Okay so i would do this with a logarithm.
5^(2x) = 28
take the log of both sides
log5^(2x) = log28 and because of this log rule: logA^n = nlogA
(2x)log5= log28
then divide by log5 on both sides
2x = log28 / log5
and divide by 2
x = (log28 / log5) /2
this can be found with a calculator:
x= about 1.03521...
round that to whatever the question says to
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>
The answer is 15. Just count the spaces between A and D.
Let the smallest integer be x
The second integer = x + 2
The third integer = x + 4
x + x + 2 = x + 4 + 24
2x + 2 = x + 28
2x - x = 28 - 2
x = 26
So the numbers are 26, 28, 30
