3(d + 11) = 6(d + 33)
So you have: 3d + 33 = 6d + 198
which is: -3d = 165
or d = -55
So you have <em>only one answer.</em>
Answer:
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The dimensions and volume of the largest box formed by the 18 in. by 35 in. cardboard are;
- Width ≈ 8.89 in., length ≈ 24.89 in., height ≈ 4.55 in.
- Maximum volume of the box is approximately 1048.6 in.³
<h3>How can the dimensions and volume of the box be calculated?</h3>
The given dimensions of the cardboard are;
Width = 18 inches
Length = 35 inches
Let <em>x </em>represent the side lengths of the cut squares, we have;
Width of the box formed = 18 - 2•x
Length of the box = 35 - 2•x
Height of the box = x
Volume, <em>V</em>, of the box is therefore;
V = (18 - 2•x) × (35 - 2•x) × x = 4•x³ - 106•x² + 630•x
By differentiation, at the extreme locations, we have;

Which gives;

6•x² - 106•x + 315 = 0

Therefore;
x ≈ 4.55, or x ≈ -5.55
When x ≈ 4.55, we have;
V = 4•x³ - 106•x² + 630•x
Which gives;
V ≈ 1048.6
When x ≈ -5.55, we have;
V ≈ -7450.8
The dimensions of the box that gives the maximum volume are therefore;
- Width ≈ 18 - 2×4.55 in. = 8.89 in.
- Length of the box ≈ 35 - 2×4.55 in. = 24.89 in.
- The maximum volume of the box, <em>V </em><em> </em>≈ 1048.6 in.³
Learn more about differentiation and integration here:
brainly.com/question/13058734
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Answer:
y= x + 1/4
Step-by-step explanation:
did the math and that should be your answer
Answer:
7) C) PQR ≈ TSR
8) B) 3
9) D) 102 ft
10) D) 21
11) C) 45
Step-by-step explanation:
7) C) PQR ≈ TSR
8) 2 * 6 = 12
so 4x = 12
x = 3
9)
24/20 = x/85
x = 24 * 85 / 20
x = 202 ft
10)
QT/35 = 24/40
QT/35 = 3/5
QT = 35 * 3 / 5
QT = 21
11)
ΔKLM ≈ΔXYZ
9/4 = 2.25
So
XY = 9 ---> given
YZ = 7 * 2.25 = 15.75
XZ = 9 * 2.25 = 20.25
Perimeter ΔXYZ = 9 + 15.75 + 20.25 = 45