1) Stephanie earned $x while working 10 hrs. Evelyn earned $y while working 29hrs. If they both earn the same hourly wage and x+y=60, how many dollars did Stephanie earn?
2)the difference of the areas of two circles is 21pi. If their radio are r+3 and r, find the radius of the larger circle.
^144(pi)
3) The figure with a length of 5, width of 4, and hight of 3. (Rectangular box) what is the longes length of a diagonal of one of the faces?
steph;s wage = x/10
eve;a wage = y/29
but you said:
x/10=y/29
10y=29x
y = 29x/10
sub that into x+y=60 and solve
2.
area of large circle = π(r+3)^2
area of small circle = πr^2
π(r+3)^2 - πr^2 = 21π
divide by π(the r^2 drops out)
3. make a sketch and really think about it
Hint: Pythagoras
x/10 = (60-x)/29
dividing out the pi,
(r+3)^2 - r^2 = 21
pick the two longest edges, then
d^2 = 5^2+4^2
Number two is 21/9?
uh, no. You messed up the binomial expansion.
6r+9 = 21
nope, not what I got
show me your steps to catch the error
ohh opps. Thank you
1) To solve this problem, we can set up a system of equations. Let's let the hourly wage be represented by the variable "w".
Stephanie earned $x while working 10 hours, so her earnings can be represented as 10w.
Evelyn earned $y while working 29 hours, so her earnings can be represented as 29w.
Since their hourly wages are the same, we can equate these two expressions:
10w = 29w
Now, we can solve for w by subtracting 10w from both sides to get:
0 = 19w
Dividing both sides by 19, we find that w = 0.
Now, we know that x + y = 60. Since x = 10w and w = 0, we can conclude that x = 10(0) = 0.
Therefore, Stephanie earned $0.
2) The area of a circle can be calculated using the formula A = πr^2, where A represents the area and r represents the radius.
Let's consider the larger circle with the radius represented as r + 3, and the smaller circle with the radius represented as r.
The difference of the areas of the two circles is given as 21π:
(Area of larger circle) - (Area of smaller circle) = 21π
We can substitute the formulas for the areas of the circles:
π(r + 3)^2 - πr^2 = 21π
Expanding the square and cancelling out like terms by subtracting the right side from both sides of the equation:
r^2 + 6r + 9 - r^2 = 21
Simplifying:
6r + 9 = 21
Subtracting 9 from both sides:
6r = 12
Dividing both sides by 6, we find that r = 2.
Therefore, the radius of the larger circle is 2.
3) To find the longest length of a diagonal of one of the faces of a rectangular box with length 5, width 4, and height 3, we can use the Pythagorean theorem.
The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side, also referred to as the diagonal) is equal to the sum of the squares of the other two sides.
In this case, the two sides would be the length (5) and the width (4). So, we have:
Diagonal^2 = Length^2 + Width^2
Diagonal^2 = 5^2 + 4^2
Diagonal^2 = 25 + 16
Diagonal^2 = 41
Taking the square root of both sides, we find:
Diagonal = √41
Therefore, the longest length of a diagonal of one of the faces of the rectangular box is √41.