Ask question

Ask question

All categories

IrinaVladis [17]

2 years ago

11

Prove that

cos 5x/sin17x-sin3x=-sin2x/cos10x" alt="cos 9x-cos 5x/sin17x-sin3x=-sin2x/cos10x" align="absmiddle" class="latex-formula">

1 answer:

maria [59]2 years ago

8 0

Cos9x - cos5x = -2sin2x.sin7x
sin17x - sin3x = 2sin7x.cos10x
so f(x) = $\frac{-2sin2x.sin7x}{2sin7x.cos10x}$ = $\frac{-sin2x}{cos10x}$

You might be interested in

8x - 3(5 + 9x what's the simplified algebraic expression p

Mashcka [7]

=8x + (-3) (5) + (-3) (9x)

=8x + -15 + -27x

=8x + -15 + -27x

(8x + -27x) + (-15)

19x +-15

19x -15

4 0

2 years ago

What is 15 to the third power

Natalka [10]

Answer:

your answer should be 45

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Solve AABC. Round your answers to the nearest hundredth, if necessary

Aloiza [94]

Answer:

$C=25^{\circ},\\a\approx 10.72,\\b\approx 11.83$

Step-by-step explanation:

The sum of the interior angles of a triangle is 180 degrees. Thus, angle C must be $180-90-65=25^{\circ}$ .

In any triangle, the Law of Sines is given by $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Therefore, we have:

$\frac{\sin 90^{\circ}}{b}=\frac{\sin 25^{\circ}}{5},\\\\b=\frac{5\sin90^{\circ}}{\sin 25^{\circ}}=11.8310079158\approx \boxed{11.83}$

$\frac{a}{\sin 65^{\circ}}=\frac{5}{\sin25^{\circ}},\\a=\frac{5\sin 65^{\circ}}{\sin25^{\circ}}=10.7225346025\approx \boxed{10.72}$

8 0

2 years ago

Gabrielle's age is three times Mikhail's age. The sum of their ages is 56, what is Mikhail’s age?

4vir4ik [10]

Gabrielle's age is 3x (in which x is mikhail's age)
the sum is 56 so 3x+x=56

3x+x=56
4x=56
x=14

since x was mikhail's age, mikhail is 14 years old

6 0

2 years ago

Dafna1 [17]

Answer:

A) 7x^3 + 13x^2 + 8x + 9

Step-by-step explanation:

(3x^2 + 2x + 7x^3) + (10x^2 + 6x + 9)

I like to line them up vertically

7x^3 +3x^2 + 2x

+ (10x^2 + 6x + 9)

-----------------------------------

7x^3 +13x^2 +8x +9

4 0

3 years ago

Read 2 more answers