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1

sin theta = 12 / 13

cos theta = -5 / 13

tan theta =  -12 / 5

               A

              40

10              

B45           D             C

Angle C =  190 - 45 - 40  = 95

Law of Sines

10 / sin 95 =  AC / sin 45

AC = 10 * sin 45  /sin 95  ≈ 7.1

[ABC ] = (1/2) 10 *  7.1 * sin 40   ≈   22.8

We can use stars and bars to solve this problem. 

We have 9 stickers in a row (stars) 

We need to split these 9 stickers into two groups. We can use one bar for this. 

We have something like this

X X X | X X X X X X

In this case, one friend gets 3, the other gets 6. 

Now, the problem becomes, how many positions can the bars be put in. 

We have \(10 \choose 1\) = 10, so there are 10 ways. 

Thanks! :)

For this problem, we must know what EPF is. 

However, there is no clear representation of what it actually is. 

In the problem, it never actually defined what P is, meaning that it is impossible to solve. 

Thanks! :)

We can use the law of cosines to solve this problem. 

First,let's say

\(AB = a = x\\ AC = b + 2x+1 \\ BC = c = 3x+5 \\ ∠BAC = 120°\)

The law of cosine states that 

From the problem, we can write \(cos(120) = \frac{(3x+5)^2+(2x+1)^2 - x^2}{2(2x+1)(3x+5)}\)

Expanding out everything and simplifying, we get \(18x^{2}+47x+31=0\)

Using the quadratic formula, we have \(x=\frac{-{47}\pm \sqrt{{47}^{2}-4\cdot {18}\cdot {31}}}{2\cdot{18}}\)

We get \(x=-\frac{47}{36}\pm \sqrt{23}\cdot (\frac{1}{36}i)\)

So \(x=-\frac{47}{36}+\sqrt{23}\cdot (\frac{1}{36}i)\)

Thanks! :)

Edit: THIS IS MY 200th answer! Not flexing...but noice :)

6000 / 2000 *  40  =  120 min

Very similar question

First up, let's isolate x for \(\frac{xy}{x+y}=2:\\ \)

We get \(x=\frac{2y}{y-2}\)

Subbing this in, we get 

\(\frac{\frac{2y}{y-2}z}{\frac{2y}{y-2}+z}=2\\ \frac{yz}{y+z}=2\)

\(\frac{2yz}{2y+z\left(y-2\right)}=2\\ \frac{yz}{y+z}=2\)

Now, we isolate y. 

\(\frac{2yz}{2y+z\left(y-2\right)}=2 \\ y=z\)

Subsituting in y=z, we get

\(\frac{zz}{z+z}=2\)

\(z=4\)

We get z=4 as our answer. 

Thanks!:)

Let's let the tangent point between Semicircle AB and the red circle as D, the tangent point between arc BC and the red circle as E, and the tangent point between the red circle and semicircle AC as F.  Label the center of semicircle AB to C1 the center of semicircle of BC to C2 and the radius of the red circle to C3.Also, let the radius be x. 

We know from the problem that \(BF=2\) so that means \(BC_3 = 2-x\)

We also know that \(C_1C_3 = 1+x.\)

Now, draw \(C_1C_3B.\) Notice this is a right triangle. 

Now, we can use the pythaogrean thereom to finish this poblem. 

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\) The x in the equation is cancelled out to get the radius of 2/3. 

2/3 is our answer. 

Thanks! :)

I don't see a graph posted for this problem. 

I'm not sure if there is supposed to be one, but this problem is impsossible without a graph. 

Thanks! :)

There are two ways to draw figure ABCD.

Law of Cosines

[ 23^2 - 25^2 -18^2] / [ -2 * 25 * 18 ] = cos (angle Q)  = 7/15

Once more  →  QM = 12.5

PM =   sqrt [18^2 + 12.5^2  - 2 [ 18 * 12.5] * [ 7/15]  ]  ≈ 16.44

Here's my attempt...

