NotThatSmart

We can use a really simple and cheeky trick to solve this problem.

Taking the product of the coeffiicents of a and b and add this to both sides, we get

\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)

Factoring the left side of the equation, we get

\((a + 4) ( b - 3) = 119\)

Now, we take the two factros with the smallest margin, 7 and 17. We have

\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)

So 3 is our answer. 

Thanks! :)

We can set variables up to solve this problem. Let's set \(d = a_2 -a_1\)

From the problem, we can write the equations

\(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\\S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\\\\\S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\)

Simplifying the first 2 equations, we get the conditions

\(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\)

Solving this gives that d is -3/250 and a_1 is 8/125. 

Thus, we have

\(S_{15} = 15a_1 + 105d = -\dfrac3{10}\)

So -3/10 is our answer. 

Thanks! :)

We can list out all numbers since there are not that many. 

5 modulo 171 means that the number leaves a remainder of 5 when divided by 171. 

First, we have 5, since we have \(171*0 + 5 = 5\)

We add 171 to every number from here on. 

So, we have

\(5+171=176\\ 176+171 = 347\\ 347+171 = 518\\ 518+171=689\\ 689+171=860\\\)

Adding anymore would result in a number greater than 1000. 

Thus, we have \(6\) numbers. 

So our answer is 6. 

Thanks! :)

I think it's just a glitch but the problem is missing some values. 

We need to know that the area of the square can be written as for p and q to solve this problem. 

I think the latex didn't go through. 

Can you please just fix the part for the equation? 

I'll try to solve it then!

Thanks! ::)

First, let's simplify this equation. Combining all like terms, we have

\(-15x^2-kx+53=0\)

Now, in order for the value of x to not be real, then the descriminant must be negative. 

Thus, we have

\({\left(-k\right)^2-4\left(-15\right)\cdot \:53} < 0\)

Now, we simplfy solve for k. We get that

\(k^{2}-3180 < 0\\ k^2 < 3180\\ -2\sqrt{795} <2\sqrt{795}\\ -56.39 < k < 56.39\)

The largest integer k can be that fits in the range given for k is 56. 

So our answer is just 56. 

Thanks! :)

We can use simple conversions and calculations to solve this problem. 

We first find the total area needed, then divide it by 400, before doing conversions. 

Thus, we have

\( [ 15 * 60 *1 / 400] = 9/4\)

Converting to meters, we have

\(\frac{9/4}{(0.3048^2)} \approx 25\)

so our answer is just 25. 

Thanks! :)

First, let's find the distance Will travels before Grace even starts rowing. 

Since Will has a 45 minute head start, we have the equation

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

This means that Will and Grace must cover

\((2800 m) – (2250 m) = 550 m \) together. 

Let's say the two row t minutes. Since we add their speeds together, we have

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

This rounds to about 6 minutes and 52 seconds. 

In total, the two took 51 minutes and 32 seconds to meet each other . So, the two will meet at \(2:51:52 pm \)

So 2:51:52 is our answer. 

Thanks! :)

Let's first note that the quadratic with roots 1 and -1 can be written in the form of

\((x-1)(x+1)=0\)

This is in the form of what we call factored form. Now, expanding this, we have

\(x(x+1) - 1(x+1)\)

Distributing the values in, we have

\(x^2+x-x-1\)

\(x^2-1\)

Thus, we have that the first blank is 0 and the second blank is -1. 

Thanks! :)