All Answers 356458 Answers

First, let's set up a quadratic equation for x. Combining some like terms, we get

\(-8x^2-kx+47=0\)

Now, in order for the problem to have no real solutions, the descriminant must be less than 0. 

Thus, let's set up the descriminant to find k. 

We get that

\(k^2+1504<0\)

Isolating k, we get

\(k^2<-1504\)

However, note that k^2 cannot be negative, meaning that k^2 cannot possibly be smaller than -1504. 

Therefore, it is impossible for the equation to have no real solutions. 

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 calculate the distance Will travels before grace even leaves. 

Since he traveled for 45 minutes, we have

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

So when Grace starts, she and Will have

\((2800 m) – (2250 m) = 550 m\) to cover. 

Let's let t be the amount of time it takes for the two to catch up to each other after Grace starts rowing. 

We have the equation

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

This is approximately 6 minutes and 52 seconds. 

So the clock time they meet is 6 min 52 sec after Grace starts.

Grace started at 2:45, so add 6 min 52 sec and the clock will read 2:51:52 pm when they meet  

Thus, our final answer is \(2:51:52\)

Thanks! :)

Here is how can we compute the answer:

f(f(f(f(1+i) = f(f(f((1+i)^2) = f(f(f(0) = f(f(0) = f(0) = 0

Therefore, the answer is 0.

(x ^ 3 + 8x ^ 2 + 21x + 18) can be factored into ( x + 2 )( x + 3 )( x + 3 ). We divide by ( x + 2 ), and they cancel. Expanding

( x + 3 ) ^ 2 gives us x ^ 2 + 6x + 9. A = 1, B = 6, and C = 9. D is the one value that x can't be. If x was -2, then we would be dividing by 0, which absolutely can't happen. So, D = -2. Adding them together, A + B + C + D = 1 + 6 + 9 + (-2) = 14.

First, let's move all terms to one side and set up a quadratic equation. We have

\(-10x^2+kx-23=0\)

Now, in order for the roots to be nonreal, the descriminant must be less than 0. Thus, we can write the equation

\(k^2-920 < 0\)

Now, we simply solve for k. We have

\(k^2<920\\ \)

Since we must square k, there are two restrictive intervals for k. We have

\(-2\sqrt{230} <2\sqrt{230}\)

So our final answer for k is

\(-2\sqrt{230} <2\sqrt{230}\)

Thanks! :)

First, we must figure out what the common difference in the arithmetric sequence is. Let's let that be d. 

From the problem, we have the equation

\(1/4 + 10d = -1/5 \\ 10d = -1/5 - 1/4 = -9/20 \\ d = -9/200\)

Now, we can use all the information we have to find the 43rd term. We have

\( -1/5 + 10d = -1/5 + 10 (-9/200) = -1/5 - 9/20 = -65/100 = -13 / 20\)

So our answer is -13/20. 

Thanks! :)

First, let's put s in terms of t so we can do some subsitutions. We get

\(s=-\frac{2\left(-4t+1\right)}{5}\) from the first equation. 

Subsituting this value of s into the second equation, we get

\(3t-6\left(-\frac{2\left(-4t+1\right)}{5}\right)=9-2t\)

Now, we solve for t. We get

\(\frac{-23t+12}{5}=9\\-23t+12=45\\t=-\frac{33}{23}\)

Now, we subsitute t back into the first equation to find s. We get

\(s=-\frac{62}{23}\)

We get 

\(s=-\frac{62}{23},\:t=-\frac{33}{23}\)

So our final answer is \((-62/23, -33/23)\)

Thanks! :)

Hi UnVerifiedTaxPayer!

ummm..just a note, but you used different variables in the problem, \(a\) and \(x\)

I'm assuming this is a typo, but if it isn't correct me. Let's just say we are trying to find the value of \(2x + 12 \pmod{7}\)

Now, since we already have the first equation in handy, we can easily do a simple process.

Let's try to turn the left side of \(x \equiv 5 \pmod{7}\) into 2x+12. 

First, let's multiply both sides of the equation by 2. This gets us

\(2x \equiv 10 \pmod7\)

Since 10 mod 7 is the same as 3 mod 7 (they both leave a remainder of 3), we have the equation \(2x \equiv 3 \pmod 7\)

Now, let's add 12 to both sides of the equation. We then get \(2x+12 \equiv 15 \pmod7\)

Since 15 mod 7 is the same as 1 mod 7, we are left with

\(2x+12 \equiv 1 \pmod7\)

Now, we can check our answer. 

Take 5 as an example. We have that

\(5 \equiv 5 \pmod7\)

We also have that \(2(5) + 12 = 22 \equiv 1 \pmod7\)

This also works for 12, 19, and 26, so our answer should be correct. 

So our answer is 1. Again, correct me if I misunderstood the typo in the problem! 

Thanks! :)

First, we need to calculate the total possible number of rolls. Since the dice is rolled 7 times, we have \(7!\) total rolls. 

However, since there are duplicates with two 6s being rolled, we must divide by 2! to avoid overcounting. 

Thus, we just have

\(7! / 2! = 2520\)

So the answer is 2520. 

Thanks! :)

First, let's make some observations about this question. 

All points 5 units away from point (2, 7) would make a circle with center (2, 7) and radius 5. 

The formula for that circle would look something like \((x-2)^2+(y-7)^2=25\)

Now, combining that with the second equation, and we get a system. 

Subsituting out y with the x values in the line equation, we get 

\((x-2)^2+(5x-35)^2=25\)

Simplifying, and we get

\(13x^{2}-177x+602=0\)

Using the quadratic equation, we find two values of x. We get

\(x=7\\ x=\frac{86}{13}\)

Plugging these x values into the second equation where we isolated y, we get that \(y=7,y=\frac{66}{13}\)

So our final two answers are (7, 7) and (86/13, 66/13)

Thanks! :)

Let's split the inequality into two different inequalities and find the overlapping numbers. 

Now, let's find an interval for x from both inequalities and find the overlapping zone.

Alright. So we have the two inequalities 

\(1-3x < x+8\space\space\space\space\space(1)\\ x + 8 \le 7-x\space\space\space\space\space\space(2)\)

From the first inequality, we have

\(1-3x -7/4\)

From the second inequality, we have

\(x+8 \leq 7-x\\ 2x \leq -1\\ x \leq -1/2\)

Thus, we have the intervals

\(x>-\frac{7}{4}\quad \mathrm{and}\quad \:x\le \:-\frac{1}{2}\)

In interval notation, this is \((-\frac{7}{4},\:-\frac{1}{2}]\)

Thanks! :)

*1500 points

We can write down two inequalities and then combine the values of x to get the interval. 

The two inequalities we have are 

\(-1 < 2x < 12 \space \space \space \space \space(1) \\ -1 < 3x - 7 < 12 \space \space \space \space \space (2) \)

Now, let's isolate x from each equation to find the values of x. 

From the first equation, we get that

\(-1/2 < x < 6 \)

From the second equation, we get that

\( 6 < 3x < 19 \\ 2 < x < 19 / 3\)

Since we must take the most restrictive interval, the overlapping values of both gets us the interval

\(2 < x < 6\) or \((2,6)\) in interval notation .

Thanks! :)