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3 years ago

Prove that the left side equals to the right side: (sin x/(1-cos x))+(sin x/(1+cos x))=2 csc x

Mathematics

2 answers:

34kurt3 years ago

8 0

sorryy I haven't learned this yet

MAVERICK [17]3 years ago

3 0

Answer:

The left side does not equal to the right side. It's not an identify

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ankoles [38]

Simplify the radical, then combine the numbers.
-131

4 0

3 years ago

A. 3 B. 10 C. 20 D. 30

Dahasolnce [82]

Answer:

Explanation:

292/100×9.9 = 2.92×9.9 = 28.908

30 is closest to 28.908

3 0

2 years ago

The length of a rectangle is 4 more than its width. The area of the rectangle is 21m^2 . What is the measure of the width?

saw5 [17]

Make the length x+4 and the width x and then make a formula x(x+4)=21
then u ll get x^2+4x-21=0 and then solve for x ..... the width is 3

5 0

3 years ago

The ratio of the numerator to the denominator of a fraction is 2 to 3. If both the numerator and the denominator are increased b

fiasKO [112]

The clues translate to

n/d = 2/3
(n+2)/(d+2) = 3/4

Solving this by cross multiplying both:

3n = 2d
3(d+2) = 4(n+2) => 3d + 6 = 4n + 8

then substituting d = 3/2 n in the second:

9/2n - 4n = 2 => 1/2 n = 2 => n = 4

if n=4 then d = 3/2*4 = 6.

So the original fraction was 4/6

5 0

3 years ago

Solve using the quadratic formula 2n^2- 15 = -4n

NeX [460]

Answer:

$\displaystyle n_1 = \frac{-2 + \sqrt{34}}{2} \approx 1.9155 \text{ or } n_2 = \frac{-2-\sqrt{34}}{2} \approx -3.9155$

Step-by-step explanation:

We are given the equation:

$\displaystyle 2n^2 - 15 = -4n$

And we want to solve using the quadratic formula.

First, isolate the equation:

$\displaystyle 2n^2 + 4n - 15 = 0$

Recall that for equations in the form ax² + bx + c = 0, the solutions are given by:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 2, b = 4, and c = -15.

Substitute and evaluate:

$\displaystyle \begin{aligned} n&= \frac{-(4)\pm\sqrt{(4)^2-4(2)(-15)}}{2(2)} \\ \\ &= \frac{-4\pm\sqrt{136}}{4} \\ \\ &= \frac{-4\pm2\sqrt{34}}{4} \\ \\ &= \frac{-2\pm\sqrt{34}}{2} \end{aligned}$

In conclusion, our two solutions are:

$\displaystyle n_1 = \frac{-2 + \sqrt{34}}{2} \approx 1.9155 \text{ or } n_2 = \frac{-2-\sqrt{34}}{2} \approx -3.9155$

6 0

2 years ago