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If you look at the powers of 10, you'll see that \(10^4<28471<10^5\) so a+b=9

:D

We can set a to be the first term and r to be the ratio.

Using the infinite geometric series formula \(S=a/(1-r)\) where S is the sum, we can plug in values to get \(80=a/(7/8)\) so a=70.

:D

A rectangle contains two circles of radius 24 and a circle of radius 16, as shown below.  Each circle is tangent to the other two circles.  Find the area of the rectangle.

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Rectangle length      L = 24 * 4 = 96

Rectangle width       W = sqrt(40² - 24²) + 24 + 16 = 72

Rectangle area        A = 96 * 72 = 6912 

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There're 3 different answers to this question!?! I understand that there could be more than one solution, but there must be only one correct answer!!! 

Promotion A - pair 1 - 30 pair 2 - 1/2*30=15 

30+15=45

Promotion B - Pair 1 - 30 pair 2 - 30-10=20 30+20=50

Promotion A will save her 5 dollars.

You're Welcome!

Find the positive value of c such that the area enclosed by the graph of y = -x^2 +c^2 and the x-axis is 36 .

Hello Guest!

\(f(x)=y=-x^2+c^2=0\\ x^2=c^2\\ \color{blue}x=\pm\ c\)

\(\int _{-c}^{+c}(-x^2+c^2)dx=\color{blue}|c^2x-\frac{x^3}{3}|_{-c}^{+c}=36\)

\(|c^2x-\frac{x^3}{3} |_{-c}^{+c}= (c^2\cdot c-\frac{c^3}{3}) - (c^2\cdot (-c)-\frac{(-c)^3}{3}) =36\)

\(c^3-\frac{c^3}{3}-(-c^3-\frac{(-c)^3}{3})=36\\ c^3-\frac{c^3}{3}+c^3-\frac{c^3}{3}=36\\ 3c^3-c^3+3c^3-c^3=3\cdot 36\\ 4c^3=108\\ c=\sqrt[3]{\frac{108}{4}}=\sqrt[3]{27}\)

\(c=3\)

  !

So first you can tell that each term after the first is mutiplied by \(1\over14\) because \(1\over7\)\(\div\)2=\(1\over14\)

Thus you can do \(2 \times \)\(1\over14\)⁴

You get \(2\times\)\(1\over38416\)

Simplified, it gives you \(1\over19208\)

Thank you so much for the explanation.

The answer is \(64\pi\).

So the area of the smaller circle is \(36\pi\) because \(6^2\pi\)

The area of the larger circle is \(100\pi \)

Thus, \(100\pi-36\pi=64\pi\)

One hundred twenty one.

Answer: 182/243

Explanation:

The common ratio is 1/3, because 1/2 * 1/3 = 1/6.

So listing them out, we get 1/2, 1/6, 1/18, 1/54, 1/162, 1/486. Adding these up gets us 364/486 which simplifies to 182/243.

You are very welcome!

:P

A trapezoid inscribed in a circle is isosceles.

Therefore, EH = GF, and arc (EH) = arc (GF).

EH is x^2-2x and FG is 48-4x.

So, x^2-2x = 48-4x => x^2+2x-48 = 0.

Factoring, we get (x+8)(x-6) = 0.

Since x = -8 wouldn't make sense, we are left with x = 6.

EH = 24, GH = 70, GF = 24.

EPF = 360-24-70-24= 212.

So I think the answer is 212.

I hope this helps!

Good job Guest! I could not have explained it better!

-Grace

You can use the technique of completing the square, so that the possible values of the box are "10, -10".

Sum of an infinite series =    first term / ( 1  -common ratio )

2009  =  328/ (1  -r)       rearrange as

2009 (1 -r)   = 328

2009  - 2009r =  328

2009 - 328  =  2009r

1681  = 2009 r

r   =   1681/2009   = 41/49

TYSM CPHILL that was helpful and correct :D

I forgot about that dang

If the second side has length x,

Then the first side has length 3x,

We know that the third side which has length 17.

Now, by the triangle inequality, we have:

x + 17 > 3x

=> 2x < 17

=> x < 8.5

Since we want the greatest perimeter,so we need the greatest integer x, and if x < 8.5.

Then x = 8

Then, the first side has length -> 3 x 8 = 24

The second side has length = 8 cm

The third side has length = 17 cm

So,perimeter = 17 + 24 + 8 = 49 cm

Cross-multiplication gives 

\(x^2+2x+3=(x+4)(x+5)=x^2+9x+20. \)

Therefore

\(0=7x+17\)

and \(x=\boxed{-\frac{17}7}\).

We  have    x^4 + bx^3  + cx^2   + dx + e

We  have  the following system......

1 + b + c + d + e   =  1    ⇒   b + c + d + e   = 0

2^4  + 8b + 4c + 2d + e  =  2   ⇒   8b + 4c + 2d + e  =  -14

3^4  + 27b + 9c + 3d + e  =  3  ⇒   27b + 9c + 3d + e  =  -78

4^4  + 64b + 16c + 4d + e  =  4 ⇒   64b + 16c + 4d + e  =  -252

b + c + d + e   = 0

 8b + 4c + 2d + e  =  -14

 27b + 9c + 3d + e  =  -78

 64b + 16c + 4d + e  =  -252

This isn't  difficult to solve, but it is tedious....I've  used some tecnology  to find the  values  of b,c,d,and e

b = -10   c   = 35    d   = -49   e  = 24

So....the polynomial is

x^4  -10x^3 + 35x^2 -49x  + 24

f(5)  =    5^4  - 10(5)^3  + 35(5)^2  - 49(5) + 24    =   29

First find the ratios of animals. The ratio of hamsters to guinea pigs to rats is 9:6:4.

This is simplified, so all we have to do is add the values.

9+6+4 is 19.

Take an example, say lev is 12 and mina is 24.

Then naomi would be 24*4/3=32 yo.

The ratio  is 12:24:32 when simplified, is 3:6:8.

Here's  my attempt

See the following image  :

There are two regions....A  and B

The diagonal line  y = x   comtains all the values  where  the two values chosen are equal

Region B contains all the  points where the first  coordinate is greater than the  second

In region   A, the second coordinate of every point  is greater  than the first coordinate.....so  the probability of landing in this area =  1/2

Let   (0,0)  be the  center  of the  circle

The lower right  vertex of the  rectangle  is   ( r -2, r - 1  )

So the distance   from  the  center  to  this point  is   r^2....so  we have

r^2 = ( r-2)^2  + (r-1)^2

r^2  = r^2  - 4r + 4  + r^2 - 2r + 1

r^2 - 6r   +  5   =   0    factor

(r - 5) ( r - 1)  = 0

The  first factor set  =  0     gifes us the radius

r - 5   =   0 

r  = 5 (cm)

THX, mobro....here's a slightly easier way

coomon ratio   = 1/5          a  = first term   = 1

Sum    =    a / (1 - r)  =  1  /( 1 -1/5)  =  1 / (4/5)  = 5/4