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We have a system of equations we need to solve. 

We have the system

\(x=4\\ x^2+y^2=25\)

Plugging in x as 4 into the second equation, we have

\(16+y^2=25\\ y^2=9\\ y= \pm 3\)

Thus, we find there are two ordered pairs and solutions to the equation. We have that

\((4, 3); (4, -3)\)

The distance between the two is just \(3-(-3) = 6\) since they share the same x value. 

Thus, 6 is our answer. 

Thanks! :)

First, let's combine all like terms for x and move all terms to one side. We have that

\(x^{2}+12x-4=0\)

Using the quadratic equation, we have that

\(x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}\)

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

Now, we can use a handy trick to solve this problem. We have that

\((2\sqrt{10}-6)^2+(-2\sqrt{10}-6)^2\)

Expanding and distributing in our values, we have

\(-24\sqrt{10}+76+4\left(\sqrt{10}\right)^{2}+12\sqrt{10}-6\left(-2\sqrt{10}-6\right)\)

All the radicals basically cancel out. In the end, we are left with 152. 

So 152 is our answer. 

Thanks! :)

The units digit of both numbers are \(3\) and \(7\), so \(3\cdot7 = 21\), where the units digit is \(1\). The units digit of \(13\cdot287\) is \(\boxed{1}\)

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! :)

The slope of the line is \(\frac{150-17}{82-3} = \frac{133}{79}\) .(This cannot be further simplified). Starting from point \((3,17)\), the next point with integer coords will be \((82, 150)\), which is the second endpoint. So there are only \(\boxed{2}\) lattice points in the line, that being the two endpoints.

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! :)

If we simplify it we get \(x^2 + 8x + 9\). The discriminant is \(b^2 - 4ac\) so we get \(64 - 4\cdot9 = 28\)

\(28\) is the discriminant.

\(\begin{align*} \dfrac{s}{2} + 5t &= 3,\\ 3t - 6s &= 9. \end{align*}\)

so ill first multiply the top equation by twelve to get 6s+60t=36

ill rearange the second equation to get -6s+3t=9

ill "add" the two equations to cancel out the s to get 63t=45

t=5/7

ill substitute t back into the first equation to get s/2+25/7=3

multiplying by two gets s+50/7=6

s=6-50/7

s=42/7-50/7

s=-8/7

so the ordered pair \((s,t)\)would be \(\boxed{(-\frac{8}{7},\frac{5}{7})}\)

\( 127 \equiv 18 \pmod{n}\)

so 109 = 0 mod n

109 is prime so the only two solutions are \(\boxed{1\space and \space 109}\)

The only one that is 2 mod 7 is 2.

(1) The answer is 50.

(2) The answer is 20.

Laverne starts counting out loud by 5's.  She starts with 2.  As Laverne counts, Shirley sums the numbers Laverne says.  When the sum finally exceeds 100, Shirley runs screaming from the room.  What number does Laverne say that sends Shirley screaming and running?   

Laverne's           Shirley's   

Counting          Summation   

     2                           2   

     7                           9   

   12                         21   

.  17                         38   

   22                         60   

   27                         87   

   32                       119    

_.

I drove to the beach at a rate of 40 miles per hour. If I had driven at a rate of 50 miles per hour instead, then I would have arrived 45 minutes later. How many miles did I drive?   

When you go faster you'll get there sooner, so the problem

must actually mean "would have arrived 45 minutes earlier."   

Distance  =  Rate • Time    

There are two cases:    (1)  D  =  (40)(t)   

                                     (2)  D  =  (50)(t –  45 / 60)   

Regarding 45 / 60  -  Since the speed is given in miles per hour, the 45 minutes   

must be converted to hours to keep units consistent.  We could have converted    

the speed to miles per minute, but that would have unnecessarily added tedium.  

Since the distance is the same in both cases, set them equal and solve for time.   

                                            (50)(t – 45 / 60)  =  (40)(t)   

                                                     50t – 37.5  =  40t    

                                                                10t  =  37.5   

                                                                    t  =  3.75   

Substitute t back into case (1)                    D  =  (40)(3.75)   

                                                                   D  =  150    

check answer     

Substitute t back into case (2)                    D  =  (50)(3.75 – 0.75)   

                                                                   D  =  (50(3.00)    

                                                                   D  =  150    

_.

Find the ordered triple (p,q,r) that satisfies the following system:   

p - 2q = 3   

q - 2r = -2 + q   

p + r = 9 + p    

from q – 2r = –2 + q    we find that r = 1   

from p + r = 9 + p        we find that r = 9    

Since r can't equal both 1 and 9    

the system has no solution.   

_.

At a cafeteria, Mary orders two pieces of toast and a bagel, which comes out to $3.15. Gary orders a bagel and a muffin, which comes out to $3.50. Larry orders a piece of toast, two bagels, and three muffins, which comes out to $8.15. How many cents does one bagel cost?  

                                           Mary          2T + B = 315 ............. —>  T = (315 – B) / 2       

                                           Gary          B + M = 350 ................—>  M = (350 – B)      

                                           Larry          T + 2B + 3M = 815  

                                           Substitute for Toast & Muffin into Larry's order  

                                           [(315 – B) / 2]  +  2B  +  [3 • (350 – B)]  =  815   

                                           [(315 – B) / 2]  +  2B  +  (1050 – 3B)  =  815   

Multiply both sides by 2       (315 – B)  +  4B  +  (2100 – 6B)  =  1630    

Combine like terms              –3B + 2415  =  1630     

                                                        –3B  =  – 785     

                                                            B  =  261 ²/₃  ¢    

_.