Well, since the graph starts at (0,50), making our y-intercept 50, we can knock off D as an option.
The next step is to find the slope. To do this, find two points and subtract the y values over the x values. (Just make sure that whichever y value you start with - start with the same x value).
So we have >>> (50, 200) and (0,50) from earlier.
200 - 50 / 50 - 0 = 150/50 or 3/1, or 3.
This makes the correct answer A since the equation is represented as y = mx + b, where m is the slope and m has a value of 3.
Answer:
1) 6x = 21
2) x + y - 3
3) x/z = y
4) 2-x = p
Step-by-step explanation:
1. The product of a number x and 6 is 21
A product is a multiplication. A product of a and b is a * b.
We then have a product of x and 6, that x * 6, which we write usually in the format 6x.
is 21: that means it's equal to 21....
so 6x = 21.
2. The sum of the quantity x- 3 and y
The sum is an addition. The sum of a and b is a + b.
In this case, the first part is x - 3, the second part is y
So, x - 3 + y, which we usually rewrite as x + y - 3
3. The quotient of x and z is y
A quotient is a division.
So, quotient of x and z is x/z.
x/z = y
4. The difference of 2 and x is p.
A difference is a subtraction.
Difference of 2 and x is 2 - x
2 - x = p
3,806,000 is the correct answer.
There are three in the millions place and 806 in the thousands. There are no numbers in the hundreds so you place zeros.
Answer:
<h2>4/3 Joules </h2>
Step-by-step explanation:
Work is said to be done when force applied to an object causes the object to move through a distance.
Work done = Force * perpendicular distance.
Given Force F = xy i + (y-x) j and a straight line (-1, -2) to (1, 2)
First we need to get the equation of the straight line given.
Given the slope intercept form y = mx+c
m is the slope
c is the intercept
m = y₂-y₁/x₂-x₁
m = 2-(-2)/1-(-1)
m = 4/2
m = 2
To get the slope we will substtutte any f the point and the slope into the formula y = mx+c
Using the point (1,2)
2 = 2+c
c = 0
y = 2x
Substituting y = 2x into the value of the force F = xy i + (y-x) j we will have;
F = x(2x) i + (2x - x) j
Using the coordinate (1, 2) as the value of s
![W = \int\limits^a_b ({2x^2 i + x j}) \, (i+2j)\\W = \int\limits^a_b ({2x^{2}+2x }) \, dx \\W = [\frac{2x^{3} }{3} +x^{2} ]\left \ x_2=1} \atop {x_1=-1}} \right.\\W = (2(1)^3/3 + 1^2) - (2(-1)^3/3 + (-1)^2)\\W =(2/3+1) - (-2/3+1)\\W = 2/3+2/3+1-1\\W = 4/3 Joules](https://tex.z-dn.net/?f=W%20%3D%20%5Cint%5Climits%5Ea_b%20%28%7B2x%5E2%20i%20%2B%20x%20j%7D%29%20%5C%2C%20%28i%2B2j%29%5C%5CW%20%3D%20%5Cint%5Climits%5Ea_b%20%28%7B2x%5E%7B2%7D%2B2x%20%7D%29%20%5C%2C%20dx%20%5C%5CW%20%3D%20%5B%5Cfrac%7B2x%5E%7B3%7D%20%7D%7B3%7D%20%2Bx%5E%7B2%7D%20%5D%5Cleft%20%5C%20x_2%3D1%7D%20%5Catop%20%7Bx_1%3D-1%7D%7D%20%5Cright.%5C%5CW%20%3D%20%282%281%29%5E3%2F3%20%2B%201%5E2%29%20-%20%20%282%28-1%29%5E3%2F3%20%2B%20%28-1%29%5E2%29%5C%5CW%20%3D%282%2F3%2B1%29%20-%20%28-2%2F3%2B1%29%5C%5CW%20%3D%202%2F3%2B2%2F3%2B1-1%5C%5CW%20%3D%204%2F3%20Joules)
