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We can solve for the x intersection points thusly

3x + a = x^2 + x

x^2 - 2x - a  = 0

x^2 - 2x  = a          complete the square on x

x^2 - 2x + 1 =  a + 1

( x - 1)^2 = a + 1

( x - 1)  = sqrt (a + 1)       or       (x  -1) = -sqrt (a + 1)

x= 1 + sqrt (a + 1)           or  x  =  1 - sqrt (a + 1)

The y values of the intersections are

3[1 + sqrt (a + 1)] + a    and     3 [ 1 - sqrt (a + 1) ] + a

Simplifying and using the square of the distance formula  we have

[ 2sqrt (a + 1) ] ^2  + [ 6sqrt (a + 1)]^2   =13

4 (a + 1) + 36 (a + 1)  = 13

40 ( a + 1)  = 13

40a + 40  = 13

40a  =  -27

a = -27 /40

BTW   ....the graph looks more like this

\( $x + y \neq -1$ \)

Add the equations

a ( x + y) + a  + b(x + y) + b + c (x + y) + c  = 11x + 11y 

(a + b + c)  + (x + y) (a + b + c)  = 11 ( x + y)

(a + b + c) (x + y + 1)  = 11 (x + y)

a + b + c  =   11 (x + y)  / ( x + y + 1) 

Add the equations

p(x + y + z) + 2p  + q(x + y + z) + 2q + r(x + y + z) + 2r  = (x + y + z) + 7 + 5 - 3 + 4 - 8

(p + q + r) ( x + y + z) + 2 (p + q + r) =  (x + y + z) + 5

(-8) (x + y + z) + 2 (-8)  = (x + y + z) + 5

-9  (x + y + z)  =  5 - 2(-8)

-9 (x + y + z)  = 5 + 16

-9 (x + y + z)  =  21

(x + y + z)  =  -21 / 9     =  -7/ 3

1/ ( x + 1) + 3 / (x +7) ≥  1/2

[ (x + 7) +  3(x + 1) ] / [(x + 1)(x+ 7)] - 1/2 ≥  0

[ 4x + 10 ]  / [ ( x + 1 ) (x + 7) ]   - 1/2 ≥  0

[ 2 (4x + 10) - (x + 1) ( x + 7) ] / [ 2 (x + 1) (x + 7) ] ≥  0

[ 8x + 20 - x^2 - 8x - 7 ] / [ 2 ( x + 1) (x + 7) ] ≥  0

[ -x^2 + 13 ] / [ (x + 1) ( x + 7) ]  ≥  0          (1)

We  have  five  possible intervals to  check

(-inf , -7)          check with x = -8   ....this interval makes (1) false

(-7 , -sqrt 13]  check with x =  -5   ....this interval makes (1) true

[-sqrt 13 , -1)  check with x = -2  ......this interval makes (1) false

(-1 , sqrt 13 ]  check with x = 0   ......this interval makes (1) true

[sqrt 13 , inf)  check with x = 5   ......this interval  makes (1) false

Solution comes from  

( -7 , -sqrt 13]  U  ( -1 , sqrt 13]

AC = sqrt (25^2  - 10^2)  = sqrt (525) 

AD = sqrt [ ( CD + 9 )^2 + 10^2 )    =  sqrt [ CD^2 + 18CD + 181 ]

AD  + DC =  AC =  sqrt (525)

sqrt [ CD^2 + 18CD + 181 ] + CD  = sqrt (525)

sqrt [ CD^2 + 18CD + 181 ]  =  sqrt [ 525 ] - CD        square both sides

CD^2 + 18CD + 181  =  525 - 2 CD*sqrt (525)  + CD^2

18CD  + 181  =  525 - 2CD * sqrt (525)

18CD + 2CD * sqrt (525)  = 525 - 181

CD ( 18  + 2sqrt (525))  = 344

CD  =  344 /  [ 18 + 2sqrt (525) ] =  172 / [ 9 + sqrt 525 ]  

AC - CD =  AD

sqrt (525)  -  172 / [ 9 + sqrt 525] =  AD

(sqrt (525) [ 9 + sqrt (525) ]  - 172) / [ 9 + sqrt (525) ] =  AD

[9sqrt (525) + 525 - 172] / [9 + sqrt 525)  = AD

[9 sqrt (525)  + 353 ] / [ 9 + sqrt (525) ] = AD  ≈  17.523

Using the Law of Sines

AD/ sin (ABC) = AB / sin (ADB)

AD/ sin 45°  = AB / sin 105°

24 * sin 105° / sin 45°  = AB    = 12 + 12sqrt (3)

Angle ACD  =angle ADC =  75°

So  AD  = AC = 24

Area of ABC = (1/2) (AC) ( AB) sin 60° =   (1/2) (24) ( 12 + 12sqrt 3) * sqrt (3) / 2  =

