Given the equation y=2x+3 and the points (-2,7), (1,1), (3,-3) show the slope is the same between these points? Show your work
1 answer:
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Answer:
According to tangents secant segments theorem,
11) x(16+x) = (x + 6)^2
x (16+x) = (x + 6)^2
16x + x^2 = x^2 + 2(6)(x) + 6^2
16x + x^2 = x^2 + 12x + 36
16x - 12x + x^2 - x^2 = 36
4x = 36
x = 36/4
x = 4
13) x(x + 5) = (x + 2)^2
x^2 + 5x = x^2 + 2x 2(2)(x) + 2^2
x^2 + 5x = x^2 + 4x + 4
x^2 -x^2 + 5x - 4x = 4
x = 4
14) x + 8( x + 8 + 32) = (3x) ^2
x + 8(x + 40) = 9x^2
x(x + 8) + 40(x + 8) = 9x^2
x^2 + 8x + 40x + 320 = 9x^2
x^2 + 48x + 320 =9x^2
48x + 320 = 9x^2 - x^2
48x + 320 = 8x^2
Dividing the whole eq. by 8,
6x = 40 = x^2
0 = x^2 - 6x - 40
0 = x^2 - 10x + 4x - 40
0 = x(x - 10) + 4(x - 10)
x - 10 = 0 OR x + 4 = 0
x = 10 OR x = - 4
length cannot be negative, so,
x = 10
15) (x + 3) ( x + 3 + 15) = (2x) ^2
(x + 3) (x + 18) = 4x^2
x(x + 18) + 3(x + 18) = 4x^2
x^2 + 18x + 3x + 54 = 4x^2
x^2 + 21x + 54 = 4x^2
0 = 4x^2 - x^2 - 21x - 54
0 = 3x^2 - 21x - 54
Dividing the whole eq. by 3,
x^2 - 7x - 18 = 0
x^2 - 7x - 18 = 0
x^2 - 9x + 2x - 18 = 0
x(x - 9) + 2 (x - 9)
x + 2 = 0 Or x - 9 = 0
x = -2 or x = 9
length cannot be negative,so,
x = 9
Answer:
If then
and
a | b | a + b (answer)
0 | 0 | 0
0 | 1 | 1
0 | 2 | 2
1 | 0 | 1
2 | 0 | 2
1 | 1 | 2
2 | 1 | 3
Step-by-step explanation:
Considering the following conditions for the real numbers:
Following the rules of these in-equations, it is possible to deduce:
Then, if the proposed statement is:
The conditions above shall comply the requirements established, but first, analyzing the statement:
If and
then
,
and
.
If and b a non negative real number, then
, but because to
, then
. Due to the commutative property of sums, the same behavior will be presented if
and a a non negative real number.
According to that, if , then
and
.