Answer:
B. 1.7 in
Step-by-step explanation:
Divide the triangle into 2 congruent right angle triangles
With hypotenuse: 2
Base: 2/2 = 1
Using pythagoras theorem
2² = 1² + h²
h² = 3
h = sqrt(3)
h = 1.732050808
Answer:
36/52 or 9/13
Step-by-step explanation:
Ace-9 for all suits; Diamond, Hearts, Spades and Clubs
Essentially when people ask you find the solution to system of equation, there asking at what x value do these to graphs intersect. The easiest way to do this is to get a graphing calculator, or desmos and type in the equation and find where they intersect. Heck, even the question says to solve it with a graph, but I'll demonstrate it algebraically.
One way you can do this is set the equation equal to each other. This is because you want to know at what x-value has the same y-value. So we get:
x^2 + 6x + 8 = x + 4
We can then combine like terms, or move everything to one side. So we get:
x^2 + 5x + 4 = 0.
Then we can use the quadratic formula to solve for x.
x=(-5 +/- sqrt(5^2 - 4(1)(4)))/(2(1)
This simplifies into:
(-5 +/- 3)/2
Finally we add and subtract:
(-5 + 3)/2 = x = -1
(-5 - 3)/2 = x = -4
And our solution is x = -1, x = -4
Answer:
The answer to your question is y - 0 = 40(x - 15)
Step-by-step explanation:
y = 40x + 600
Convert to y - y₁ = m(x - x₁)
Process
1.- Factor 40 in the right side of the equation
y = 40(x - 15)
2.- Give the form of the point slope form
y - 0 = 40(x - 15)
where
y₁ = 0
m = 40
x₁ = 15
Answer: 
Step-by-step explanation:
Given
Position of the particle moving along the coordinate axis is given by

Speed of the particle is given by

Acceleration of the particle is

velocity can be negative, but speed cannot
