Let
A(4,0)
B(12,5)
C(20,10)
we know that
The x-intercept of the line is the value of x when the value of y is equal to zero
The y-intercept of the line is the value of y when the value of x is equal to zero
The point A(4,0) is the x-intercept of the line
<u>Find the slope AB</u>
m=(y2-y1)/(x2-x1)
mAB=(5-0)/(12-4)--------> mAB=5/8
<u>with the slope m and the point A(4,0) find the equation of the line</u>
y-y1=m*(x-x1)----------> y-0=(5/8)*(x-4)-------> y=(5/8)x-2.5-----> y=0.625x-2.5
<u>Find the y-intercept</u>
For x=0
y=0.625*0-2.5---------> y=-2.5
the y-intercept is the point (0,-2.5)
therefore
<u>the answer is</u>
the x-intercept is the point (4,0)
the y-intercept is the point (0,-2.5)
see the attached figure to better understand the problem
Answer:
49 cartons
Step-by-step explanation:
1: we need to find how many cartons the roses will fill so
1368/28 = 48.857 cartons.
2: We can't have .857 of a carton, so we need 49 cartons to hold all the roses.
Answer:
mode is the most occuring number which is in this case 11
you get the median by putting them in order from least to greatest and them marking them out until you get to the middle but if there are two numbers in the middle you add them and then divide by two and that is your median but in this case your median is 11
you get the mean by adding all the numbers together and dividing by how many numbers there are so in this case you are dividing by 9 and when you add oall of these numbers together you get 98 and 98/9 is 10.8888888889 which you would round and get 10.9
hope this helps
Part A:
Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4
The perimeter of the square is given by 4(x + 4) = 4x + 16
The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12
For the perimeters to be the same
4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2
The value of x that makes the <span>perimeters of the quadrilaterals the same is 2.
Part B:
The area of the square is given by

The area of the rectangle is given by 2(3x + 4) = 6x + 8
For the areas to be the same

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.
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