All Answers 356462 Answers

We could do \(ab^4/ab^2=48/4\)

So the A will cancel out and we will be left with \(b^2=12\)

So the answer is \(2 \sqrt{3} \) or\(-2\sqrt{3}\)

(a + ab) + (a - ab) ==> 250 + (-240)

2a ==> 10

a=5

solve it from here by plugging in a=5 into the equation. 

I know this is stupid, but I'm kinda dumb, so here's the entire polynomial

(This took a while)

\(729y^{12}+2916y^{11}-2430y^{10}-19980y^{9}+135y^{8}+59976y^{7}+6364y^{6}-99960y^{5}+375y^{4}+92500y^{3}-18750y^{2}-37500y+15625\)

The answer is 375...

Thanks! :)

We can use subsitution to our advantage. 

Now, first, let's put a terms of b. From the second equation, we get that

\(a=\frac{4}{b^2}\)

Subsituting this value of x into the first equation, we get

\(\frac{4}{b^2} \cdot b^4 = 48\\\)

Now, the top and the bottom cancel out, so we have

\(4b^2 = 48\\ b^2=12\\ b=2\sqrt{3},\:b=-2\sqrt{3}\)

Thus, we find two values of b. We have

\(b=2\sqrt{3},\:b=-2\sqrt{3}\)

Thanks! :)

We could isolate a from the first equation and get \(a=\frac{250}{1+b}\) before plugging it into the second equation, but let's do something simpler. 

Adding the first and the secone equation, we get

\(2a=10\\ a=5\)

Plugging this in for b, we get

\(a=5,\:b=49\)

So, our final answer is 5. 

Thanks! :)

*1200 points

First, let's isolate a^3 as much as possible. 

We have

\(\frac{1}{64a^3+7}=7\\448a^{3}+49=1\\ 448a^3=-48\\ a^3=-48/448 \)

From this, we get that

\(a=\sqrt[3]{-28/3}\)

This is actually the only real factor for a, so it is our final answer. 

Thanks! :)

First, moving all terms to one side other than the constants, we get

\(x^2+y^2-6x-6y=56\)

Completing the square for both x and y, we get

\(\left(x-3\right)^2+\left(y-3\right)^2=\left(\sqrt{74}\right)^2\)

Thus, the radius is

\(r=\sqrt{74}\)

Using the circle equation, we have

AREA OF CIRCLE = \(\sqrt{75}^2\pi = 74\pi\)

Thus, 74pi is the answer. 

Thanks! :)

First, let's simplify the equation we are given. We get

\(10x^2 + 10x - 1\\ ab = -1/10\\ 2ab = -2/10 = -1/5\)

Now, take a + b  =  -1

Squaring both sides, we get

\(a^2 + 2ab + b^2  = 1 \\ a^2 - 1/5 + b^2  = 1 \\ a^2 + b^2  = 1 + 1/5  =  6/5\)

\((a -4) / ( b -4) + ( b -4) / (a -4) = [ (a - 4)^2 + ( b -4)^2] ] ' [ (a -4) ( b-4) ] = \\ [ a^2 - 8a +16 + b^2 - 8b + 16 ] / [ ab - 4(a + b) +16 ] = \\ [ (a^2 + b^2) - 8(a + b) + 32 ] / [ ab - 4 (a + b) + 16 ] =\\ [ 6/5 - 8 (-1) + 32 ] [ -1/10 - 4 (-1) + 16 ] = \\ [ 6/5 + 40 ] / [ 20 - 1/10] = 20497 / 25\)

The answer is \(20497 / 25\)

Thanks! :)

Simplifying the equation, we have \(x^2 + 6x - 2  = 0 \)

In the form mx^2 + nx + p  = 0, the Product of the roots ab = -n/m =   -2  

So 2ab = -4

Sum of the roots a + b =  p/m =  -6     

Squaring both sides, we have 

\(a^2 + b^2 + 2ab = 36 \\ a^2 + b^2 - 4 =36 \\ a^2 + b^2  = 40 \)

\(1/a^3  + 1/b^3  = \\(a^3 + b^3) / (ab)^3   \)        Note  :  a^3 + b^3 = (a + b) (a^2 + b^2 - ab)

So we have

\((a + b) ( a^2 + b^2 -ab)  / (ab)^3  = \\ (-6) (40 - -2) / (-8)   =  \\ (6) (42 / 8)  =   31.5 \)

Thus, the answer is 31.5

Thanks! :)

First, let's calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

\((45 min)(50 m/min) = 2250 m \)

So when Grace starts, she and Will have

\((2800 m) – (2250 m) = 550 m \) to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

\( (50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875 \)

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is \(2:52:52\)

Thanks! :)

