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3x^2 + 12x + 4 - 2x^2 + 11x - 7      simplify

x^2 + 23x -3                                  complete the square for 'x'

(x + 11.5)²  - 135.25      You can finish !    (hint: a =1)

t^2 +2t + 9       in simplified form     is a bowl shaped parabola   the minumum     

                              will occur at 't' value = - b/2a    where b = 2   a = 1        

                                     Use this value of 't' in the equation to calculate the minumum value fo the quadratic

8 8 12

9 9 10

10 10 8

11 11 6

12 12 4

13 13 2

Edit: Nevermind, I got it now! All I had to do was change the signs (+ to -) and - to +! Thank you so much for your help!! 

Hello! Thank you for your great explanation! It really helped me understand this better! However, the question said that it was wrong because it wants the format to be in a > or >= format. How would I change x+3y+8<=0 to >=? Thank you so much again!

The y intercept isn't too clear in this picture.

Since you've already found the slope, I would recommend using the point-slope formula, y - y_1 = m(x - x_1).

One point on the graph is (1, -3) 

y - -3 = -1/3(x - 1)

y + 3 = -1/3x + 1/3

y  = -1/3x - 8/3

3y = -x - 8

x + 3y + 8 = 0

Since the yellow goes down the y axis. 

x + 3y + 8 <= 0

=^._.^=

The answer is 15.

The answer is sqrt(5)/2.

The largest possible value of a is 4.

\(f(x) = \begin{cases} ax^2 & \text{if } x \geq a,\\ ax +2a& \text{if } x

\( f(x) = \begin{cases} ax^2 & \text{if } x \geq a,\\ ax +2a& \text{if } x

The first digit is 5. 

5 _ _

There are 2 cases.

5 9 a, where a is not 9, or 5 b 9, where b is not 9. 

9 options for a, 9 options for b. 9 + 9 = 18

=^._.^=

Yikes, sorry, HT.

Hey there, Guest!

\(x^2+10x+25=8+17\)

Step 1: Simplify both sides of the equation.

\(x^2+10x+25=25\)

Step 2: Subtract 25 from both sides.

\(x^2+10x+25−25=25−25\)

\(x^2+10x=0\)

Step 3: Factor the left side of the equation.

\(x(x+10)=0\)

Step 4: Set factors equal to 0.

\(x=0\) or \(x+10=0\)

x=0 or x=−10

Therefore your answer is -10.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

Let's find the critical points of the inequality:

\(x^3+4x=5x\)

\(x^3+4x−5x=5x−5x\) (Subtract 5x from both sides)

\(x^3−x=0 \)

\(x(x+1)(x−1)=0\) (Factor left side of the equation)

\(x=0\) or \(x=-1\) or \(x=1\)

Check intervals in between critical points. (Test values in the intervals to see if they work.)

\(x<−1\) (Doesn't work in original inequality)

\(−1  (Works in original inequality)

\(0  (Doesn't work in original inequality)

\(x>1\) (Works in original inequality)

Therefore the answers to your equation are: 

−1  or  x>1

Hope this helped! :)

( ﾟдﾟ)つ Bye

1/   Semiperimeter       s = 1/2(a + b + c)

2/   Area      A = sqrt[s (s - a)(s - b)(s - c)]

3/   Inradius     r = A / s  

Re-write as   x^2 + 10x = 0

                        x(x+10) = 0        x = 0   or  -10        

sine(x + pi/2) = cos(x)

In this problem, we can ignore the 2x. 

Since the functions have a period of 2pi, n = 2pi*x + pi/2, how many integer x values make n inbetween 0 and 100?

=^._.^=

The center is (a, 8) since it is tangent to the y-axis at the point (0,8). 

Since (12, 0) and (0, 8) both touch it, the center is the same distance from them. 

sqrt((12-a)^2 + (8-0)^2) = sqrt((a-0)^2 + (8-8)^2)

144 - 24a + a^2 + 64 = a^2 + 0

a = 26/3

(26/3, 8) is the center of the circle. 

The radius is the distance from the center to one of the points, (26/3, 8) to (0, 8) has a distance of 26/3.

=^._.^=

Image below

Use cosine law     BC^2 = 8^2 + 13^2 - 2 ( 8)(13) cos 30         You can finish.... remember : this is to find BC²

5x^2 - kx + 8 = 4x^2 has no real solution if the discriminant is negative. 

5x^2 - kx + 8 = 4x^2

x^2 - kx + 8 = 0

b^2 - 4ac < 0

(-k)^2 - 4(1)(8) < 0

(-k)^2 < 32

-sqrt(32) < k < sqrt(32)

-4sqrt(2) < k < 4sqrt(2)

=^._.^=

re-write as

x^2 - kx + 8 = 0          discriminant of Quadratic Formula must be < 0 for no real roots

b^2 - 4ac < 0

k^2 - 4(1)(8) < 0

k^2 < 32

k  <   sqrt 32         sqrt 32 = 5.65    so largest integer would be 5

yup, but if I remember correctly, it was a triangle with one angle being 30, one side being 13, and the other... being... uhhh... 8? I think.

Taking a really long time to download the diagram........