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3 years ago

How do I do an exponent with a letter variable in it?

Mathematics

1 answer:

WARRIOR [948]3 years ago

7 0

First you transform the expression and then evaluate the power

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Ata Sets

Fynjy0 [20]

Describe the data:

Mean=12

Interquartile range=7

Range=17

Do Not describe:

Range-23

Median=10

Interquartile range=12

6 0

2 years ago

1-cos^2 A / sec^2 A - tan^2 A

tiny-mole [99]

I hope this helps you

✔cos^2A+sin^2A=1

✔1-cos^2A=sin^2A

✔cos2A=cos^2A-sin^2A

✔sin2A=2.sinA.cosA

secA=1/cosA

tgA=sinA/cosA

sin^2A/1/cos^2A-sin^2A/cos^2A

sin^2A.cos^2A/cos2A

2.sin^2A.cos^2A/cos2A

sin2A.2.sin2A/cos2A

tg2A.2.sin2A

6 0

3 years ago

Which statement is true

Lorico [155]

Answer:

Step-by-step explanation:

(7/8)(-1/6)(-7)(-1/14)=

-14(-7)(-1/14) =

98(-1/14)=

-7

6 0

3 years ago

Which dot plot represents this data set?

zmey [24]

I think it is B is ur answer

4 0

3 years ago

Can someone please explain how to do this!!!

DedPeter [7]

Hi there!

So we are given that:-

tan theta = 7/24 and is on the third Quadrant.

In the third Quadrant or Quadrant III, sine and cosine both are negative, which makes tangent positive.

Since we want to find the value of cos theta. cos must be less than 0 or in negative.

To find cos theta, we can either use the trigonometric identity or Pythagorean Theorem. Here, I will demonstrate two ways to find cos.

Using the Identity

$\large \displaystyle{ {tan}^{2} \theta + 1 = {sec}^{2} \theta}$

Substitute tan theta = 7/24 in.

$\large \displaystyle{ {( \frac{7}{24}) }^{2} + 1 = {sec}^{2} \theta }$

Evaluate.

$\large \displaystyle{ \frac{49}{576} + 1 = {sec}^{2} \theta} \\ \large \displaystyle{ \frac{625}{576} = {sec}^{2} \theta}$

Reminder -:

$\large \displaystyle{ sec \theta = \frac{1}{cos \theta} }$

Hence,

$\large \displaystyle{ \frac{576}{625} = {cos}^{2} \theta} \\ \large \displaystyle{ \sqrt{ \frac{576}{625} } = cos \theta} \\ \large \displaystyle{ \frac{24}{25} = cos \theta}$

Because in QIII, cos<0. Hence,

$\large \displaystyle \boxed{ \blue{cos \theta = - \frac{24}{25} }}$

Using the Pythagorean Theorem

$\large \displaystyle{ {a}^{2} + {b}^{2} = {c}^{2} }$

Define c as our hypotenuse while a or b can be adjacent or opposite.

Because tan theta = opposite/adjacent. Therefore:-

$\large \displaystyle{ {7}^{2} + {24}^{2} = {c}^{2} } \\ \large \displaystyle{49 + 576 = {c}^{2} } \\ \large \displaystyle{625 = {c}^{2} } \\ \large \displaystyle{25 = c}$

Thus, the hypotenuse side is 25. Using the cosine ratio:-

$\large \displaystyle{cos \theta = \frac{adjacent}{hypotenuse}}$

Therefore:-

$\large \displaystyle{cos \theta = \frac{24}{25} }$

Because cos<0 in Q3.

$\large \displaystyle \boxed{ \red{cos \theta = - \frac{24}{25} }}$

Hence, the value of cos theta is -24/25.

Let me know if you have any questions!

8 0

3 years ago