I just want to add on that we can represent the answer \(10\sin(75) =  10\tan(15) = 10\tan(\frac{30}{2}).\)
Now we can use the trigonometric half-angle identity \(\tan(\frac{\alpha}{2}) = \frac{\sin(\alpha)}{\cos(\alpha)+1}\).
We can input our normal values: \(10\tan(\frac{30}{2}) = 10(\tan(\frac{30}{2})) = 10(\frac{\sin(30)}{\cos(30)+1}) = 10(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}+1) = 10(\frac{1}{\sqrt{3}+2}) = 10(2-\sqrt{3})= 20 - 10\sqrt{3}.\)
Therefore \(BE = \sqrt{(20 - 10\sqrt{3})^2+10^2} = \sqrt{400 + 300 - 400\sqrt{3} + 100} = \sqrt{800 - 400\sqrt{3}} = \sqrt{400(2-\sqrt{3})} = 20\sqrt{2-\sqrt{3}}\)

First, let's simplify the equation. We have

\(x^2 - 6x + 13\)

Since the constant is equal to uv, we have the equation

\(uv = 13 \\ 2uv = 26\)

The coefficient of x is equal to u+v, so we have

\(u + v = 6\)

Squaring both sides of the equation, we have the equation

\(u^2 + 2uv + v^2 = 36 \\ u^2 + 26 + v^2 = 36 \\ u^2 + v^2 = 10\)

Now, it's time for the fun part of this problem. We can simplify what we want to figure out to

We already know all these terms! We can easily find them! We have

\(u^3 + v^3 = ( u + v) ( u^2 + v^2 - uv) = (6)(10 - 13) = -18 \\ [uv]^2 = [ 13]^2 = 169\)

So, The answer is \(u/v^2 + v/u^2 = [ u^3 + v^3 ] / [ uv]^2 = -18 / 169\)

Thanks! :)

We get 3a + 2b = 2

6a + 2b = k - 3a - ab

Then we get 9a + (2 + m) * b = k

Equation first is equivalent to

9a + 6b = 6

There are an infinite number of solutions if second equation and third equation are equivalent. That is true only if the coefficients of b in both equations are the same and the constants in both are the same.

2+m = 6 --> m = 4
k = 6

ANSWERS: k = 6; m = 4

The answer is k = 17 and m = -10.

1. Our most obvious step is to draw the points \(O, P, Q\) and label their lines' respective gradients: \(OP_m = 4, OQ_m = 5\). We are trying to find out \(PQ_m\). Where m stands for the gradient. \([y = mx + c]\)
2. Let's label point Q(a, 5a) and P(b, 4b). Since OP = OQ, that means we can equate their sum of squares together.

3. \(b^2 + (4b)^2 = a^2 + (5a)^2\)

  i) \(b^2 + 16b^2 = a^2 + 25a^2\)

  ii) \(17b^2 = 26a^2\)

  iii) \(b\sqrt{17} = a\sqrt{26}\)

  iv) Now we can express b in terms of a and produce \(b = a\sqrt{\frac{26}{17}}\)

4. To find \(PQ_m\), we need to use our gradient formula which is \(\frac{y_2-y_1}{x_2-x_1} = \frac{4b-5a}{b-a}\)

5. for simplification we see that \(\frac{4b-5a}{b-a} = \frac{4b-4a-a}{b-a} = \frac{4(b-a)-a}{b-a} = 4 - \frac{a}{b-a}\)

6. Now we can express a in terms of b which turns \(\frac{a}{b-a} = \frac{a}{a\sqrt{\frac{26}{17}}-a} \)

7. \(\frac{a}{a(\sqrt{\frac{26}{17}}-1)} = \frac{1}{\sqrt{\frac{26}{17}}-1)},\) which we rationalize by multiplying both the numerator and denominator by\( \sqrt{\frac{26}{17}}+1\) which gives us \(\frac{\sqrt{442}+17}{9}\)

8. Therefore our gradient is \(4 - \frac{\sqrt{442}+17}{9} = \frac{19-\sqrt{442}}{9}\)
 

I'm pretty sure this slope is negative so I'm confident this may be the right answer. However, please do check for any calculation errors I've made :)

First of all, let's move almost all the variables to one side of the equation and combine all the like terms. 

We get \(x^2 - 21x + y^2 + 13y = 25\). 

Now, we complete the square for both x and y on the right side of the equation. 

\(x^2 - 21x + 441/4 + y^2 + 13y + 169/4 = 25 + 441/4 + 169/4\)

\((x - 21/2)^2 + (y + 13/2)^2 = 355 / 2\)

Wait! This looks like the equation for a circle! 

Indeed, we have a circle at center  \((21/2 , -13/2)\) with radius \(\sqrt{355 / 2}\). 

The value of the largest possible x is just the radius added onto the x value of the center. 

\(x = (21/2) + \sqrt{355 / 2}\)

Thanks! :)

We can use variables to complete this problem.

