NotThatSmart

First, let's simplify the equation. 

Combining all like terms and reeaaranging the left, we get the equation

\(y=-5x^{2}-6x+10\)

Now, we know the x value of the vertex can be written in the form \(-\frac{b}{2a}\) for quadratics in the form \(y=ax^2+bx+c\)

Plugging in what we have, we get \(-\frac{6}{10}=-0.6\)

Subsituting this back to find y, we get \(y=11.8\)

Thus, our final answer is \((-0.6,11.8)\)

Thanks! :)

Let's calculate the number of ways we can seat the children and adults seperately. 

First, there are \(4! = 24\) ways to seat the children and \(4! = 24 \) ways to seat the adults. 

We now multiply these values together to get

\(24*24=576\)

So 576 is our answer. 

Thanks! :)

Since we have \(a \equiv 5 \pmod{7}\), then we have

\(a + 1 \equiv 5+1 \pmod{7}\)

So 6 is the answer. 

There actually is no minimum value. 

The larger x gets, the smaller the fraction becomes, therefore meaning there is no minimum value. 

The range is \(0 <1\) , meaning no minimum value is present. 

Here's an image

The horizontal asymtote is \(y=0\)

Thanks! :)

First, let's convert 1/288 into base 13. 

Since 1 is the smae in every base, and we have

\((192)_{13} = (1 × 13^2) + (9 × 13^1) + (2 × 13^0) = (288)_{10} \), we have the equation

\(\frac{1}{192}_{13} = \frac{1}{288}_{10}\)

Now, we complete base division. The answer is approximately \(0.0078229A44BC67118933... \)

I don't think the number is repeating, although I am not sure. 

If I had to geuss with the first 20 digits, it isn't repeating, but maybe it is. 

Thanks! :)

We can solve this problem by looking at the last digit of the numbers. 

A power of 5 always ends with 5, so 5^19 ends with 5. 

We also know that  \(7^{13}=96889010407\) which ends with 7. 

So, we we add up the last digits, we get \(7+5+3 = 15\)

Since it ends with 5, the number must be divisble by 5. 

We also know that it can't be divisble by 2 and 3 since neither of \(5^{19} , 7^{13} , 23\) are divisble. 

So our final answer is 5. 

Thanks! :)

Let's identify all the ways we can get a terminal 0. 

First, since 40 has a terminal 0, we already have 1. 

Since \(75*12\) also has a terminal 0, which makes it 2. 

Lastly, \(6*75\) has one terminal 0, making 3. 

We could also multiply everything out. We have

\(40 \cdot 6 \cdot 75 \cdot 12=216000\)

As shown, there are 3 terminal 0s. 

So 3 is our answer. 

Thanks! :)

To solve this problem, let's first put y in terms of x. 

From the first equation, we have \(y=25-x\)

Subsituting this value of y into the expression we are trying to compute, we get

\(3x+75\)

Since x cannot be negative, the smallest possibvle x number is 0. 

When x is 0, we have \(3(0)+75=75\)

So 75 is our final answer. 

Thanks! :)

We can write an equation to solve the problem. Let's let x be the total number of buns he made. 

Using the problem, we have the equation

\(\frac{60}{100}(x-120) = 1/3x\)

Solving this equation for x, we find that

\(x=270\)

This means in the afternoon, he sold \(\frac{60}{100}(150) = 90\)

Thus, he sold 90 buns. 

So 90 is our answer. 

Thanks! :)