<img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B3%7D%20x%20%5Btex%5D%5Cfrac%7B4%7D%7B15%7D" id="TexFormula1" title="\frac{2}{

All categories

DaniilM [7]

3 years ago

3} x [tex]\frac{4}{15}" alt="\frac{2}{3} x [tex]\frac{4}{15}" align="absmiddle" class="latex-formula">

Mathematics

1 answer:

igomit [66]3 years ago

4 0

Answer: 8/45
Just times 2 by 4 and 3 by 15

You might be interested in

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

Whats the answer???

Novay_Z [31]

X is -7. :)

Hope that helped :)

6 0

3 years ago

!!! Help ASAP please!!!!In DEF, sin D = 24/26 What is cos E?

Georgia [21]

Answer:

Step-by-step explanation:

Cosine is the ratio of "adjacent" to "hypotenuse"

With respect to the Angle E, the adjacent side is FE and the hypotenuse (always opposite side to 90 degree angle) is DE.

We are given DE = 26, but we need FE. We will solve for FE using Pythagorean Theorem, which tells us:

leg^2 + another leg^2 = hypotenuse^2

So, we will have:

10^2 + FE^2 = 26^2

Now, we solve for FE:

$10^2 + FE^2 = 26^2\\100+FE^2=676\\FE^2=576\\FE=24$

So,

Cos(E) = adj/hyp = 24/26

Correct answer is A

7 0

3 years ago

Find y' by implicit differentiation: xy + 2x + 3x^2 = 4

NeX [460]

(xy)' + (2x)' + (3x^2)' = (4)'

y + xy' + 2 + 6x = 0

xy' = -y -2 -6x

y' = [-y -2 -6x] / x

Now solve y from the original equation and substitue

xy + 2x + 3x^2 = 4 => y = [-2x - 3x^2 + 4] / x

y' = [(-2x - 3x^2 +4) / x - 2 - 6x ] / x

y' = [-2x - 3x^2 + 4 -2x -6x^2 ] x^2 = [ -4x - 9x^2 + 4] / x^2 =

= [-9x^2 - 4x + 4] / x^2

3 0

3 years ago

The graph of h is a translation 4 units right and 1 unit down of the graph of f(x) = x2.

Galina-37 [17]

Answer:

Step-by-step explanation:

3 0

3 years ago