1 5 x x x x  and the other numbers can  be arranged in 4!  ways

1 6 x x x x  and the other numbers can be arranged in 4! ways

x x x x 5 1  and the other numbers can be arranged in 4! ways

x x x x 6 1 and the other numbers can be arranged in 4! ways

5 1 6 x x x   ....the  5 1 6  can occupy any of 4 positions and the other numbers can be arranged in 3! ways

6 1 5  x x x  ... the  6 1 5  can occupy  any of 4 positions and the other numbers can be arranged in 3! ways

So 

4 *4!   + 2* 4*3!   =   144 ways

The diameter of AB is \(9 + 16 + 9 = 34.\) This means the radius is \(34/2 = 17\)

However, let's note something. 

\(QD = \sqrt{9(9 + 16)} = 15\), but in the problem, it says the square sidelenghth is 16. This means that PQRS is NOT a square, but a rectange with sidelengths 16 and 15. 

Like I said, the diameter is 34. We already know the height of the trapezoid, which we got earlier as 15. DC is 16, since PQRS is a rectangle. 

We have everything we need to solve for the area of the trapezoid! 

The area of a trapezoid is \(\frac{(b1+b2)h}{2}\) where b1 and b2 are the bases and h is the height. 

We have \(\frac{(25+16)15}{2} = (31)15/2 = 232.5\)

so the area of the trapezoid is 232.5. 

Thanks! :)

We want to put \( x^2+y^2=2x-6y+16+14x-4y+20\) in the form of a circle's equation. 

Moving all terms to one side, we get \(x^2-16x+y^2+10y=36\)

Completing the square for both x and y, we get \((x-8)^2+(y+5)^2=125\)

\(\left(x-8\right)^2+\left(y+5\right)^2=\left(5\sqrt{5}\right)^2\)

From this, we get that \(5\sqrt5\) is the radius. 

The area is \(\pi r^2 = \pi (125) = 125\pi\)

125pi is our answer

Thanks! :)

First, we have the fact that the axis of symmetry as 2. 

Since the distance (4, 7) from x = 2 is the same as (?, 7), we can easily find another point. 

We find that (4, 7) is two units away from the line x = 2, so we want to find another point that has y value of 7 and is also 2 units away. 

We find that (0, 7) matches all these characterisitics, thus being our answer!

Thanks! :)

We have the vertex form

y = a(x - h)^2  + k         where the vertex  (h,k)  = (3,1)

We  know that  (0,7)  is on the graph..so....

7  = a (0 - 3)^2  + 1

6 =  9a

a = 6/9  = 2/3

So

y = (2/3) ( x - 3)^2 + 1

y = (2/3) (x^2 - 6x + 9) + 1

y = (2/3)x^2 - 4x +6 + 1

y = (2/3)x^2 - 4x + 7

a*b*c =  (2/3) (-4) (7)  =  -56 / 3

First, let's write this in vertex form. 

We have \(y = a(x-h)^2 + k\) where \((h,k) \) denotes the vertex. From the graph, we can write, \(y=a(x-3)^2+1\) since the vertex is at (3, 1)

Now, we have to find a. We can use the y intercept to help. We have 

\(7 = a (0 - 3)^2 + 1 \\ 6 = 9a \\ a = 6/9 = 2/3\)

We find that when \(y=2/3(x-3)^2+1\), we have a y intercept at (0, 7), meaning we found our equation. 

Now, we simply have to expand it. Expaning, we get \(y = \frac{2}{3} x^ 2 − 4 x + 7\). 

This means \(abc = 2/3(-4)7 = -56/3\)

-56/3 is our answer!

Thanks! :)

I'm not sure if it's just me, but I don't see a graph. 

The equation of a circle is \((x – h)^2 + (y – k)^2 = r^2\) where \((h, k)\) is the coordinates of the center of the circle and r is the radius. 

Thanks! :)