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150   = ( 1 + x) ^4         take the  positive/negative 4th root

± (150)^(1/4)  =  1 + x    subtract 1 from both sides

± (150)^(1/4) -  1   =  x

The two (real)  answers  are

x = (150)^(1/4) -  1  ≈ 2.499

Or

x  = - (150)^(1/4) -  1  ≈  -4.499

Since CB is tangent to the circle at B, it  will have a negative reciprocal slope to radius AB

Slope AB  =   [ 4 -2] / [ 6 - 3 ] =    2/3

So.....the slope of BC  =    -3/2

It’s not that basic, Geno. There are two solutions.

The formula to find the area of a regular polygon given a radius is:  Area  =  [ r² · n · sin( 360 / n ) ] / 2 

where  r = radius  and  n = number of sides.

Area  =   [ 5² · 30 · sin( 360 / 30 ) ] / 2  =  78  (approximately)

It is 1/3 times the area of the circle times the perpendicular height.

150  =  (1 + x)⁴

Take the fourth root of both sides; the fourth root of 150 is approximately 3.4996.

3.4996  =  1 + x

Subtract 1     --->     x  =  2.4996     (approximately)

This isn’t advanced algebra. It’s basic algebra.

1a) Finding f(2) means that you have to substitute 2 for the variable x. So,

      when you substitute, the equation becomes f(2)=-3(2)+7. Then just do the math:

f(2)=-6+7, and then: f(2)=1. Voila!

1b) Just do the same thing: substitute for x, then do the math.

Um, no, CPhill is correct, unless the problem was mistyped...

Nope, but the * sign will

I'm pretty sure it won't, but the * will come up as a multiplication sign (on a keyboard, it's above the 8).

0.00408163257

Oh wow, I feel really dumb! My real problem was that this is a final study guide and I forgot that non-right angled triangles have law of cosine/sine instead of just using regular trigonometry. It's been a long year, thank you so much for your help!

201.0619298297467673, which is approximately equal to 201.1

First you need to draw the trinagle/s.  Do you now how to do that?

Question 1) In triangle CDE, the measure of angle C is equal to 52, the measure of angle D is equal to 17, and e is equal to 28.6. Find c to the nearest tenth.

You will need to find angle E and then use the sine rule.

Question 2) In triangle PQR, p is equal to 56, r is equal to 17, and the measure of angle Q is equal to 110. Find q to the nearest tenth.
Draw the pic and decide if you can use the sine rule or the cosine rule.

you do not need to find anything else first.  

REmember:

The sine rule uses two sides and their opposite angles.  You need to have three of these and then you can find the 4th one.

The cosine rule uses  3 sides and 1 angle.  You must have three of these and then you can find the 4th.

Perhaps your real problems is that you do not know how to label / draw the triangles.

Do you need me to help you with that?

Here is a video for you to watch :) 

so like 360 degrees multiplied by 0.11

Hey Melody, how do you enter aswers in that format with fractions and exponents?

Since it decided to delete some of my numbers:

Question 1) In triangle CDE, the measure of angle C is equal to 52, the measure of angle D is equal to 17, and e is equal to 28.6. Find c to the nearest tenth.

Question 2) In triangle PQR, p is equal to 56, r is equal to 17, and the measure of angle Q is equal to 110. Find q to the nearest tenth.
Don't know what happened with the original post, sorry about that!

We have a ghost on this site. "The ghost of web2"

He has been playing games again :)

whats 50% times 30 and why?

50% is 50/100 which simplifies to 1/2

half of 30 is 15

That's weird, I had all the numbers in there? I'll edit it to put them back in I guess I've never seen it do that

Your question/s are incomplete.

(7t^5+14t^4-28t^3+35t^2)/7t

\(\frac{(7t^5+14t^4-28t^3+35t^2)}{7t}\\ =\frac{7t^5}{7t}+ \frac{14t^4 }{7t}+ \frac{-28t^3 }{7t}+ \frac{35t^2} {7t} \\ = t^4 + 2t^3-4t^2+5t\)