Answer:
The answer is y=-x+9.
The first thing to do is find where the line intercepts the y-axis, which is at positive 9 so you'll add a +9 to the end of your equation. Next you find the slope and since it's going down left to right, you know it is a negative. Now slope is found by change in y divided by change in x. And since both change and y and change in x are both 1, 1/1 is 1. And since you know it's negative that means the slope is -x. Altogether you get y= -x+9
the fractions are set up so that each unit cancels out until its only meters/hour
you can multiply all the numerators together to get 72000 and all the denominators to get 1
72000 meters/1 hour
72000 meters in 1 hour
hope this helped
Answer:
25
Step-by-step explanation:
We require to solve for n, hence
n(n + 1) = 325
multiply both sides by 2 to eliminate the fraction
n(n + 1) = 650
n² + n = 650
subtract 650 from both sides to have equation in standard form
n² + n - 650 = 0 ← in standard form
(n + 26)(n - 25) = 0 ← in factored form
equate each factor to zero and solve for n
n + 26 = 0 ⇒ n = - 26
n - 25 = 0 ⇒ n = 25
however, n > 0 ⇒ n = 25
Answer:
The answers to the first question are A,C,D
The answer to the second question is YZ=16
Step-by-step explanation:
(1st Question)
Since <K and <M Are equal, and both <L's are equal, KL and ML are congruent (Answer choice) because of the ASA postulate.
You need to create the following equation to find the length of KN and MN 7x-4=5x+12
(Get the "x" variable to one side)
2x-4=12
(Isolate the variable)(Remove the 4)
2x=16
(Divide the 2 by itself to remove it from the x, remember to divide both sides by 2)
x=8 (Answer Choice)
Plug in the x value into each equation
KN= 7(8)-4
KN= 56-4
KN= 52
MN= 5(8)+12
MN= 40+12
MN= 52 (Answer Choice)
MN=KN
(Second Question)
Since XWY(20) is half of XWZ(40), ZWY also equals 20.
This now proves the triangle is congruent by the AAS postulate.
Since the triangles are congruent, if XY = 16, YZ also equals 16.
The nature of a graph, which has an even degree and a positive leading coefficient will be<u> up left, up right</u> position
<h3 /><h3>What is the nature of the graph of a quadratic equation?</h3>
The nature of the graphical representation of a quadratic equation with an even degree and a positive leading coefficient will give a parabola curve.
Given that we have a function f(x) = an even degree and a positive leading coefficient. i.e.
The domain of this function varies from -∞ < x < ∞ and the parabolic curve will be positioned on the upward left and upward right x-axis.
Learn more about the graph of a quadratic equation here:
brainly.com/question/9643976
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