GYanggg

2. 

Solve by graphing. 

https://www.desmos.com/calculator/x1mgab6pen

We get \(5^x+12^x=y\)

Y-intecept = 2. 

Only real solution is 2. 

I hope this helped,

gavin

1. 

If we set the other 4 test scores as high as possible, the 5th one must be the lowerst possible score. 

We can assign the lowest possible score as x. 

Then we have the equation: 

\(\frac{4\cdot100+x}{5}=88\\ 400+x=440 \\ x=40\)

The lowest possible score is 40. 

I hope this helped

Gavin

3. 

Since the sum of two absolute value equation is equal to 0, each of the absolute value equations is equal to zero. 

This is because \(|x|\ge0\).

We can write the systems of equation: 

\(x + y + 7=0\)

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

Solve by adding:

\(x+2x+y-y+7+2=0 \\ 3x = -9 \\ x=-3\)

\(y+7-3=0 \\ y=-4\)

\( \boxed{(-3,-4)}\)

I hope this helped,

Gavin

Hey wiskdls!

If there are 9 people in a group, there are \(\binom94\) = 126 ways to choose a 4 person group. 

Since the 4 person group is already chosen, we have 5 people left. 

To choose a 3 person group, we do \(\binom53\) = 10

Now there is 2 people left and to choose a 2 person group, there is only 1 way to do it. 

\(126\cdot10=\boxed{1260}\)

Note: The order which you choose groups does not matter. You can first choose the 3 person group or first choose the 2 person group. The answer is still going to be 1260.

I hope this helped,

Gavin.

What we need to do in this problem is to use the information that is given in the problem to find the expression for f(x). We are given the horizontal asymptote, the vertical asymptote, and the x-intercept.

To find the vertical asymptote, all you need to do is set the denominator equal to 0 and solve for x, but since we already know the vertical asymptote, we could work backwards to find the value of c.

Since the vertical asymptote is 3, we can plug in 3 for x and solve for c. 3+c=0

After solving this equation, we have c=-3 

Our updated expression for f(x) becomes \(f(x)=\frac{ax+b}{x-3}\).

Finding the horizontal asymptote is different from finding the vertical asymptote.

There are three cases. Case 1: If the degree of numerator is less than the degree of the denominator, the horizontal asymptote is y=0. 

Case 2: If the degree of numerator is equal to the degree of the denominator, you take the coefficients of the variable and you divide them.

Case 3: If the degree of the numerator is more than the degree of the denominator, there will be no horizontal asymptote. In this case, we see that the degree of the numerator and denominator is the same, they are both equal to 1, so we must divide the coefficients of the variable.

In the numerator, the coefficient of x is a, and in the denominator, the coefficient of x is 1.

So we can write the equation, \(a\div1=-4\)

After solving the equation, we have a=-4.

Now, our new updated expression is \(f(x)=\frac{-4x+b}{x-3}.\)

To find b, all we need to do is plug in the values for x and y the problem already gives us.

Plugging in 1 for x and 0 for y, we have: \(0=\frac{-4\cdot1+b}{1-3}\)

Solving the equation, we have $b=4.$ Now we have all four variables, the final expression for f(x) is: \(\boxed{f(x)=\frac{-4x+4}{x-3}} \)

For Part B, there are two methods.

Method 1: We can approach it the same way we did in Part A.

Once again, to find the vertical asymptote, we set our denominator equal to zero, and plug in 3 for x.

We have: \(2\cdot{3}+t=0\) Solving for t, we have t=-6

Also, the degree of the numerator is the same as the denominator, so you divide the coefficients of the x.

Since the horizontal asymptote is y=-4, we can write the equation: \(r\div2=-4.\)

Solving for r, we have r=-8

Our updated expression of f(x) is: \(f(x)=\frac{-8x+s}{2x-6}\).

Plugging in 1 for x and 0 for y, we have: \(\frac{-8\cdot{1}+s}{2\cdot{1}-6}=0\) Solving for s, we have s=8.

Our final expression for f(x) is \(\boxed{f(x)=\frac{-8x+8}{2x-6}}\)

Method 2: All we need to do to get the 2x in the denominator is multiply both the numerator and the denominator by 2 in the answer in Part A.

Our final expression for \(f(x) = \frac{2\cdot{-4}x+2\cdot{4}}{2\cdot{x}-2\cdot{3}}\) After simplifying, we have: \(\boxed{f(x)=\frac{-8x+8}{2x-6}}\)

I hope this helped,

Gavin.

I think this is false. Here is my reasoning:

When you look at the 2D version of the sphere, a circle,

it is obvious that you can split the circle into equal slices if you cut through the center. 

You split the circle into infinitesimally small triangles that is used to contruct a rhombus, hence the proof of the circle's area. 

You cannot do the same which a sphere. Split the sphere into infinitesimally small cones from the center, to construct the sphere. But the curved sides of the cone cannot be fitted together, it must be infinitesimally small pyrimaids. 

Therefore, a sphere cannot be split into cones, but only pyramids. 

I hope this helped,

Gavin

We can imagine the problem like this:

There are "x" toppings to choose from. 

There are \(\binom{x}{2}\) ways to choose the toppings. 

Therefore \(\binom{x}{2}\) = 106 

We can rewrite this as:

\(\frac{x\cdot(x-1)}{2}=106 \)

Then we have to add the choice of "no" toppings. 

\(\frac{x\cdot(x-1)}{2}+1=106 \\ x^2-x-210=0 \\ (x+14)(x−15)=0 \\ x_1=-14,\ x_2=15 \)

Since there cannot be negative toppings, the answer is 15. 

We can double check this.

If there are 15 toppings and you can have no toppings, how many ways can we "top" our ice-cream. 

\(\binom{15}{2}+1=106\), it works!

I hope this helped,

gavin

Hey Guest!

\(\begin{align*} N &\equiv 6 \pmod{12}, \\ N &\equiv 6 \pmod{18}, \\ N &\equiv 6 \pmod{24}, \\ N &\equiv 6 \pmod{30}, \\ N &\equiv 6 \pmod{60}. \end{align*}\)

LCM [12, 18, 24, 30, 60] = 360

N = 360k + 6. 

6 is your answer.

Guest is right, my previous answer was wrong. 

I hope this helped,

gavin

To find 30_n we use formula

S_n=n/2⋅(2a₁+(n−1)⋅d)

In this example we have a₁=2 , d=2 , n=30. After substituting these values into the above equation, we

obtain:

S₃₀  = 930

i hope this helped,

gavin