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You have a 0.2 probability 35 times in a row.

\( 0.2^{35} = 3.436*10^{-25}\)

This is almost zero.

Not a good way to take a test

Since  two out of three in mrs. Karopkin's class own a cell phone, that means that 2/3 of her students own a cell phone. Therefore if we use the same ratio for the entire school of 150 students, then:

2/3 X 150=100 students will own a cell phone.

2/3 * 150   = 100 students in total own a cell phone

Because you have tow sides, use Pythagoras's Theorem:

Broken part=Hypotenuse

H^2=19^2 + 70.908^2

H^2=5,388.94 take the sqrt of both sides

H=73.41 feet-Length of the broken portion.

Sorry, my software can handle it!. It gives these very limited results:

lim_(x->0)  integral_0^2 e^x/x dx = (integral does not converge)

lim_(x->0^+)  integral_0^2 e^x/x dx =  integral_0^2 e^x/x dx(one-sided limit)

Draw  AO, BO, AB and  BP.......label the intersection of TO and AB as point F

With the above additions, I believe the image below is what you describe :

By SSS, triangle ATO congruent to triangle BTO.....thus angle ATO = angle BTO

Thus, by SAS, triangle ATF congruent to triangle BTF

Thus, FA  = FB

And a radius bisecting a chord does so at right angles... so, OP is perpendicular to AB

Thus, by LL, triangle PAF congruent to triangle PBF

Thus angle PAF  = angle PBF

And, by Euclid, m< TAP  = 1/2 minor arc PA

But, angle PBF =  angle PBA .....and PBA is an inscribed angle = 1/2 minor arc PA

But, it was shown that angle PBF = angle PAF......thus, angle PAF also = 1/2 minor arc PA

So.......m<TAP  = m<PAF = m<PAB

So....m<TAP = m<PAB, and AP bisects TAB

Solving for "a" only gives the broken flag pole length, so what would I do next to find the length of the broken portion?

You can online and use this very simple fractions calculator:

http://www.calculatorsoup.com/calculators/math/fractions.php

If 30, 30, 90 are SIDES, then you cannot use Pythagoras's Theorem, because:

90^2≠ 30^2 + 30^2

During a windstorm, part of the top of a flagpole breaks off. The broken top portion touches the ground at an angle of 75 deg 19 feet from its base.
 

Tan(75)=Height of broken pole/19

3.732=H/19

H=19 X 3.732

H=70.908 feet-Length of the pole before windstorm.

Compute the definite integral:
integral_0^1 x log(x+1) dx
For the integrand x log(x+1), substitute u = x+1 and  du =   dx.
This gives a new lower bound u = 1+0 = 1 and upper bound u = 1+1 = 2:
  =   integral_1^2 (u-1) log(u) du
Expanding the integrand (u-1) log(u) gives u log(u)-log(u):
  =   integral_1^2 (u log(u)-log(u)) du
Integrate the sum term by term and factor out constants:
  =   integral_1^2 u log(u) du- integral_1^2 log(u) du
For the integrand u log(u), integrate by parts,  integral f dg = f g- integral g df, where
f = log(u),     dg = u  du,
df = 1/u  du,     g = u^2/2:
  =  1/2 u^2 log(u)|_1^2-1/2 integral_1^2 u du- integral_1^2 log(u) du
Evaluate the antiderivative at the limits and subtract.
1/2 u^2 log(u)|_1^2 = 1/2 2^2 log(2)-1/2 1^2 log(1) = log(4):
  =  log(4)-1/2 integral_1^2 u du- integral_1^2 log(u) du
Apply the fundamental theorem of calculus.
The antiderivative of u is u^2/2:
  =  log(4)+(-u^2/4)|_1^2- integral_1^2 log(u) du
Evaluate the antiderivative at the limits and subtract.
(-u^2/4)|_1^2 = (-2^2/4)-(-1^2/4) = -3/4:
  =  -3/4+log(4)- integral_1^2 log(u) du
For the integrand log(u), integrate by parts,  integral f dg = f g- integral g df, where
f = log(u),     dg =   du,
df = 1/u  du,     g = u:
  =  -3/4+log(4)+(-u log(u))|_1^2+ integral_1^2 1 du
Evaluate the antiderivative at the limits and subtract.
(-u log(u))|_1^2 = (-(2 log(2)))-(-log(1)) = -log(4):
  =  -3/4+ integral_1^2 1 du
Apply the fundamental theorem of calculus.
The antiderivative of 1 is u:
  =  -3/4+u|_1^2
Evaluate the antiderivative at the limits and subtract.
u|_1^2 = 2-1 = 1:
Answer: | =  1/4

You have to have at least ONE ANGLE and ONE SIDE.

Multiply all terms together and then take the nth root of their product, n being the number of terms: Example: 2 X 3 X 4 X 5 X 6 X 7 X 8 X 9=362,880

Now, because you have 8 terms, you take the 8th root of their product thus:

(362,880)^(1/8)=4.95416..........etc. That is your geometric mean.

Solve for y:
3 y+3 (2 y-4) = y

3 (2 y-4) = 6 y-12:
6 y-12+3 y = y

Grouping like terms, 6 y+3 y-12 = (3 y+6 y)-12:
(3 y+6 y)-12 = y

3 y+6 y = 9 y:
Answer: | 9y-12 = y=3/2

Not to be mean but i'm 100% sure that that is not right

You would start by getting off of this site so that people who would like to use it for what it's meant to be used for can do so. The point of this site is for us to gain knowledge, not for you to drive it out of our heads with your stupid questions. :)

What is the square root of your mum=square root of your mum!!

5*35=175

   [0.3+0.5i]°  ????????

A initial investment of $10,000 grows at 11% per year. What function represents the value of the investment after t years

f(t)=10,000 X 1.11^t

Volume=π.r^2.h

V=3.141592 X 14^2 X 7.5

V=4,618.14 cubic units.

Write the following as an algebraic expression. Then simplify.

The perimeter of the square with side legth y.

The answer is: 4y

That is it!.

Judy wants to make a small patio in her garden. She plans to use pavers. The area of the patio is 2p2 + 30p + 108 square meters, and the area of each paver is p2 + 5p − 6 square meters. The number of pavers Judy will need is given by the expression:

Hi Diva, 

Welcome to the forum :)

Do you have any idea how to start this?

I'll hep :)

Say the area of the pavement was 20m^2 and the pavers were 4m^2 then how many pavers would you need?

How did you put 20 and 4 together to get this answer?

If you know how to do it with little numbers then you know what you are supposed to do with the rediculously complicated expressions that you have there.

SO

What do you have to do ?   and then.........How might you be able to do it?

A commercial jet liner hits an air pocket and drops 277 feet. After climbing 175 ​feet, it drops another 105 feet. What is its overall vertical​ change?

Unless there is more to it than I think, but this is what think:

-277 + 175 - 105=-207 feet drop before it hit the air pocket, which is the net vertical change.

3k>8k-10

8k-3k<10

5k<10 divide both sides by 5

k<2