, and tangent of ∠F

Naddika [18.5K]

Answer:

1. cos(F) = 2. 5/13

tan (F) = 3. 12/5

sin (F) = 1. 12/13

2. Sin (R) = 1. 60/68 = 30/34 = 15/17

cos (R) = 2. 32/68 = 16/34 = 8/17

tan (R) = 3. 60/32 = 30/16 = 15/8

Step-by-step explanation:

1 picture:

sin = Opp/Hyp

cos = adj/hyp

tan = opp/adj

cos(F) = 2. 5/13

tan (F) = 3. 12/5

sin (F) = 1. 12/13

2 Picture:

Sin (R) = 1. 60/68 = 30/34 = 15/17

cos (R) = 2. 32/68 = 16/34 = 8/17

tan (R) = 3. 60/32 = 30/16 = 15/8

8 0

3 years ago

Write the equation from the graph in slope intercept form

Oxana [17]

Answer:

y=2/3x−4

Step-by-step explanation:

5 0

3 years ago

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