solve for c, express your answer in simplest and exact form
a=9(b+c)
To solve for c, we need to isolate it on one side of the equation. Let's start by distributing 9 to (b+c):
a = 9b + 9c
Next, subtract 9b from both sides of the equation to move it to the right side:
a - 9b = 9c
Now, divide both sides of the equation by 9 to solve for c:
c = (a - 9b)/9
Therefore, the solution for c is (a - 9b)/9.
solve for u
Q=1/8(u+w+v)
To solve for u, we need to isolate it on one side of the equation. Let's start by multiplying both sides of the equation by 8 to get rid of the fraction:
8Q = u + w + v
Next, subtract w and v from both sides of the equation to isolate u:
8Q - w - v = u
Therefore, the solution for u is 8Q - w - v.
In a small rectangular wallet photo, the width is 6cm less than the length. If the border (perimeter) of the photo is 20cm, find the length and width.
Part 1 of 2
The length of the rectangular photo is
.
Part 2 of 2
The width of the rectangular photo is
.
Part 1 of 2:
Let's denote the length of the rectangular photo as L. According to the problem statement, the width is 6cm less than the length, so the width can be expressed as L - 6.
Part 2 of 2:
The border, which represents the perimeter of the photo, is given as 20cm. The formula for the perimeter of a rectangle is 2L + 2W, where L represents the length and W represents the width. We can set up the equation as follows:
2L + 2(L - 6) = 20
Simplifying the equation:
2L + 2L - 12 = 20
4L - 12 = 20
4L = 32
L = 8
Therefore, the length of the rectangular photo is 8cm.
Substituting this value into the width formula L - 6:
W = 8 - 6
W = 2
Therefore, the width of the rectangular photo is 2cm.
The perimeter of a triangle is 20ft. One side is 5ft longer than the shortest side. The third side is 3ft longer than the shortest side. Find the lengths of all the sides.
Part 1 of 3
The length of the shortest side is
.
Part 2 of 3
The length of the largest side is
.
Part 3 of 3
The length of the remaining side is
.
Part 1 of 3:
Let's denote the length of the shortest side as x.
Part 2 of 3:
According to the problem, one side is 5ft longer than the shortest side. So, the length of this side can be expressed as x + 5.
Part 3 of 3:
The third side is 3ft longer than the shortest side. So, the length of the third side can be expressed as x + 3.
The perimeter of a triangle is the sum of the lengths of its sides. In this case, the perimeter is given as 20ft.
Therefore, we can set up the equation as follows:
x + (x + 5) + (x + 3) = 20
Simplifying the equation:
3x + 8 = 20
3x = 12
x = 4
So, the length of the shortest side is 4ft.
Substituting this value into the expressions for the other two sides:
The length of the largest side = x + 5 = 4 + 5 = 9ft.
The length of the remaining side = x + 3 = 4 + 3 = 7ft.
Therefore, the lengths of all the sides are:
Shortest side: 4ft
Largest side: 9ft
Remaining side: 7ft.
Two angles are complementary. One angle is 8 degrees less than the other angle. Find the measures of the angles.
Let's denote one angle as x degrees.
Since the two angles are complementary, the other angle can be expressed as 90 degrees minus x degrees.
According to the problem statement, one angle is 8 degrees less than the other angle, so we can set up the following equation:
x = (90 - x) - 8
Now we can solve for x:
x = 90 - x - 8
2x = 82
x = 41
Therefore, one angle measures 41 degrees.
To find the measure of the other angle, we can subtract 41 from 90:
90 - 41 = 49
So, the other angle measures 49 degrees.
In summary, one angle measures 41 degrees and the other angle measures 49 degrees.