Answer:
In the first basket, Callie can make 15 apples, 10 oranges and 5 bananas.
In the second basket, she can make 12 apples, 8 oranges and 4 bananas and
in the third basket, she can make 9 apples, 6 oranges and 3 bananas.
Step-by-step explanation:
The largest basket contains 30 pieces of fruit. So, it contains
1/2 of 30 = 15 apples
1/3 of 30 = 10 oranges and
1/6 of 30 = 5 bananas.
The second and third baskets should contain at least 12 pieces and at most 30 pieces (since the largest basket contains 30 pieces). Also, the number should be divided by 2, 3 and 6. So, it must be a multiple of 6.
Clearly, the number of multiples of 6 between 12 and 30 are 18 and 24.
Hence, the second basket contains 24 pieces of fruits and it contains
1/2 of 24 = 12 apples
1/3 of 24 = 8 oranges and
1/6 of 24 = 4 bananas
The third basket contains 18 pieces of fruits and it contains
1/2 of 18 = 9 apples
1/3 of 18 = 6 oranges and
1/6 of 18 = 3 bananas.
I am not sure if I am right, but this is the best I can answer.
Useing reason, and logic, my guess would be if you take 125, and divided it by 4, you would get 31.25. that should be the answer... if i am wrong, i am truely sorry. I havent been in the subject in a while...
Hope this helps.
Answer:
15/17
Step-by-step explanation:
Sine is
, and we see that the angle opposite of A is 15 while the hypotenuse is 17. Plug that into your formula and you get
.
Problem 1)
AC is only perpendicular to EF if angle ADE is 90 degrees
(angle ADE) + (angle DAE) + (angle AED) = 180
(angle ADE) + (44) + (48) = 180
(angle ADE) + 92 = 180
(angle ADE) + 92 - 92 = 180 - 92
angle ADE = 88
Since angle ADE is actually 88 degrees, we do NOT have a right angle so we do NOT have a right triangle
Triangle AED is acute (all 3 angles are less than 90 degrees)
So because angle ADE is NOT 90 degrees, this means
AC is NOT perpendicular to EF-------------------------------------------------------------
Problem 2)
a)
The center is (2,-3) The center is (h,k) and we can see that h = 2 and k = -3. It might help to write (x-2)^2+(y+3)^2 = 9 into (x-2)^2+(y-(-3))^2 = 3^3 then compare it to (x-h)^2 + (y-k)^2 = r^2
---------------------
b)
The radius is 3 and the diameter is 6From part a), we have (x-2)^2+(y-(-3))^2 = 3^3 matching (x-h)^2 + (y-k)^2 = r^2
where
h = 2
k = -3
r = 3
so, radius = r = 3
diameter = d = 2*r = 2*3 = 6
---------------------
c)
The graph is shown in the image attachment. It is a circle with center point C = (2,-3) and radius r = 3.
Some points on the circle are
A = (2, 0)
B = (5, -3)
D = (2, -6)
E = (-1, -3)
Note how the distance from the center C to some point on the circle, say point B, is 3 units. In other words segment BC = 3.