Fergus has 3 1/4 cups of oats
= 13/4 cups of oats
Oatmeal bars = 2 2/3
= 8/3 cups of oatmeal used
Therefore cups of oats left = 13/4 - 8/3
= 39/12 - 32/12
= 7/12 cups of oats left
She borrowed 1 cup of oats from neighbour
7/12 + 1/1
= 7/12 + 12/12
= 19/12 cups of oats
Blueberry oats muffin = 1 3/4 cups of oats
= 7/4 cups of oats
Total she will have now = 19/12-7/4
= 19/12 - 21/12
= - 2/12
= -1/6
Which means she needs 1/6 cup of oats more
Therefore its option 4
Must click thanks and mark brainliest
Look at the graph below carefully
Observe the results of shifting ={2}^{x}f(x)=2x
vertically:
The domain, (−∞,∞) remains unchanged.
When the function is shifted up 3 units to ={2}^{x}+3g(x)=2x +3:
The y-intercept shifts up 3 units to (0,4).
The asymptote shifts up 3 units to y=3y=3.
The range becomes (3,∞).
When the function is shifted down 3 units to ={2}^{x}-3h(x)=2 x −3:
The y-intercept shifts down 3 units to (0,−2).
The asymptote also shifts down 3 units to y=-3y=−3.
The range becomes (−3,∞).
Answer:
a. 12 feet b. 12 feet 0.5 inches c. 8.33 %
Step-by-step explanation:
a. How far out horizontally on the ground will it protrude from the building?
Since the rise to run ratio is 1:12 and the building is 12 inches off the ground, let x be the horizontal distance the ramp protrudes.
So, by ratios rise/run = 1/12 = 12/x
1/12 = 12/x
x = 12 × 12
x = 144 inches
Since 12 inches = 1 foot, 144 inches = 144 × 1 inch = 144 × 1 foot/12 inches = 12 feet
b. How long should the ramp be?
The length of the ramp, L is gotten from Pythagoras' theorem since the ramp is a right-angled triangle with sides 12 inches and 144 inches respectively.
So, L = √(12² + 144²)
= √[12² + (12² × 12²)]
= 12√(1 + 144)
= 12√145
= 12 × 12.042
= 144.5 inches
Since 12 inches = 1 foot, 144.5 inches = 144 × 1 inch + 0.5 inches = 144 × 1 foot/12 inches + 0.5 inches = 12 feet 0.5 inches
c. What percent grade is the ramp?
The percentage grade of the ramp = rise/run × 100 %
= 12 inches/144 inches × 100 %
= 1/12 × 100 %
= 0.0833 × 100 %
= 8.33 %
