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2 years ago

6

What's 19 to the power of 63? Brain fart-

2 answers:

iVinArrow [24]2 years ago

6 0

Answer:

2.5782963e+29

Step-by-step explanation:

basically 19 times itself 63 times

sertanlavr [38]2 years ago

6 0

Answer:

19^63 , 3.6431483821774E+80

Step-by-step explanation:

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This is what we call a ratio with unlike quantities

astra-53 [7]

It is called a rate.

Your Welcome

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3 years ago

If (-1, y) lies on the graph of y=2^2x then y=

zubka84 [21]

Answer:

y is 1/4 when x is -1 if I did interpret your equation right. Please see that I did.

I believe the equation to be $y=2^{2x}$ . If you meant $y=2^2x$ , please let me know.

Thanks kindly.

Step-by-step explanation:

If (-1,y) lies on the graph of $y=2^{2x}$ , then by substitution we have:

$y=2^{2\cdot -1}$ .

We just need to simplify to determine y:

$y=2^{2 \cdot -1}$

Multiply the 2 and -1:

$y=2^{-2}$

Use reciprocal rule for exponents to get rid of the negative:

$y=\frac{1}{2^2}$

$2^2$ just means $2(2)$ :

$y=\frac{1}{2(2)}$

$y=\frac{1}{4}$ .

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2 years ago

How many faces does the solid made from the net below have?<br>

nikdorinn [45]

Answer:We need to see the net.

Step-by-step explanation:

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3 years ago

Directions: Find the indicated trigonometric ratio as a fraction in simplest form. 1. Sin L =

Rainbow [258]

Answer:

$\sin L = 0.60$

$tan\ N = 1.33$

$\cos L = 0.80$

$\sin N = 0.80$

Step-by-step explanation:

Given

See attachment

From the attachment, we have:

$MN = 6$

$LN = 10$

First, we need to calculate length LM,

Using Pythagoras theorem:

$LN^2 = MN^2 + LM^2$

$10^2 = 6^2 + LM^2$

$100 = 36 + LM^2$

Collect Like Terms

$LM^2 = 100 - 36$

$LM^2 = 64$

$LM = 8$

Solving (a): $\sin L$

$\sin L = \frac{Opposite}{Hypotenuse}$

$\sin L = \frac{MN}{LN}$

Substitute values for MN and LN

$\sin L = \frac{6}{10}$

$\sin L = 0.60$

Solving (b): $tan\ N$

$tan\ N = \frac{Opposite}{Adjacent}$

$tan\ N = \frac{LM}{MN}$

Substitute values for LM and MN

$tan\ N = \frac{8}{6}$

$tan\ N = 1.33$

Solving (c): $\cos L$

$\cos L = \frac{Adjacent}{Hypotenuse}$

$\cos L = \frac{LM}{LN}$

Substitute values for LN and LM

$\cos L = \frac{8}{10}$

$\cos L = 0.80$

Solving (d): $\sin N$

$\sin N = \frac{Opposite}{Hypotenuse}$

$\sin N = \frac{LM}{LN}$

Substitute values for LM and LN

$\sin N = \frac{8}{10}$

$\sin N = 0.80$

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2 years ago

What is the measure of angle G?

Allushta [10]

Answer:

28

Step-by-step explanation:

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2 years ago