Ask question

Ask question

All categories

I am Lyosha [343]

3 years ago

9

I’m having trouble. Help me please!!!

1 answer:

Lady_Fox [76]3 years ago

8 0

Hopes this helps:

Answer: No Solution

You might be interested in

PLEASE HELP ME . how can the converse of the Pythagorean theorem help you determine whether the roped off area is in the shape o

trapecia [35]

Step-by-step explanation:

From the Pythagorean theorem we have:

$a^2+b^2=c^2$ <em>subtract b² from both sides</em>

$a^2=c^2-b^2\to a=\sqrt{c^2-b^2}$

Same:

$b=\sqrt{c^2-a^2}$

4 0

3 years ago

What percent of 25 is 30

Thepotemich [5.8K]

0.833333333% is the answer

3 0

2 years ago

if a right angle has a hypotenuse of 19 units and a leg with the length of 7 units, find the length of the other leg

anyanavicka [17]

Square root of 19^2-7^2=17.66
Answer 17.6

8 0

2 years ago

A recipe calls for 3/4 cups of flour . Estevan wants to make 1/3 of the recipe how much flour will he need

romanna [79]

1/3 × 3/4 = 1/4 :) hope this helped!

7 0

3 years ago

Read 2 more answers

If tanA=a <br>then find sin4A-2sin2A/ sin4A+2sin2A

anygoal [31]

Answer:

The value of the given expression is

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

Step by step Explanation:

Given that $tanA=a$

To find the value of $\frac{sin4A-2sin2A}{sin4A+2sin2A}$

Let us find the value of the expression :

$\frac{sin4A-2sin2A}{sin4A+2sin2A}=\frac{2cos2Asin2A-2sin2A}{2cos2Asin2A+2sin2A}$ ( by using the formula $sin2A=2cosAsinA$ here A=2A)

$=\frac{2sin2A(cos2A-1)}{2sin2A(cos2A+1)}$

$=\frac{(cos2A-1)}{(cos2A+1)}$

$=\frac{(-(1-cos2A))}{(1+cos2A)}$ (using $sin^2A+cos^2A=1$ here A=2A)

$=\frac{-(sin^2A+cos^2A-(cos^2A-sin^2A))}{sin^2A+cos^2A+(cos^2A-sin^2A)}$ (using $cos2A=cos^2A-sin^2A$ here A=2A)

$=\frac{-(sin^2A+cos^2A-cos^2A+sin^2A)}{sin^2A+cos^2A+(cos^2A-sin^2A)}$

$=\frac{-(sin^2A+sin^2A)}{cos^2A+cos^2A}$

$=\frac{-2sin^2A}{2cos^2A}$

$=-\frac{sin^2A}{cos^2A}$

$=-tan^2A$ ( using $tanA=\frac{sinA}{cosA}$ here A=2A )

$=-a^2$ (since tanA=a given )

Therefore $\frac{sin4A-2sin2A}{sin4A+2sin2A}=-a^2$

6 0

2 years ago