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3 years ago

Find the value when x= 2 and y= 3. x^-3y^-3

Mathematics

1 answer:

Bas_tet [7]3 years ago

4 0

Answer:

1/216

Step-by-step explanation:

x^(-3)y^(-3)

(2)^(-3)=1/2^3=1/8

(3)^(-3)=1/3^3=1/27

(1/8)(1/27)=1/216

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I need help with this

Stells [14]

Answer:

AB = 21

Step-by-step explanation:

So we have two triangles (AEB and ADC), and they're similar by AA

Now, you can find the ratio of similitude by checking AE/AD which is 14/26 = 7/13

AB/AC = 7/13

Take AB as x aight

x/x+18 = 7/13

x=21

5 0

3 years ago

PLEASE HELP ME I GIVE BRAINLIEST

mariarad [96]

Decimal Form: 11.5

Fraction Form: 11 1/2

7 0

2 years ago

Read 2 more answers

Which ratio is equivalent to the rate<br> 8 meters<br> 1 second<br> ?

KIM [24]

A unit rate where one is in the Denometer, so you would want to divide the bottom by 10.

5 0

2 years ago

Yall im almost done pls HELP

S_A_V [24]

Answer:

61 degrees

Step-by-step explanation:

==>Given ∆MNO,

MO = 18,

MN = 6

m<O = 17°

==>Required:

Measure of <N

==>SOLUTION:

Use the sine formula for finding measure of angles which is given as: Sine A/a = Sine B/b = Sine C/c

Where,

Sine A = 17°

a = 6

Sine B = N

b = 18

Thus,

sin(17)/6 = sin(N)/18

Cross multiply

sin(17)*18 = sin(N)*6

0.2924*18 = 6*sin(N)

5.2632 = 6*sin(N)

Divide both sides by 6

5.2632/6 = sin(N)

0.8772 = sin(N)

sin(N) = 0.8772

N = sin^-1(0.8772)

N ≈ 61° (approximated)

8 0

3 years ago

Considering only the values of β for which sinβtanβsecβcotβ is defined, which of the following expressions is equivalent to sinβ

-Dominant- [34]

Answer:

$\tan(\beta)$

Step-by-step explanation:

For many of these identities, it is helpful to convert everything to sine and cosine, see what cancels, and then work to build out to something. If you have options that you're building toward, aim toward one of them.

${\tan(\theta)}={\dfrac{\sin(\theta)}{\cos(\theta)}$ and ${\sec(\theta)}={\dfrac{1}{\cos(\theta)}$

Recall the following reciprocal identity:

$\cot(\theta)=\dfrac{1}{\tan(\theta)}=\dfrac{1}{ \left ( \dfrac{\sin(\theta)}{\cos(\theta)} \right )} =\dfrac{\cos(\theta)}{\sin(\theta)}$

So, the original expression can be written in terms of only sines and cosines:

$\sin(\beta)\tan(\beta)\sec(\beta)\cot(\beta)$

$\sin(\beta) * \dfrac{\sin(\beta) }{\cos(\beta) } * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) } {\sin(\beta) }$

$\sin(\beta) * \dfrac{\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} * \dfrac{1 }{\cos(\beta) } * \dfrac{\cos(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}} {\sin(\beta) \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{---}}$

$\sin(\beta) *\dfrac{1 }{\cos(\beta) }$

$\dfrac{\sin(\beta)}{\cos(\beta) }$

Working toward one of the answers provided, this is the tangent function.

The one caveat is that the original expression also was undefined for values of beta that caused the sine function to be zero, whereas this simplified function is only undefined for values of beta where the cosine is equal to zero. However, the questions states that we are only considering values for which the original expression is defined, so, excluding those values of beta, the original expression is equivalent to $\tan(\beta)$ .

8 0

2 years ago