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The answer is 32.

i dont think you are correct -_-'

https://web2.0calc.com/questions/help-algebra_31592

If we drew the graph for all the points 5 units away from (2, 7), we would get a circle with radius \(\sqrt5\) and center (2, 7).

A circle like that would have the equation \((x - 2)^2 + (y -7)^2 = 25\). 

Now, we just have to find the intersection points with the graph of \(y = 5x - 28\). 

We get a system of equations like this, and we just have to find (x, y). Luckily, y is already isolated in the second equation, so we just have to sub that into the first equation. We get

\((x -2)^2 + (5x -28 - 7)^2 = 25 \\ (x - 2)^2 + (5x - 35)^2 = 25\)

\(x^2 - 4x + 4 + 25x^2 - 350x + 1225 = 25 \\ 26x^2 - 354x + 1204 = 0\)

From this, we get\(x=86/13, 7\)

Plugging these back for y, we get \(y = 66/13, 7\). 

This means our two points are \((86/13, 66/13) & (7, 7)\)

Thanks! :) 

We can easily just graph out the numbers since there are not a lot of terms we have to go through.  

This means when Laverne says 12, Shirley goes off running!

Thanks! :)

Not too bad of a problem!

Let's let n be the constant we add to the 3 numbers. Lets try to figure out the 3 new numbers first. 

Since all terms form a geometric sequence, we can write

\(\frac{120+n}{60+n}=\frac{160+n}{120+n}\\ (120+n)^2=(60+n)(160+n)\\ 14400+n^2+240n=9600+n^2+220n\\ 4800=-20n\\ n=-240\)

Now, I know the n is negative, when if we apply it, we dot get -180, -120, and -80, which do form a geometric series. 

-180/-120 = 3/2, so 3/2 is the answer. 

I'm not sure if this is the correct answer, but it does work. 

Thanks! :) 

I'm going to do this the hard way and expand eveyrhing out. 

Expanding the awful polynomial, which I DEFINTELY NOT reccomending, we have

\(x^{7}-7x^{6}+10x^{5}-13x^{4}+8x^{2}-9x+14\). 

As you can seem the coefficient is just -9. 

If you want to simplify the process, just note this. 

In the expansion, the only two terms that matters is 1*-23x amd x*14x. Adding these 2 we get -9. 

Thanks! :) 

We can solve this by realizing that 50¢ can be represented as the sum of 32¢ and 18¢.

For the 32¢ portion, Bina has two 32¢ coins, so there's only 1 choice (use both 32¢ coins).

For the remaining 18¢, Bina can use:

One 16¢ coin and one 2¢ coin (2 choices)

Two 8¢ coins (1 choice)

Nine 2¢ coins (1 choice)

So in total, there are 1 (32¢) + 2 (16¢+2¢) + 1 (8¢+8¢) + 1 (2¢+2¢+...+2¢) = 5 different combinations that add up to 50¢.

To complete this problem, we must make an important observation.

NOTE: 

\(3 + 4 + 5 = 3 *4 \\ 13 + 14 + 15 = 3 * 14\)

Now, we just have to figure out \(3 [ 4 + 14 + 24 + ....... + 994 ] = \)

There are 100 terms in the sequence, so we have

\(3 [ 994 + 4 ] [ 100 / 2 ] = 149,700\). 

Thanks! :)

First, let's multiply it out. \(x^3(x^2+tx) = x^5+tx^4\). 

As you can see, there is no x^2 term in this expansion. 

This means t can be pretty much anything we want as long as it doesn't contain x^-2 in it!