Let's let P = (x,y)

Plugging this value of P into what we know, we have

\(PA^2 + PB^2 + PC^2   = \\ (x-7)^2 + (y + 11)^2  + (x-10)^2 + (y-13)^2 + (x-18)^2 + ( y + 22)^2\)

\(3 x^2 - 70 x + 3 y^2 + 40 y + 1247 \)
\(3 [ x^2 - 70/3 x + y^2 + 40/3 y ] + 1247 \)

Now, we must complete the square for x and y. Doing this, we get

\(3[ x^2 - 70/3 x + 4900/36 + y^2 + 40/3y + 1600/36 ] + 1247 - 4900/12 -1600/12 \\ 3 [ (x - 70/6]^2 + (y + 40/6)^2 ] +  2116 / 3 \\ Q = (70/6, -40/6 ) =  (35/3 , - 20/3)\)

Thus, we have k = 2116 / 3

So our final answer is 2116/3. 

Thanks! :)

First, we know the equation of a sliope is \(\frac{y_2-y_1}{x_2-x_1}\)

In this problem, we know the two points we have are (a, a^2) and (b, b^2). 

Plugging these two points in, we get

\( [ b^2 - a^2 ] / [ b - a ]\)

Using the difference of squares equation for the numerator, we have

\( [ (b -a) (b + a) ] / (b -a)\)

Notice the b - a cancel each other out. 

This leaves us with

\( b + a \)

Since the slope is 2, we simplfy have

\( b + a = 2\)

So the answer is 2. 

Thanks! :)

Let's assign the values of the dimensions of the prims as \(p, q, r.\) 
For the surface area: \(2(pq + pr + qr) = 15\)
For the sides: \(p + q + r = \frac{17}{4}\)
To find the diagonal we have to know: \sqrt{p^2 + q^2 + r^2}

With some basic algebraic manipulation, we know that \((p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + pr + qr)\), therefore:

\(p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + pr + qr)\). We can substitute the values we know and write, 
\(p^2 + q^2 + r^2 = (\frac{17}{4})^2 - (15), p^2 + q^2 + r^2 = \frac{289}{16} - \frac{240}{16},  p^2 + q^2 + r^2 = \frac{49}{16}, \)

therefore \(\sqrt{p^2 + q^2 + r^2} = \sqrt{\frac{49}{16}} = \frac{7}{4}\). 

Now we found out that the diagonal of the prism is \(\frac{7}{4}\) :)

Let's set some variables to solve this question. 

Now, since AD bisects  BAC

Let's let BD  = 3 - x

Let's let  CD =  x

Now, we have the formula

\( BD / AB = CD / AC \\ (3 - x) / 4 = x / 5\\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

Thus, we finally can caclulate ADC as

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3 \)

Thus, The answer is \(10/3\)

Thanks! :)

Let's make some obersvations. We have that
\(L =   \sqrt {92^2  + (33-25)^2} =   \sqrt{ 92^2 + 8^2} =  4\sqrt {533} \\ R = 33 \\ r =25\)

Surface area = \(\pi L (R + r) + \pi (R^2 + r^2)  = \pi [ 4\sqrt{533} ](58) + \pi ( 33^2 + 25^2)  ≈ 22211.49 units^2 \)

\(S1  =  \pi * 33^2    \\       S2  = \pi * 25^2    \\   H  = 92 \)

Volume  = 

\( H/3 * ( S1 + S2 +  \sqrt  { S1 * S2 })  =  \\ 92/3  *  ( \pi (33^2 + 25^2) +  \sqrt{\pi*33^2 * \pi* 25^2} ) ≈ 244612.78 units^3\)

Thus, The answer is \( 244612.78\)

Thanks! :)

First, let's find the radius of the circle. We simply have

\(25 \pi = \pi *r^2 \\ 25 = r^2 \\r = 5 \)

Let's let R be the radius of the sphere. 

\(R^2 = [ 5^2 + r^2 ] = [ 5^2 + 5^2 ] = 50 \)

Now, we can calculate the surface area of the sphere. We have

\(4 \pi R^2 = \\ 4 \pi * 50 = \\ 200 \pi\)

So, the answer is \(200 \pi \)

Thanks! :)

WAIT a second...

"The length of $\overline{AB}$ is $4,$ and the circle has a radius of $5.$  Find the length $AB.$"

I think we know the answer is 4 now 😁

The length of AB is \(4\)

Thanks! :)

Let's set up variables to complete this problem. 

Since AD bisects  BAC

Lets let  BD  = 3 - x

Lets let  CD =  x

Now, we can easily solve the problem with an equation. We have

\(BD / AB = CD / AC\\ (3 - x) / 4 = x / 5\\ 5 (3 - x) = 4x\\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

Thus, taking this value of CD and plugging it into the area equation, we find

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3 \)

thus, The answer is \(10/3\)

Thanks! :)