First: work out the difference (increase) between the two numbers you are comparing. Increase = New number - Original Number
Then: divide the increase by the original number and multiply the answer by 100. % Increase = Increase ÷ Original Numberｘ100<span>
30 - 22 = 8
8</span> ÷ 22ｘ100 = 36.3636363636
Answer:
339 bricks.
Step-by-step explanation:
We have the weight of each brick and what the truck can support. Therefore what we must do is pass all to the same unit of measurement to calculate the quantity of bricks.
In this case we will pass everything to pounds.
We have that a 1 pound is 16 ounces, therefore 14 would be:
14 ounces * 1 pound / 16 ounces = 0.875 pounds
In addition we have that 1 ton is 2204.62 pounds, therefore 3/4 would be:
3/4 ton * 2204.62 pounds / 1 ton = 1653.467 pounds
Therefore, in total the brick weighs 4,875 pounds (4 + 0.875) and the truck can support 1653,467 pounds, the number of bricks would be:
1653.467 / 4.875 = 339.17
In other words, it can support about 339 bricks.
Answer:
T = 96
Step-by-step explanation:
From the question given above, we were told that the time, T (seconds) it takes for a pot of water to boil is inversely proportional to the cooker setting, H. This can be written as:
T ∝ 1/H
T = K/H
Cross multiply
K = TH
Next, we shall determine the value of K. This can be obtained as follow:
H = 8
T = 120
K =?
K = TH
K = 120 × 8
K = 960
Finally, we shall determine the value of T when H = 10. This can be obtained as illustrated below:
H = 10
K = 960
T =.?
T = K/H
T = 960/10
T = 96
Answer:
14 rows. Hope this helped :) !
a. 
× 
A
b. 
× 
A
c. 
A
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
<h3>How to determine the current</h3>
The formula for finding current
I = V/R
Where v = voltage
R = resistance
A. V = 12V
R = 4. 7 + 15 = 19. 7 kΩ = 19700 Ω in series
× A
B. V = 9V
R = 4. 7 + 1 = 4. 7 kΩ = 4700Ω in series
× A
C. V= 12V
1/R = 
= 
×
R = 
= 310. 56 Ω
A
It is important to note that the brightness of a bulb depends on both current and voltage depending on whether the bulb it is in parallel or series.
The LED which would glow brightest is LED C with the greatest current and voltage
The LED which would be the most dim is LED B with low voltage and consequently low current.
Learn more about Ohms law here:
brainly.com/question/14296509
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