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3 years ago

Can 39/121 be simplified and if so what is it??

Mathematics

1 answer:

BabaBlast [244]3 years ago

3 0

Hi there!

Unfortunately, the beautiful fraction you've given us is simplified enough, and it cannot go any lower than what it is. So basically, you're left with $\frac{39}{121}$

Hope this helps!

Message me if you need anything else! I'd be happy to help! :D

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Use the image below to answer the following question. Find the value of sin x° and cos y°. What relationship do the ratios of si

kozerog [31]

The ratios of sin x° and cos y° are 15/8.

We have given that,

Use the image below to answer the following question.

We have to determine to find the value of sin x° and cos y°.

By applying Pythagoras theorem in the triangle given in the picture,

<h3>What is the Pythagoras theorem?</h3>

(Hypotenuse)² = (leg 1)² + (leg 2)²

PO² = (15)² + 8²

PO² = 225 + 64

PO = √289

PO = 17

By applying the sine rule in the given triangle,

sin(x°) = Opposie side/hypotenous

= 15/17

cos(y°) = Adjucent side/hypotenous

= 8/17

The relation between the ratio of sin(x) and cos(x) will be,

$\frac{sinx}{cosx} =\frac{\frac{15}{17} }{\frac{8}{17} }$

$=\frac{15}{8}$

Therefore the ratios of sin x° and cos y° are 15/8.

To learn more about the trigonometric function visit:

brainly.com/question/24349828

#SPJ2

5 0

2 years ago

A right triangle has an area of 2k^2-k-15 One of the legs has a length of k^2-9 What is the ratio of the area of a triangle to t

Darya [45]

<h3>Given</h3>

A = 2k² -k -15

L = k² -9

A/L

<h3>Solution</h3>

$\displaystyle\frac{A}{L}=\frac{2k^2-k-15}{k^2-9}=\frac{(k-3)(2k+5)}{(k-3)(k+3)}\\\\=\frac{2k+5}{k+3}$

4 0

3 years ago

Can someone please explain this to me? Thanks!

Makovka662 [10]

Answer: Choice D

$\displaystyle F\ '(x) = 2x\sqrt{1+x^6}\\\\$

==========================================================

Explanation:

Let g(t) be the antiderivative of $g'(t) = \sqrt{1+t^3}$ . We don't need to find out what g(t) is exactly.

Recall by the fundamental theorem of calculus, we can say the following:

$\displaystyle \int_{a}^{b} g'(t)dt = g(b)-g(a)$

This theorem ties together the concepts of integrals and derivatives to show that they are basically inverse operations (more or less).

So,

$\displaystyle F(x) = \int_{\pi}^{x^2}\sqrt{1+t^3}dt\\\\ \displaystyle F(x) = \int_{\pi}^{x^2}g'(t)dt\\\\ \displaystyle F(x) = g(x^2) - g(\pi)\\\\$

From here, we apply the derivative with respect to x to both sides. Note that the $g(\pi)$ portion is a constant, so $g'(\pi) = 0$

$\displaystyle F(x) = g(x^2) - g(\pi)\\\\ \displaystyle F \ '(x) = \frac{d}{dx}[g(x^2)-g(\pi)]\\\\\displaystyle F\ '(x) = \frac{d}{dx}[g(x^2)] - \frac{d}{dx}[g(\pi)]\\\\ \displaystyle F\ '(x) = \frac{d}{dx}[x^2]*g'(x^2) - g'(\pi) \ \text{ .... chain rule}\\\\$

$\displaystyle F\ '(x) = 2x*g'(x^2) - 0\\\\ \displaystyle F\ '(x) = 2x*g'(x^2)\\\\ \displaystyle F\ '(x) = 2x\sqrt{1+(x^2)^3}\\\\ \displaystyle F\ '(x) = \boldsymbol{2x\sqrt{1+x^6}}\\\\$