Simply you can substitute theta=45 degrees (or any angle,not zero) in left hand side and right hand side.

or

using a^2 - b^2 =(a+b)(a-b)

= sin(2x+x)sin(2x-x)
= sin 3x sin x

3 0

3 years ago

2 1/4 x2 1/4 in mixed number

maks197457 [2]

Answer:

$5 \frac{1}{16}$ is the answer.

Step-by-step explanation:

Given:

$2 \frac{1}{4} \times 2 \frac{1}{4}$

Step 1: Converting the mixed fraction into fraction:

Multiply 4 by 2 add 1 to the result.

$2\frac{1}{4}$ = $\frac{4 \times 2+1}{4}$

= $\frac{8+1}{4}$

= $\frac{9}{4}$

Step 2: Performing the multiplication using the fraction

= > $\frac{9}{4} \times \frac{9}{4}$

=> $\frac{81}{16}$

Step 3: Converting the product back to mixed fraction

divide 81 by 16 and find the remainder

=> $quotient \frac{remainder}{divisor}$

Dividing 81 by 16 we get 5 as quotient and 1 as remainder

Thus the result in mixed fraction will be

=> $5 \frac{1}{16}$

4 0

3 years ago

1) -24, -4, 16, 36, ... Find a23

viktelen [127]

We're given the Arithmetic Progression -24, -4, 16, 36 ... .

We know that a term in an AP is generally represented as:

$\bf a_n\ =\ a\ +\ (n\ -\ 1)d$

where,

a = the first term in the sequence
n = the number of the term/number of terms
d = difference between two terms

We need to find $\sf a_2_3$ .

From the given progression, we have:

a = -24
n = 23
d = (-24 - (-4) = -20

Using these in the formula,

$\sf a_2_3\ =\ a\ +\ (n\ -\ 1)d\\\\\\a_2_3\ =\ -24\ +\ (23\ -\ 1)\ \times\ (-20)\\\\\\a_2_3\ =\ -24\ +\ 22\ \times (-20)\\\\\\a_2_3\ =\ -24\ -\ 440\\\\\\\bf a_2_3\ =\ -464$

Therefore, the 23rd term in the AP is -464.

Hope it helps. :)

7 0

2 years ago

In the figure below, segment AC is congruent to segment AB:

stepladder [879]

Answer: A. "Segment AD bisects angle CAB." is the right answer.

Step-by-step explanation:

Given : In ΔABC ,AC≅AB.

⇒∠ACB=∠CBA....(1) (∵ angles opposite to equal sides of a triangle are equal )

Now in ΔACD and ΔABD

AD=AD (common)....(2)

Here we need one more statement to prove the triangles congruent that is only statement (A) fits in it.

If AD bisects ∠CAB then ∠CAD=∠BAD..(3)

Now again Now in ΔACD and ΔABD

∠ACB=∠CBA [from (1)]

AD=AD [common]

∠CAD=∠BAD [from (3)]

So by ASA congruency criteria ΔADC≅ΔABD.

7 0

3 years ago