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3 years ago

8,330,000,000 written in scientific notation?

2 answers:

6 0

The answer should be $8.33 * 10 ^{9}$ .

All you have to do is to convert constants into decimals and then count the number of zeroes and decimals.

Lubov Fominskaja [6]3 years ago

5 0

8.33x10^9 is the answer I believe

You might be interested in

Find the greatest common factor of 14 and 33.

Paraphin [41]

Answer:

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

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

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago

First person to comment gets brainliest

bearhunter [10]

Answer:

Step-by-step explanation:

6 0

2 years ago

2x − 5y = 16 <br>3x + 2y = 5 <br>can you also explain how you did it. Thanks :)

Reptile [31]

Answer:

x=3

y=-2

Step-by-step explanation:

2x - 5y = 16

3x + 2y = 5

Solve for x in the first equation.

2x - 5y = 16

2x = 16 + 5y

x = (16 + 5y)/2

x = 8 + 5/2y

Put x as 8 + 5/2y in the second equation and solve for y.

3(8+5/2y) + 2y = 5

24+15/2y + 2y = 5

24 + 19/2y = 5

19/2y = 5-24

19/2y = -19

y = -2

Put y as -2 in the first equation and solve for x.

2x - 5(-2) = 16

2x +10 = 16

2x = 16-10

2x = 6

x = 3

6 0

3 years ago

Read 2 more answers

Simplify (9 + 7i)/(-10 + 8i)

Vera_Pavlovna [14]

Multiply numerator and denominator by the conjugate of the denominator

Denominator: -10 + 8i
Conjugate of the denominator: -10 - 8i

Operation:

[9 +7i] / [-10 + 8i] * [-10 -8i] / [-10 - 8i] = [-90 - 72i - 70i +56] / [100 + 64] =

[-34 -142i] / 164 = [-17 -71i] / 82

3 0

3 years ago