Answer: 1 cup of sugar
Step-by-step explanation:
A recipe for banana bread requires 3 cups of bananas for every 1 1/2 cups of sugar used.
Converting 1 1/2 cups of sugar to improper fraction becomes 3/2 cups of sugar.
If 3 cups of bananas is required for 3/2 cups of sugar,
x cups of bananas will require 1 cup of sugar
3x / 2 = 3
3x = 6
x = 6/3 = 2
Therefore,
2 cups of bananas will require 1 cup of sugar.
We are looking for the number of cups of sugar that would require 2 cups of bananas.
We already got the answer.
1 cup of sugar would be used if 2 cups of bananas are used.
Answer:
Step-by-step explanation:
1/6
Answer:
6 units squared
Step-by-step explanation:
To solve this problem you can solve for the entire area of the square and subtract the area of the unfilled areas for the area of the shaded region..
Area of entire square is 4*4=16
Now for the sections that are unfilled:
4*3/2= 6
1*1/2= 0.5
2*1 = 2
1*1 = 0.5
2*1/2 = 1
Total area of Unfilled region = 6 + 0.5 + 2 + 0.5 +1 = 10
Area of filled region = 16 - 10 = 6 units squared
Hope this helped!
Answer:
y = 8
x = 8√5
z = 4√5
Step-by-step explanation:
Using the formula (which is a geometric mean):
a y
---- = -----
x a
Where a is the altitude (y), x is a projection (16), and y is the other projection (4):
y 4
---- = -----
16 y
Cross multiply:
y² = 16 (4)
y² = 64
√y² = √64
y = 8
To find x and z, use the Pythagorean Theorem:
16² + y² = x²
256 + 8² = x²
256 + 64 = x²
320 = x²
√320 = √x²
√64√5 = √x²
8√5 = x
x = 8√5
4² + y² = z²
16 + 64 = z²
80 = z²
√80 = √z²
√16√5 = √z²
4√5 = z
z = 4√5
Answer:
![[B]=2](https://tex.z-dn.net/?f=%5BB%5D%3D2)
Step-by-step explanation:
![B=\left[\begin{array}{ccc}3x-5&-3+2y\\11-3x&2y-5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3x-5%26-3%2B2y%5C%5C11-3x%262y-5%5Cend%7Barray%7D%5Cright%5D)
With 
and 
![B=\left[\begin{array}{ccc}3(3)-5&-3+2(4)\\11-3(3)&2(4)-5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%283%29-5%26-3%2B2%284%29%5C%5C11-3%283%29%262%284%29-5%5Cend%7Barray%7D%5Cright%5D)
![B=\left[\begin{array}{ccc}9-5&-3+8\\11-9&8-5\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D9-5%26-3%2B8%5C%5C11-9%268-5%5Cend%7Barray%7D%5Cright%5D)
![B=\left[\begin{array}{ccc}4&5\\2&3\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%265%5C%5C2%263%5Cend%7Barray%7D%5Cright%5D)
solving the determinant of matrix B
![[B]=\left[\begin{array}{ccc}4&5\\2&3\end{array}\right]=(4.3-5.2)=(12-10)=2](https://tex.z-dn.net/?f=%5BB%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%265%5C%5C2%263%5Cend%7Barray%7D%5Cright%5D%3D%284.3-5.2%29%3D%2812-10%29%3D2)