Thanks! :)

Haven't bothered to count them but the path must start at C and  end at D   or vice-versa (beginning and  ending at vertexes with odd valences)

Let these two lines intersect a cirlce with a radius of 1

To find the point on y= x  that lies on the circle

x^2 + x^2  =1

2x^2 =1

x^2= 1/2

x =1/sqrt 2     and y= 1/sqrt 2

To find the point on y= 2x that lies on the circle

x^2 + (2x)^2 = 1

5x^2 =1

x^2 = 1/5

x = 1/sqrt5      y = 2/sqrt 5

The midpoint  between these two points is [  (1/sqrt 2 + 1/sqrt 5) / 2 , (1/sqrt 2 + 2/sqrt 5)/2 ] =

[ (sqrt 2 + sqrt 5)  / (2sqrt 10) ,  (2sqrt 2 + sqrt 5) / (2sqrt 10) ]

The slope of the line drawn from the origin through this point = m  =

[ 2sqrt 2 +sqrt 5 ] / [ sqrt 2 + sqrt 5] =  

[ sqrt 5 - sqrt 2] [ sqrt 5 + 2sqrt 2] / 3  =

[ 5 + sqrt 10 - 4 ] /  3  =

[ 1 + sqrt 10 ] /  3

First off, let's move all the terms to one side of the equation before combining like terms!

\(x^2 - 4x + y^2 - 2y = 0 \)

Completing the square for x and y, we get

\(x^2 - 4x + 4 + y^2 - 2y + 1 = 4 +1 \\ (x- 2)^2 + ( y - 1)^2 = 5\)

Wait! This is the equation for a circle. 

Using the equations for circles, this circle has a radius of \(\sqrt{5}\) and center (2, 1). 

The largest value of x would be \(x = 2 +\sqrt 5\).

The reason for this is because the largest value of x would just be the radius plus the x value of the center!

Thanks! :)

First off, let's move all terms to one side and combine all like terms. 

We get \(2x^2+20x+18=0\). Simplifying, we get \(x^2+10x+9 = 0\). 

Now, let's note that this quadratic can be factored! Putting it into factored form, we get \((x+1)(x+9)=0\). 

This gets us the values of \(x = -1, -9\). 

Check your answers and make sure they work. These do if we plug them back in!

Thanks! :)

After Dirk finally finishes the race, Edith has ran 14 km while Foley has ran 9 km. 

This essentially means that Foley is 9/14 as fast as Edith. 

This means when Edith has ran 24 km, Foley has ran \(24 (9/14) ≈ 15.43 \) km. 

So Foley is \(24 - 15.43 ≈ 8.57\)km behind Edith!

Thanks! :)

x^2 + (-4)^2  = 20

x^2 + 16 =20

x^2 = 4

x= 2    or x = -2

A= (-2,4)   B = (2,-4)

AB  =  [ 2 - -2 ]  = 4

Slope = (4 -  - 5)  / ( 5 - - 1)  = 9/6 =  3/2

Eq of line

y = (3/2) (x -5) + 4

2y = 3(x -5) + 8

2y = 3x -7

y = (3/2)x - 7/2

(3/2)x - y = 7/2           mult through by 2/7

(3/7)x - (2/7)y = 1

x / (7/3) - y / (7/2) = 1

r = .036 / .4  =  36 / 400 =  9 /100 =  .09

Sum =   .4 / [ 1 -.09] =  .4 / .91  = 40 /91

(x + 1 ) / (x - 1)  *   ( x + 2) / (x )  * ( x + 3) /( x +1)  = 10    simplify

(x + 2) (x + 3)  = 10 (x)(x -1)

x^2+ 5x + 6 = 10x^2 -10x

9x^2 - 15x - 6 =  0

3x^2 - 5x - 2  = 0

(3x +1) (x -2)  = 0 

x =  2

The first fraction is    3 / 1

                     A    E 75°

       B                  D                   C15°

              10                   10

Since ED is a  perpendicular bisector... BD / CD = BE / CE 

BD = CD   so   BE = CE   

CE  = DC / sin DEC

CE  = 10 / sin (75°)

CE = 10 / sin (30 + 45)

CE  =10 / [ sin30 cos 45 + sin45 cos 30 ] 

CE = 10 [ (1/2)(sqrt 2/2  + sqrt2/2sqrt 3/2]

CE =10 [ sqrt 2/4  + sqrt 6/4 ]

CE =  40 / [ sqrt 2 +sqrt 6]

CE = 40 [ sqrt 6 -sqrt 2 ] / [ 6 -2]

CE =  10 [ sqrt 6 -sqrt 2]  ≈ 10.35 =  BE

https://web2.0calc.com/questions/bisectors_18

https://web2.0calc.com/questions/triangle_49107#r1