All Answers 358514 Answers

4/3 x 5/8  

                                         4             5                20                             5  

                                     ———  x  ———   =   ———   reduces to  ———  

                                         3             8                24                             6  

_.

< What is Sophie's favorite number? >    

It's –49.

You can go to  https://web2.0calc.com/questions/algebra_49748 

to see how I obtained that solution a couple of days ago.  

Copy the link then paste it in to the URL field on any computer page.  

_.

What do you mean "I can't get this" ? Did you try?

What happens if you divide one by the other like this:

[ab^4] / [ab] = 48 / 72

b^3 = 2 / 3

b=[2/3]^(1/3)

First, you must bring variables to one side and constants to one side, this makes 2/3m + 14 = 0 the same as 2/3m = -14, then we must divide 2/3m by 2/3 to isolate m, but what we do to one side, we must do to the other, dividing by a fraction is also multiplying by its reciprocal which means 2/3m (3/2) = -14(3/2) which gives us m = -14(3/2), can you solve the rest? If you need any hep, just ask :)

To get your result it would seem to be a whole lot easier to simply remove a factor 3 from the second bracket and to move that into the first bracket.

However the result would then be

\(\displaystyle (162u^{2}v+54uv^{2})(3u+v),\)

which is different from your result, and not much of an improvement on the original.

I think that they are looking for something different.

Notice that 18uv is a common factor of the two terms in the first bracket, remove that and see where that leads.

This question was posted a few days ago, but in that version CX = 4.

I'll sketch out a solution for that one. (This new version is marginally easier.)

For a triangle ABC, the sine rule says that

\(\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B} =\frac{c}{\sin C}= 2R,\)

where R is the radius of the circumscribing circle (of the triangle).

Applying that to the triangles ACX and ABX, we have,

\(\displaystyle \frac{AX}{\sin C}=2R_{1} \quad \text{ and } \quad\frac{AX}{\sin B}=2R_{2}\)

and since the two radii are equal, it follows that the angles B and C are equal meaning that the triangle ABC is isosceles.

Drop a perpendicular from A to BC, meeting BC at Y, say, then XY = 3, and calculate AY using Pythagoras.

The area of ABC will then be equal to (1/2)*BC*AY.

The line will touch the parabola at just one point where the line is tangent to the parabola, at which point their slopes are equal.

Slope of straight line = 1

Slope of parabola = 2x/3

So the x-coordinate is given by  2x/3 = 1  or x = 3/2

At this point the value of y is y = (1/3)*(3/2)^2  or y = 3/4

So, using the straight line equation:    3/4 = 3/2 + c    or    c = -3/4

Hello,

To simplify the expression (54u^2v + 18uv^2)(9u + 3v), we can use the distributive property of multiplication.

What I found was c = –0.75  

I don't know how to solve this algebraically, so

I went to Desmos and graphed both curves, and  

Desmos let me add a slider to vary the value of c. 

The line is tangent to the parabola about (1.33, 0.60)  

which I estimated by eyeballing it.  

.

The area of a circle is πr^22, so the areas of the circles are 1, 9, 25, and 49, respectively. The black area is 9 + 25 = 34, so the ratio of black area to white area is 34/16 = 17/8​​.

[12 C 4]  +  [11 C 4]  +  [10 C 4] ==495  +  330  + 210==1,035 - total number of solutions.

Thank you for helping but that wasn't right.

We can solve this problem using the angle bisector theorem and some basic geometry:

First, since PQR is an isosceles triangle with PR = QR, we know that the angle bisector of angle P also bisects QR, so X is the midpoint of QR.

Next, since XY is perpendicular to PR, we can use the Pythagorean theorem to find its length. Let's call the length of XY h. Then, using the fact that QX = RX = QR/2 = 4.5, we have:

h^2 = PX^2 - QX^2          (by the Pythagorean theorem)
     = (PQ*QR/PQ + QR)*QR/4 - QR^2/4  (by the angle bisector theorem)
     = 81/4 - 81/4
     = 0

Therefore, h = 0, which means that XY coincides with PR. So, the length of XY is equal to the length of PR, which is 9. 

To simplify this expression, we can use the properties of exponents, including the power rule, product rule, and quotient rule. 

First, we can simplify the terms inside the parentheses:

(pq^2)^3 = p^3 * q^(2*3) = p^3 * q^6
(4pq^2)^(-3) = (4^(-3)) * (p^(-3)) * (q^(2*(-3))) = (1/64) * (1/p^3) * (1/q^6) = 1/(64p^3q^6)
(2pq)^2 = 2^2 * p^2 * q^2 = 4p^2q^2
(2p^2*q^3)^2 = 2^2 * p^(2*2) * q^(3*2) = 4p^4q^6

Next, we can substitute these simplified expressions back into the original expression and simplify further:

(pq^2)^3 * (4pq^2)^(-3) * (2pq)^2 * (2p^2*q^3)^2
= (p^3*q^6) * (1/(64p^3q^6)) * (4p^2q^2) * (4p^4q^6)
= (p^3*q^6) * (4p^2q^2) * (4p^4q^6) * (1/(64p^3q^6))
= (16p^9*q^14) / (64p^3*q^6)
= 1/4 * p^6 * q^8

Therefore, the simplified expression is 1/4 * p^6 * q^8.

Assuming that x neq 1/2, simplify (12x^2 - 6x)*(4x - 2).  

                                                         (12x² – 6x) * (4x – 2)  

factor 6x out of the first term             6x * (2x – 1) * (4x – 2)  

factor 2 out of the second term         6x * (2x – 1) * 2 * (2x – 1)  

combine 6x and 2 and  

combine both (2x – 1) terms              12x * (2x – 1)²  

_.

The problem can be broken down into 3 cases:

Case 1: x=0

In this case, we need to find the number of solutions to the equation u+v+w=8, where u,v,w are nonnegative integers. This is a standard stars and bars problem, and the number of solutions is c(10,2)=45.

Case 2: x=1

In this case, we need to find the number of solutions to the equation u+v+w+x=8, where u,v,w,x are nonnegative integers. This is also a standard stars and bars problem, and the number of solutions is c(11,3) =165.

Case 3: x=2

In this case, we need to find the number of solutions to the equation u+v+w+x=8, where u,v,w,x are nonnegative integers. This is also a standard stars and bars problem, and the number of solutions is c(12,4)=495.

Therefore, the total number of solutions is 45+165+495=705​.

The parabola and the line will intersect at most once if the line does not pass through the vertex of the parabola. The vertex of the parabola is at (0,0). The line will pass through the vertex if c=0. Therefore, the maximum value of c such that the graph of the parabola y=x^2/3 has at most one point of intersection with the line y=x+c is 0​.

The answer is 6*sqrt(5).

The possible values of a_{10} are -5 and 2.

x = 9/5.

C ==Full tank

1/2C - 1/6C ==1/3C

1/3C == 2

C ==2  x  3 ==6 gallons - full capacity of the tank.