Answer:
<h2><em>
2ft by 2ft by 1 ft</em></h2>
Step-by-step explanation:
Total surface of the cardboard box is expressed as S = 2LW + 2WH + 2LH where L is the length of the box, W is the width and H is the height of the box. Since the cardboard box is without a lid, then the total surface area will be expressed as;
S = lw+2wh+2lh ... 1
Given the volume V = lwh = 4ft³ ... 2
From equation 2;
h = 4/lw
Substituting into r[equation 1;
S = lw + 2w(4/lw)+ 2l(4/lw)
S = lw+8/l+8/w
Differentiating the resulting equation with respect to w and l will give;
dS/dw = l + (-8w⁻²)
dS/dw = l - 8/w²
Similarly,
dS/dl = w + (-8l⁻²)
dS/dw = w - 8/l²
At turning point, ds/dw = 0 and ds/dl = 0
l - 8/w² = 0 and w - 8/l² = 0
l = 8/w² and w =8/l²
l = 8/(8/l² )²
l = 8/(64/I⁴)
l = 8*l⁴/64
l = l⁴/8
8l = l⁴
l³ = 8
l = ∛8
l = 2
Hence the length of the box is 2 feet
Substituting l = 2 into the function l = 8/w² to get the eidth w
2 = 8/w²
1 = 4/w²
w² = 4
w = 2 ft
width of the cardboard is 2 ft
Since Volume = lwh
4 = 2(2)h
4 = 4h
h = 1 ft
Height of the cardboard is 1 ft
<em>The dimensions of the box that requires the least amount of cardboard is 2ft by 2ft by 1 ft</em>
<h3>
Answer: A) 20+10h</h3>
Work Shown:
It grows 2h inches for 4 weeks, so 2h*4 = 2*4h = 8h is the amount it grows for those 4 weeks.
Then it grows h inches per week, for 2 more weeks, giving h+h = 2h additional inches.
Overall, 8h+2h = 10h is the amount the plant grows.
It starts off at 20 inches so its final height is 20+10h
The answer is <span>4 3/4 yards of the fabric.</span>
This can be calculated using the proportion.
If the 3 1/6 yards of the fabric is enough for <span>1 1/3 shirts, how many yards are necessary of 2 shirts:
</span>
Let's express 3 1/6 as 19/6:
Similarly, 1 1/3 = 4/3:
Now, let's use this in the proportion:
<span>
</span>
Crossing the products:
⇒
⇒
⇒
Therefore, Mary needs in total 4 3/4 yards of the fabric.
Answer:
B/ x=4
Step-by-step explanation:
Just take 2x to another side to be neg-, the same thing with 4 , then divide it both sides by 4 to get the answer.
Let's look at numbers with the same digit in different places and see if we can determine some relationship.
Consider the number 20.
Now, consider the number 200, which has the 2 in the location just to the left of where it is in 20. You're expect to observe that the number 200 is <em>ten times</em> the number 20.
Consider the number with the 2 in the position to the right of where it is in 20. That number is 2. You are expected to observe that the number 2 is <em>one-tenth</em> the number 20.
The place-value of a digit increases by a factor of 10 when moved one place left, and is reduced by a factor of 1/10 when moved one place right.
_____
This is what makes a place-value number system work. In Roman Numerals, for example, the value of a character is changed by ...
- putting it ahead of or after a higher-value character: IV, VI
- changing the character: I, V, X, L, C, D, M
Place-value number systems don't have to have 10 as their base. We use 60 for the base in (minutes):(seconds), both for time and angle measures. We use 2, 8, or 16 as the base in the binary, octal, and hexadecimal numbers used by computer systems. These other place-value systems have the same characteristic: the value of a digit is increased by a factor of the base when moved to the left, and decreased by a factor of the base when moved to the right. (The hexadecimal value A7C0 has 16 times the value of A7C, for example, and 1/16th the value of A7C00.)
