The rule that has been given means that x coordinate of some point we increase by 8 and y coordinate of some point we increase by 5.
Coordinates of K point on original Figure are:
(-2,-3)
once we implement rule on this we get K':
K' ( -2+8,-3+5)
or
K' ( 6,2)
Answer is third option.
If you have a calculator with statistical functions, that's the way to go.
On my TI-83, I typed in invNorm(0.88) and got the result z = 1.17.
88% of the area under the normal curve is to the left of z = 1.17.
Answer:
x = -10; x = 7
Step-by-step explanation:
|2x + 3| - 6 =11
Add 6 to each side.
|2x + 3| = 17
Apply the absolute rule: If |x| = a, then x = a or x = -a.
(1) 2x + 3 = 17 (2) 2x + 3 = -17
Subtract 3 from each side
2x = 14 2x = -20
Divide each side by 2
x = 7 x = -10
<em>Check:
</em>
(1) |2(7) + 3| - 6 = 11 (2) |2(-10) + 3| - 6 = 11
|14 + 3| - 6 = 11 |-20 + 3| - 6 = 11
|17| - 6 = 11 |-17| - 6 = 1
1
17 - 6 = 11 17 - 6 = 11
11 = 11 11 = 11
The answer would be 7/24.
So distribute using distributive property
a(b+c)=ab+ac so
split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24
the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24
