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∑]1 / 7^n, n, 1, ∞]==It converges to 1/6

a/b ==1 / 6

a + b ==1 + 6 ==7

The smallest integer is -17.

The common ratio is 11/3.

Find the equation of the line that is tangent to both of the parabolas y^2 = 4x and x^2 = -20y.

Hello Guest!

\(f(x)=y=\pm2\sqrt{x}\\ f'(x)=\pm\ x^{-\frac{1}{2}}\\ g(x)=y=-\frac{x^2}{20}\\ g'(x)=-\frac{x}{10}\\ f'(x)=g'(x)\\ -\frac{2x}{10}=\pm \ x^{-\frac{1}{2}}\\ \color{blue}x = \pm 4,641588833408518\\ © Arndt\ Br\ddot unner\)

\(y_f=2\cdot \sqrt{4,641588833408518}=4.30886937996\\ P_f(4,641588833408518,4.30886937996)\\ y_g=-\frac{x^2}{20}=-1.07721734492\\ P_g(-4,641588833408518,-1.07721734492)\)

\(m=\dfrac{y_f-y_g}{x_f-x_g}=\dfrac{4.30886937996-(-1.07721734492)}{4,641588833408518-(-4,641588833408518)}\\ \color{blue}m=0.580198604205\)

Point-direction equation of the straight line.

\(y=m(x-x_1)+y_1\\ y=0.580198604205(x-4,641588833408518)+4.30886937996\\ The\ equation\ of\ the\ line\ that\ is\ tangent\ to\ both\ of\ the\ parabolas\\ y^2 = 4x\\ and\\ x^2 = -20y\\ is\\\color{blue}y=0.5802x+1.692\)

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\((e_1, e_2) = (4,6)\)

How many fewer hats did Jake collect than Troy?

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\(t=4j\\ d=2j\\ t+j+d=21\\ 4j+j+2j=21\\ \color{blue}j=3\\ t=12\\ \color{blue}t-j=12-3=9 \)

9 fewer hats did Jake collect than Troy.

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How many baseball cards did each have at first?

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\(j=n\\ j-49=5(n-85)\\ n-49=5n-425\\ 4n=425-49\\ n=94\\ j=94\)

94 baseball cards did each have at first.

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lol you forgot 8*7*-8

p^2+4p+4

There are 9 choices for the hundreds digit. 

The only possible cases for the tens and units digits are: (0, 5 - 9), (1, 7 - 9), (2, 9)

This makes for 9 cases for the units and tens digits. 

Thus, there are \(9 \times 9 = \color{brown}\boxed{81}\) cases. 

We have: \(3a + 2 = b + 3\)

Rearrange as: \(3a = b + 1\)

We want a to be as small as possible, so \(a = 4\), meaning \(b = 11\)

This means that the sum is \(4 + 11 = \color{brown}\boxed{15}\)

Rearrange as: \( (y^2 + 2y) +(x^2 +2x+ 2xy) + 4 \)

Consider this as 2 separate equations (\(x^2 +2x+ 2xy\) and \(y^2 + 2y\)) and a constant. Our goal is to minimize the equations.

Start with \(y^2 + 2y\). The minimum value of this is -1, and it occurs when \(y = {-b \over 2a} = {-2 \over 2} = -1\) 

Now, subbing in \(y = -1\) into the second equation gives us \(x^2 + 2x - 2x = x^2\). 

The minimum value of this, quite obviously, is 0, and it occurs when \(x = 0\).

This means the minimum value is: \(0 - 1 + 4 = \color{brown}\boxed{3}\)

Just multiply the three negative numbers together to get your answer

Here :  https://web2.0calc.com/questions/help-with-geometry_86

Find t if the expansion of the product of x^3 - 4x^2 + 5x - 5 and x^2 + tx - 7 has no x^2 term.

what are the x^2 terms

-4x^2*( -7)    +    5x(tx)   +    -5(x^2)

so what is the coefficient of x^2 in terms of t  ?

(37 + 20sqrt (3)) (37 + 20 sqrt (3) )  =

1369  + 2*37*20sqrt (3)  +  400 * 3  =

1369 + 1480sqrt (3) + 1200  =

2569 + 1480sqrt (3)

(a, b) =   (2569 , 1480)

Usig the square of the distance formula

(2a - 4)^2  + (a - 4 - 1)^2 =   (2sqrt (10) ) ^2

4a^2 - 16a + 16  +  (a - 5)^2  =  4*10

4a^2 - 16a + 16 + a^2 - 10a + 25 = 40   simplify

5a^2 - 26a  + 1  =  0

By Vieta.....the product of all possible values for a = 

1 / 5  

.30 * 2  =  .6 liter is acid

.20 * 3 =  .6 liter is acid

.10 * 5  =  .5 liter is acid

% total acid =      ( .6 + . 6 + .5 ] ( 2 + 3 + 5)  =   1.7  /  10 = 17 /100 =      17%

tan A + sec A =   3            {  tan^2 A + 1  =  sec^2 A  →    tan^2 A = sec^2 A - 1  → tan A = sqrt [ sec^2 A - 1]  }

sqrt [ sec^2A -1 ] + sec A  = 3

sqrt [sec^2A  - 1 ]  = 3  - sec A             square both sides

sec^2 A   - 1  =  9 - 6sec A + sec^2 A     rearrange as

6sec  A  =  10

sec  A  = 10   / 6     =    5/ 3

cos A  =  1 /sec A  =  1  /(5/3) =   3 / 5

2sin A =  5 cos A     rearrange as

sin A / cos A  = 5 / 2

r =   sqrt ( 5^2 + 2^2) =  sqrt (29)

sin A  =  y / r   =      5 / sqrt (29)

The largest  angle is opposite the longest side

Let A  be that angle

Using the Law of Cosines

8^2    = 3^2 + 7^2  - 2 (3)(8) cos A

cos A   =    [  8^2 - 3^2 - 7^2 ] / [ -2 * 3 * 8 ] 

arccos  [ ( 8^2 - 3^2 - 7^2 )  / ( -2 * 3 * 8 )  ]   = A   =    97.18°  

shorter side^2  +  longer side^2 = hyp ^2

x^2 + (x + 8)^2  = (x + 18)^2

x^2 + x^2 + 16x + 64 =  x^2  +  36x  + 324    simplify

x^2 - 20x - 260  =  0  

Using the Q formula

                    20  + sqrt  [ 20^2  - 4 (1) (-260 ]                 20  + sqrt [ 1440]

x =            ____________________________  =       _______________  =

                                         2 * 1                                               2

20  +  12 sqrt (10)

______________  =        10  + 6sqrt (10)

          2

The sides are

10 + 6sqrt (10)

18 + 6 sqrt (10)

28 + 6sqrt (10)

A8 ==98 

3B ==35

98  x  35 ==3,430

A==9,   B==5,    C==3

The right triangle   has a base = 19  - 13   = 6

By the Pythagorean Theorem, the height =  sqrt (10^2 - 6^2) =  sqrt (100 - 36)  = sqrt (64)  = 8

This is also the height of the  rectangle on the left

So....the perimeter =   13 + 8 + 19 + 10  =    50