The slope is 4/3x.
Since the y-intercept is (0,-3), you go up and see that the next point is (1,1).
The paper is 7 cm by (x + 8) cm.
A photo is 5 cm by (x + 6) cm, and is printed in the middle of the paper.
The area of the border/space around the photo is 50 cm².
To find the area of the border around the photo, you find the area of the paper and subtract it by the area of the photo.
The area of the paper:
A = 7 · (x + 8)
The area of the photo:
A = 5 · (x + 6)
Your equation to find the area of the border around the photo is:
(7 · (x + 8)) - (5 · (x + 6)) = A
[area of paper - area of photo = area of border]
Since you know the area of the border, you can plug it in:
(7 · (x + 8)) - (5 · (x + 6)) = 50
Multiply the 7 into (x + 8), and multiply the 5 into (x + 6)
(7x + 56) - (5x + 30) = 50
Distribute/multiply the - to (5x + 30)
7x + 56 - 5x - 30 = 50
Combine like terms
2x + 26 = 50
Subtract 26 on both sides
2x = 24
Divide 2 on both sides
x = 12
Answer:
7
Step-by-step explanation:
σ = 4 ; μ =?
8.52 to the left of X
.
P(X < 8.52) = 64.8%
P(X < 8.52) = 0.648
Using the Z relation :
(x - μ) / σ
P(Z < (8.52 - μ) / 4)) = 0.648
The Z value of 0.648 of the lower tail is equal to 0.38 (Z probability calculator)
Z = 8.52 - μ / 4
0.38 = 8.52 - μ / 4
0.38 * 4 = 8.52 - μ
1.52 = 8.52 - μ
μ = 8.52 - 1.52
μ = 7
Answer:
Use Desmos.com/calculator
Step-by-step explanation:
Remember, y=mx+b
y is equal to any given y point
x is equal to any given x point
m is equal to slope
b is equal to the y-intercept, or where x = 0 and the line crosses the horizon line.
In order to graph the line correctly, you have to isolate y.
y+x=-3
y=-x-3 would be equal to y=mx+b format
slope is negative 1
y intercept is negative 3
start on the y line, go to (0,-3) and start your line.
slope is negative 1, so you go down one and right one.
This is a really interesting question! One thing that we can notice right off the bat is that each of the circles has the same amount of area swept out of it - namely, the amount swept out by one of the interior angles of the hexagon. Let’s call that interior angle θ. We know that the amount of area swept out in the circle is proportional to the angle swept out - mathematically
θ/360 = a/A
Where “a” is the area swept out by θ, and A is the area of the whole circle, which, given a radius of r, is πr^2. Substituting this in, we have
θ/360 = a/(πr^2)
Solving for “a”:
a = π(r^2)θ/360
So, we have the formula for the area of one of those sectors; all we need to do now is find θ and multiply our result by 6, since we have 6 circles. We can preempt this but just multiplying both sides of the formula by 6:
6a = 6π(r^2)θ/360
Which simplifies to
6a = π(r^2)θ/60
Now, how do we find θ? Let’s look first at the exterior angles of a hexagon. Imagine if you were taking a walk around a hexagon. At each corner, you turn some angle and keep walking. You make 6 turns in all, and in the end, you find yourself right back at the same place you started; you turned 360 degrees in total. On a regular hexagon, you’d turn by the same angle at each corner, which means that each of the six turns is 360/6 = 60 degrees. Since each interior and exterior angle pair up to make 180 degrees (a straight line), we can simply subtract that exterior angle from 180 to find θ, obtaining an angle of 180 - 60 = 120 degrees.
Finally, we substitute θ into our earlier formula to find that
6a = π(r^2)120/60
Or
6a = 2πr^2
So, the area of all six sectors is 2πr^2, or the area of two circles with radii r.