12 * 12  ( 1 + sqrt (3)) sqrt (3)  / 2  =

72 ( 1 + sqrt 3) (sqrt 3)  =

72 ( 3 + sqrt 3)

x^2 - y^2 = 5100

(x + y) ( x - y)  = 5100

Divisors of 5100  = 

1 | 2 | 3 | 4 | 5 | 6 | 10 | 12 | 15 | 17 | 20 | 25 | 30 | 34 | 50 | 51 | 60 | 68 | 75 | 85 | 100 | 102 | 150 | 170 | 204 | 255 | 300 | 340 | 425 | 510 | 850 | 1020 | 1275 | 1700 | 2550 | 5100 (36 divisors)

x             y       

1276    1254

 428       422

 260       250

 100         70

   92         58

   76         26

[Intro: Lexy Medgaus & Jack Stauber]
Can I have a peppermint?
You can have a peppermint
That's too hot for me!
Bleh! Hey! Aaah!
I don't like the peppermint
You don't like the peppermint? (No)
It's too spi-icy

[Outro: Lexy Medgaus & Jack Stauber]
Well, why'd you ask me for a mint?
I don't know
When did the spiciness begin?
I don't know
Is it all because of me?
I don't know
Am I sleeping on the couch?
Yeah

Two men stand back-to-back and walk in opposite directions for $40$ yards each.  Each of them then turns left and walks another $40$ yards each.  In yards, how far are the two men from one another?  

square root of 80² + 80²  =  sqrt(12,800)  =  113.14 yards  

Draw it.  Remember, the men's lefts are opposite directions.      

The men are 80 yds north & south  

 and another 80 yds east & west. 

Just imagine that it's a big box, and solve for the diagonal.  

edited to add:  I calculated the diagonal by using the Pythagorean Theorem.  

Since it's a big square, we could have multiplied a side by the square root of 2.  

_.

We will use Pythagoras' Theorem to solve for AB.  

Actually, we won't even find AB, all we need is AB².  

Then we will use it again to solve for BC.  

AB²  =  17² – 7²  =  289 – 49  =  240    

BC²  =  AB² + AC²  =  240 + 3²  =  249   

BC  =  sqrt(249)  =  15.78    

_.

What is the area of a circle with circumference 16pi?  

Since the circumference is π • d, we can divide by π to get the diameter.  

The diameter, then, equals 16, which makes the radius half of that, i.e., 8.  

Area = π • r² so, substituting values, Area = 3.1416 • 8²  =  201.06 units²    

_.

We can use the pir^2 formula for the circle with radius r.  The area is (16/4)^2*pi = 16*pi.

We are given that PS bisects ∠QPR, so ∠PSQ=∠PSR.

If we add the condition that PQ=PR, then we have SAS congruence, so △PQS≅△PRS.

The other conditions are not enough to guarantee congruence. For example, if SQ=SR, then △PQS and △PRS are similar, but not congruent.

Therefore, the answer is A​.

Absolutely, Keihaku is a notorious homework cheater. They have a long history of lazy plagiarism and should never be trusted to complete academic tasks on their own. The numbers in the question have been altered to prevent Keihaku from finding a convenient answer on the web, which forces them to actually use their brain for once. Unfortunately, they've demonstrated an incredible lack of critical thinking. Not only is Keihaku a shameless cheater, but they also lack basic problem-solving skills and would be well-served by acquiring those skills through diligent study and practice.

See the answer here. Modify it if necessary: https://web2.0calc.com/questions/modular-number-theory#r6

Starting with the "2" add 1, then add 2, then add 4, then add 8, then add 16 .....   

so the next number will be 33.  To continue, keep doubling the amount that you add.   

_.

Since PS = QR and PQ || RS, quadrilateral PQRS is a parallelogram. Therefore, diagonals PR and QS bisect each other at X, so PXQ = RXS and SPR = RQS. Also, SPQ = QRS. So, all the conclusions must be true. The answer is

A,B,C.

Adding the three congruences, we get [5ab + 5bc + 5ca \equiv 11abc \equiv 0 \pmod{13},]so ab+bc+ca≡0(mod13). Subtracting the second congruence from the first, we get [ab + 3bc + 3ca \equiv 0 \pmod{13},]so ab+bc+ca≡−3abc≡0(mod13). Subtracting the third congruence from the first, we get [4ab + 4bc + 4ca \equiv -8abc \pmod{13},]so ab+bc+ca≡−4abc≡0(mod13).

Since ab+bc+ca≡0(mod13), we can write ab+bc+ca=13m for some integer m. Then −4abc=13m, so abc=−3m. Since m is an integer, abc must be divisible by 3. However, a,b,c are positive integers less than 13, and the only multiples of 3 less than 13 are 3, 6, and 9. But abc cannot be 3, because −3m is negative. Also, if abc=6, then m=−2, which is not an integer. Therefore, abc=9, and a+b+c=12​.

Hello Guest!

To solve the equation 5x + 2 - (2x + 1) = 61, you can simplify it step by step:

\(5x - 2x + 2 + 1 = 61\)

\(3x + 3 = 61\)

\(3x + 3 - 3 = 61 - 3\)

\(3x = 58\)

\((3x)/3 = 58/3\)

\(x = 58/3\)

~asinus

!