All Answers 356758 Answers

Simple;

PQR is an isosceles triangle, and Q and P are reflected over the one line of symmetry, which happens to be PM if M is the midpoint of PR.

The possible lengths are 3 and \(\sqrt{31}\) .

Reason: 3² + 4² = 5² & 4² + 5² = 31.

1/3 + 2/3^2 + 3/3^3 + 4/3^4 + 5/3^5 +..........+k/3^k

sum_(n=1)^k 3^(-n) n = 1/4 3^(-k) (-2 k + 3^(k + 1) - 3) =converges to 3/4. See here:

https://www.wolframalpha.com/input/?i=1%2F3+%2B+2%2F3%5E2+%2B+3%2F3%5E3+%2B+4%2F3%5E4+%2B+5%2F3%5E5+%2B..........%2Bk%2F3%5Ek

The coordinates of C are  (2c, d)

The coordinates of D are  ( c, 0 )

The slope of AB and DC = 

[ (b + d) - b] / [ (a + c) - a ]  =  [ (d - 0)] / [ 2c - c]  =   [ d / c ]

The slope of  AD and BC =

[ ( b - 0)] /  [(a - c} =  [ (b + d) - d ] / [  (a + c) - 2c]  =  [ b / (a - c) ]

This might help you!
 

This may help you!

D. You can map △ABC onto △A′B′C′ by rotating it 180° about the origin and translating it 2 units right, which is a sequence of rigid motions.

Camille dilates a digital photograph by a factor of 1.25 to create a new digital photograph similar to her original photograph.

x^3 -2x^2 - 3x < 0

Let's solve this

x^3 - 2x^2 - 3x  = 0      factor

x( x^2 - 2x - 3)  = 0

x( x - 3) ( x + 1)  = 0

Set   each factor to 0   and solve for x   and the solutions are

x = 0 , x = 3   and x = -1

We have four possible solution intervals

(-inf, -1), ( - 1, 0) , (0, 3)  and (3, inf )

This graph shows that the intervals in red are the correct solution intervals :

https://www.desmos.com/calculator/z8f6gksyuq

35x^2 - 51x + 18 = 0

The solutions are  

x= 3/5  and x = 6/7

Letting  a = 3, b= 5, c = 6 and d = 7   we have that

a (b + d)  + c ( b + d)  =

(a + c) (b + d) =

( 9) ( 12)   =  108

P.S. -  Letting  a = 6, b = 7, c = 3 and d = 5   produces the same result  .........

(x^2 + 3x - 1)^2  < 9

Let us solve this

(x^2 + 3x - 1)^2  = 9

For this to be true, either

x^2 + 3x - 1  =  3              or            x^2 + 3x - 1  =  -3          so

x^2 + 3x - 4  = 0                              x^2 + 3x + 2  = 0

(x + 4) ( x - 1)  = 0                            (x + 1) ( x + 2)  = 0

x = -4, x= 1                                          x = -1, x = -2

So  we have  5  possiible intervals to test

(-inf ,  -4) ,  (-4, -2) , (-2, - 1), (-1, 1) and (1, inf) 

Picking a test point in each interval, the intervals highlighted in red are the solution intervals

Here's the graph to prove this : https://www.desmos.com/calculator/znkgxcywlo

And..as saseflower would say....the sum of the x values that solve this system is just  0

QR is the hypotenuse of this right triangle....and we have that.....

tan R  = 4 / sin R

sinR * tanR  = 4

sin^2R / cosR =  4

(1 - cos^2R) / cosR  = 4

1 - cos^2R = 4cosR

cos^2R + 4cosR - 1  = 0

Let x = cosR  ........ so....

x^2 + 4x - 1   = 0

Solving this for x  gives  

x = - √5 - 2       or   x  =   √5 - 2

However....since R  is acute......the second value will only be good for the cosine

So x =  cos R  =  √5 - 2

So

Cos R  =  PR / QR

√5 - 2  =  PR / 20

So     ....    PR   =  20 ( √5 - 2) 

x^4 - 10x^2 + 9

Taking the derivative, we have that

4x^3 - 20x  = 0    factoring, we have that

4x ( x^2 - 5)  = 0

Setting each factor to 0  and solving for x produces

x = 0,  x = √ 5,   x = - √ 5

Taking the second derivative, we have

12x^2 - 20

Putting  0  into this produces a negative...so this is a relative max at x =0

Subbing the othertwo values into the second derivative produces positives...so....we have absolute minimums  at  x  = ±√5 

3. 

1. Given

2. Measure of Angle 1 + Measure of Angle 2  = 90°

3. BD bisects ADC

4. Defintion of Bisection

5. A bisected angle forms a pair of angles whose measures is equal

6.  Measure of Angle 1 + Measure of Angle 3 = 90°

7.Angle 1 is complementary to Angle 3     ....   Definition of Complementary

2.     Scale factor (5/3)  =   center of dilation is (2, -2)

Translate  the center of dilation to the origin < -2, 2 >

This gives us   < x - 2, x + 2 >

Apply the scale factor

5/3 < x - 2, x + 2 >  =   < (5/3)x - 10/3, 1.6x + 10/3 >

Shift back again < (5/3)x - 10/3 + 2,  (5/3)x + 10/3 - 2 >

So....the rule is  ( (5/3)x - 4/3, (5/3)x  + 4/3)

S remains the same = S'

T' =  ( (5/3)(-1) - 4/3, (5/3)(-2) + 4/3)  =  ( -3, - 2)

R'  =  ((5/3)(2) - 4/3 ,  (5/3)(2) + 4/3)  =  (2, 14/3 )

Thank you so much! I fixed it

You made a small mistake:

x = +or-​ 8sqrt(2)/3

thanks

3y - 49 = 0

x^2 + y^2 - 281 = 0

First I would isolate the variable (y) in the first equation

3y - 49 = 0

Add 49 on both sides

3y = 49

Divide by 3 on both sides

y = 49/3

Then take your new equation and plug it into the second equation

x^2 + (49/3)^2 - 281 = 0

Simplify

x^2 + (2401/9) - 281 = 0

Subtract 281 from the fraction

x^2 + (-128/9) = 0

Add (128/9) on both sides

x^2 = (128/9)

Root on both sides

|x| = \(\sqrt{\frac{128}{9}}\)

Simplify (the x is a multiplication)

|x| = \(\frac{\sqrt{64 x 2}}{3}\)

Simplify again

|x| = \(\frac{8\sqrt{2}}{3}\)

Then the absolute value turns the other side into ±

x = ±\(\frac{8\sqrt{2}}{3}\)

And there's your answer! :) Hope this helps

Here is the graph of it: https://www.desmos.com/calculator/ox7g6zpagq

Notice that it comes down to a point twice, meaning it has 2 relative minimums.

Just divide:

(2 4/6) / (4/5) =

(16/6) x (5/4)  =

80 / 24 = 3 1/3 - Pounds of potatoes to fill the entire container.

 (4x^2 + 6x + 4)(4y^2 - 12y + 25) = 28

(2x^2 + 3x + 2) ( 4y^2  - 12y + 25)  =  14

WolframAlpha shows the two solutions as 

x = -3/4    y = 3/2

Thank you so much! :)