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2 years ago

5

Give me a good explanation please! I don't really understand what to do when exponents aren't the same xD

2 answers:

ohaa [14]2 years ago

7 0

Answer:

1.5

Step-by-step explanation:

First I convert it using the negative exponent rule

so 10^-7 becomes

1

------

10^7

(Dont ask how the negative exponent rule works i have no clue to be honest)

Now we have to do the 10^7

10^7 = 10000000

So the fraction is now

1

------------------------

10000000

What we have above equals 0.00000010

So we got that part now we have to do

5 * 0.00000010 = 0.00000050

So now we have (0.00000050) x ( 3 x 10^6)

Its fairly easy from here

first we do

10^6 = 1000000

then we do

3x1000000 = 3000000

Now we have

(0.00000050) * (3000000)

The parentheses are useless at this point so we can just remove them

0.00000050 * 3000000 which gets us to.. 1.5

I hope this is better than my last answer lol.. this hurt my head

nevsk [136]2 years ago

6 0

$(5*10^{-7} )*(3*10^{6} ) = (5*3) *(10^{-7} *10^{6} ) = 15*(10^{(-7) + 6} ) = 15 * 10^{-1} =0.15*10^{1}$

Hope that helps!

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What number equals 2 4/5?<br> O 0.28<br> O 0.357<br> O 2.45<br> O 2.8

Travka [436]

I think its 2.8.

Reason why:

4/5 = 0.8

Then the 2 would a whole number...

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i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

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