All Answers 355975 Answers

We can use the POP formula. 3(3+5)=x(x+6) so 24=x^2+6x meaning x^2+6x-24=0. We can use the quadratic formula to get x=-6+-(sqrt132)/2 which means x=sqrt33-3 because the length can't be negative.

:D

You're Welcome!

I think you want  CE , not DC    (DC  = 3)

We  have  the secant-secant theorem

CE (CA)    =  CD (CB)

x ( 6 + x)  =  3 ( 3 + 5)

6x + x^2  =  3 * 8

x^2 + 6x  =  24    complete the square on x

Take (1/2) of 6 = 3  

Square it =  9   and add to both sides

x^2 + 6x  + 9 =  24 +  9

(x + 3)^2  =  33      take the  positive  root

x +  3   =sqrt (33)

x =  CE  =   sqrt (33)  -  3   ≈  2.7446

Thanks!! 

The answer is 9.

9.27/1.03=9

I just had some stray typing  at the bottom....the solution is  8sqrt(8)

(2)  Again.....I'm not totally sure...but....

I think the answer is  y =  sec x

Look at the  graph here : https://www.desmos.com/calculator/v7cuzaovi5

As we approach  0  from the left side....the slope is  negative   and  on the other  side of  0 the slope is positive

Also....the other clue is that  on each side of the asymptotes on the secant, the slopes occur in "pairs"  negative-negative  and positive-positive....notice that this occurs in the slope field, as well

Hope this is correct   !!!

(1).....really reaching back into my memory banks  here

I believe  that the answer is   y  = tan x +  C

Note  that the slopes shown in the  field  are always positive...and the slope  of the tangent is always positve (where defined).........also.....the other  cliue is that near 0,  the  slope of the tan  decreases   on  both sides of  0  and then increases dramatically   away from 0  to  pi/2  (and - pi/2)

Look at the graph here  :  https://www.desmos.com/calculator/uceau9xjx7

I hope that I haven't mis-lead you

I can't see parts of the solution CPhill.

So how would I do this one

The answer is \(50\over49\)

First you can write it as \(3\over2.94\)

If you mutiply the numerator and denominator by 100 you get \(300\over294\)

Simplifed you get \(50\over49\)

Draw  perpendiculars from the centers of  both  circles  to  a  side of the  triangle

Using similar triangles  we  have  that

(h - 5)  / 1  =  ( h - 2) /2

2 (h - 5) = 1 ( h -2)

2h  -10 =   h - 2

h = 8

Let the center of the top circle =  D   the apex  = A   and the  point where the radius of this circle meets the  side of the triangle, E....angle ACB   =  90.....and   AD = 3    DE  = 1

So

AE = sqrt ( AD^2 + DE^2)  =  sqrt( 3^2- 1^2)  = sqrt(8)

tan  DAE  =  DE/  AE   =  1/sqrt (8)

So

tan DAE  =  (base/2) / height

1/sqrt (8)   =base/2 * 8

8/sqrt (8)   =  base/2  =sqrt (8)

So   [ ABC ]    =   ( base / 2)  * height   =  8 sqrt (8)

The answer is 106.

Reason

Triangle Exterior Angle Theorem

The exterior angle of a triangle is equal to the opposite interior angles.

So 140=34+x

x=106

LOL!

Okay so 90x9 equals to 810. Here is how I did it...

(90x10)-(90x1)=900-90=810

6x14 equals to 84. The steps are here...

(5x14)+(1x14)=70+14=84

Hope this helps : )

$\frac{a}{1-r} = 80$

$\frac{a}{7/8} = 80$

$a = 70$

ok I fixed it

The expression evaluates to 49.

I'm sorry Guest!

I can't seem to access the file, perhaps change sharing restrictions?

=^._.^=

Hi Guest!

Background

To solve this question, we should use the quadartic equation. 

If you don't know what that is, here's a wiki page: https://en.wikipedia.org/wiki/Quadratic_equation

Whether or not a quadratic has a real answer is based on it's discriminant, b^2 - 4ac. 

If the discriminant is negative, then the solutions won't work since you can't square root a negative number. 

Equation

6x - x^2 = k

x^2 - 6x + k = 0

Discriminant

b^2 - 4ac

(-6)^2 - 4(1)(k)

36 - 4k 

Solving

So, we're looking for the greatest k value where 36 - 4k is non-negative. 

The smallest non-negative number is 0. 

36 - 4k = 0

36 = 4k

k = 9

Answer

Thus, our answer is 9. 

I hope this helped. :))))

=^._.^=

$$12/3 = 4, 12/9 ≈ 1.$$

There are 5 powers of 3 in $12!$

$$15/3=5, 15/9 ≈1.$$

There are 6 powers of 3 in $15!$

$$18/3 = 6, 18/9 =2.$$

There are 8 powers of 3 in $18!$

You can likewise figure out that there are 9 in $21!$ and 10 in $24!.$ Thus, as there are 5 numbers, there will be no multiplying of 3 within non-3-divisible numbers, so the answer will relate to the sum $3^5 + 3^6 + 3^8 + 3^9 + 3^{10}.$ This is just $3^5(1 + 3 + 27 + 81 + 243,)$ or $3^5(355),$ so the answer is $\boxed{5}.$

We might also take the discriminant, b^2 - 4ac.

We have that (-a)^2 - 4 * 2016 * 1 > 1, so a^2 - 8064 > 0. Thus, a^2 > 8064, and a > ±√8064, so a> about 89.8 and a< about -89.8. However, we would like to find the integer solutions, which means a must be an integer, which means that a ≤ -90.

(By the way CPhill, you have a typo on the first line)

This problem is unanswerable, because you have not included a diagram.