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sergiy2304 [10]
3 years ago
8

Write the standard number for 800,000 + 4,000 + 600 + 40 + 2.

Mathematics
1 answer:
murzikaleks [220]3 years ago
4 0

Answer:

804,642

Step-by-step explanation:

800,000 + 4,000 + 600 + 40 + 2.

Add the terms

804,642

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Solve the inequality <br> x-2&gt;15
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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
Ratio is 4:7, smaller volume is 1, 331, find larger volume
viva [34]
If the smaller volume ratio is 4 to 1331, divide 1331 by 4 to get the multiplying factor.
1331/4 = 332.75
Now we multiply 332.75 by 7 to get our larger volume.
332.75 x 7 = 2329.25
2,329 is our larger volume.
I hope this helps!
3 0
3 years ago
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