For
|a|>15
assume
a>15
a<-15
so
x+6>15
minus 6
x>9
x+6<-15
minu 6
x<-21
B is answer
Hello :)
3x+y=27
-3x+4y= -42
+____________
(3x -3x) + (y+4y) = (27-42)
0 + 5y = -15
y = -3
Put "-3" for "y" in the system:
3x + y =27
3x -3=27
3x = 30
x = 10
Answer = x:10 y: -3
Have a nice day :)
Answer: The correct answer is H.
Step-by-step explanation: The correct order of the weights from largest to smallest is shown in choice H. The correct order is 2.25, 2.234, 2.205.
Answer:
x = -4037/127
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Equality Properties
Step-by-step explanation:
<u>Step 1: Define</u>
14x + 18 + x = -112(x + 36) + 13
<u>Step 2: Solve for </u><em><u>x</u></em>
- Distribute -112: 14x + 18 + x = -112x - 4032 + 13
- Combine like terms: 15x + 18 = -112x - 4019
- Add 112x on both sides; 127x + 18 = -4019
- Isolate <em>x</em> term: 127x = -4037
- Isolate <em>x</em>: x = -4037/127
1.
the x value of the vertex in form
ax^2+bx+c=y
is
-b/2a
so
-2x^2+8x-18
x value of vertex is
-8/(2*-2)=-8/-4=2
plug it in to get y value
-2(2)^2+8(2)-18
-2(4)+16-18
-8-2
-10
vertex is at (2,-10)
or you could complete the square to get into y=a(x-h)^2+k, where the vertex is (h,k)
so as follows
y=(-2x^2+8x)-18
y=-2(x^2-4x)-18
y=-2(x^2-4x+4-4)-18
y=-2((x-2)^2-4)-18
y=-2(x-2)^2+8-18
y=-2(x-2)^2-10
vertex is (2,-10)
5.
vertex is the time where the speed is the highest
at about t=10, the speed is at its max