The volume of the tower is 224 in³, which is not listed among the given options.
To find the volume of the rectangular block, we need to multiply its length, width, and height. Therefore, the volume of the single block is:
8 x 3.5 x 2 = 56 cubic inches.
Since we are stacking four blocks to create a tower, we need to multiply the volume of a single block by 4.
Thus, the volume of the tower is:
56 x 4 = 224 cubic inches.
The volume of a rectangular block can be found using the formula,
V = length × width × height.
In this case, the length is 8 inches, the width is 3.5 inches, and the height is 2 inches.
V = 8 × 3.5 × 2 V
= 28 × 2 V
= 56 in³
Now that we know the volume of one block, we can find the volume of the tower created by stacking four blocks. Tower volume = block volume × number of blocks Tower volume = 56 in³ × 4 Tower volume = 224 in³
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Answer and Step-by-step explanation:
Let x and y be two positive integers and their sum is 14:
X + y = 14
And the sum of square of this number is:
f = x2 + y2
= x2+ (14 – x)2
Differentiate with respect to x, we get:
F’(x) = [ x2 + (14 – x)2]’ = 0
2x + 2(14-x)(-1) = 0
2x +( 28 – 2x)(-1) = 0
2x – 28 +2x = 0
2x + 2x = 28
4x = 28
X = 7
Hence, y = 14 – x = 14 -7 = 7
Now taking second derivative test:
F”(x) > 0
For x = y = 7,f reaches its maximum value:
(7)2 + (7)2 = 49 + 49
= 98
F at endpoints x Є [ 0, 14]
F(0) = 02 + (14 – 0)2
= 196
F(14) = (14)2 + (14 – 14)2
= 196
Hence the sum of squares of these numbers is minimum when x = y = 7
And maximum when numbers are 0 and 14.
To find two positive integers such that their sum is 14, and the sum of their squares is minimized, we need to consider all possible pairs of positive integers and calculate their sums of squares. The pair (6, 8) has the minimum sum of squares of 100. To find two positive integers such that their sum is 14, and the sum of their squares is maximized, the pairs (1, 13) and (2, 12) both have the maximum sum of squares of 170. Since we need to find two positive integers, the pair (1, 13) is the answer.
To find two positive integers such that their sum is 14 and the sum of their squares is minimized, we need to consider all possible pairs of positive integers that add up to 14 and calculate their sums of squares. Let's list all the pairs:
From the list, we can see that the pair (6, 8) has the minimum sum of squares, which is 100.
Similarly, to find two positive integers such that their sum is 14 and the sum of their squares is maximized, we need to again consider all possible pairs and calculate their sums of squares. Let's list the pairs:
From the list, we can see that the pair (1, 13) and the pair (2, 12) both have the maximum sum of squares, which is 170. Since we need to find two positive integers, the pair (1, 13) is the answer.
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72
Answer:
72 = {2,2,2,3,3}
Step-by-step explanation:
72
89
2433
22
Answer:
y<3
Step-by-step explanation:
Let's solve your inequality step-by-step.
5y+9<24
Step 1: Subtract 9 from both sides.
5y+9−9<24−9
5y<15
Step 2: Divide both sides by 5.
5y/5 < 15/5
y<3
Answer:
Step-by-step explanation:
Total number of tickets sold = 3388
Total number of coach tickets = 3069
Total number of first-class tickets = Total number of tickets sold - Total number of coach tickets
= 3388 - 3069
= 319
Ratio of the number of first-class tickets to the total number of tickets = 319:3388
Answer:
Step-by-step explanation:
Given, Total no. of tickets sold = 3388
Total no. of coach tickets = 3069
Then, No. of first class ticket:
= 3388 - 3069
= 319
We need to find the ratio of first-class tickets to the total number of tickets: 319:3388
b. you add 6000 kilograms of water to the pool. what is the depth of the water in the pool? write your answer as a fraction. the water is about meters deep.
Answer:
1). Mass of water present= 18000 kg
2).4/3 meters deep
Step-by-step explanation:
Area of the rectangle= 6*3= 18m²
Volume of water in the pool
= Deepness of water*area of rectangle
= 1*18
= 18 m³
density of water is about 1 gram per cubic centimeter
In kg per m³= 1000 kg/me
Mass of water present= density*volume
Mass of water present= 1000*18
Mass of water present= 18000 kg
2)6000 kilograms of water is added to 18000 of
Total mass present= 6000+18000
Total mass present=24000 kg
If density= 1000kg/m³
Volume present= mass/density
Volume present= 24000/1000
Volume present= 24 m³
Area of the rectangle= 18 m²
deepness of the pool= volume/area
deepness of the pool= 24/18
deepness of the pool= 4/3 meters deep