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3 years ago

Select the closest answer. *

Mathematics

1 answer:

Eduardwww [97]3 years ago

8 0

Answer:

D. 221.7 cm

Step-by-step explanation:

Surface area of a sphere is: $SA=4\pi r^2$

'r' - radius

We are given the radius of 4.2 centimetres.

$SA=4\pi 4.2^2\\4 * \pi * 4.2^2\\\rightarrow 4.2^2 =17.64\\\text {Using 3.14 for pi: }\\4 * 3.14 *17.64\\12.56*17.64\\\boxed {221.5584}$

221.5584 ≈ 221.6

The closest answer is Option D, therefore it should be the correct answer.

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Answer:

Probability of getting at least one 2 equals 0.5177

Step-by-step explanation:

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3 years ago

PLEASE HELP!!! BRAINLIEST TO CORRECT COMPLETE ANSWER!

notka56 [123]

Answer:

y=16x+1

Step-by-step explanation:

You want to find the equation for a line that passes through the two points:

(1,17) and (2,33).

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

First, let's find what m is, the slope of the line...

The slope of a line is a measure of how fast the line "goes up" or "goes down". A large slope means the line goes up or down really fast (a very steep line). Small slopes means the line isn't very steep. A slope of zero means the line has no steepness at all; it is perfectly horizontal.

For lines like these, the slope is always defined as "the change in y over the change in x" or, in equation form:

So what we need now are the two points you gave that the line passes through. Let's call the first point you gave, (1,17), point #1, so the x and y numbers given will be called x1 and y1. Or, x1=1 and y1=17.

Also, let's call the second point you gave, (2,33), point #2, so the x and y numbers here will be called x2 and y2. Or, x2=2 and y2=33.

Now, just plug the numbers into the formula for m above, like this:

33 - 17/ 2 - 1

or...

m= 16/ 1

or...

m=16

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=16x+b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(1,17). When x of the line is 1, y of the line must be 17.

(2,33). When x of the line is 2, y of the line must be 33.

Because you said the line passes through each one of these two points, right?

Now, look at our line's equation so far: y=16x+b. b is what we want, the 16 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (1,17) and (2,33).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!.

You can use either (x,y) point you want..the answer will be the same:

(1,17). y=mx+b or 17=16 × 1+b, or solving for b: b=17-(16)(1). b=1.

(2,33). y=mx+b or 33=16 × 2+b, or solving for b: b=33-(16)(2). b=1.

See! In both cases we got the same value for b. And this completes our problem.

The equation of the line that passes through the points

(1,17) and (2,33) is y=16x+1

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2 years ago

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.

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1 year ago

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