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tekilochka [14]
3 years ago
12

Math test assesment chapter 5 answer #9

Mathematics
1 answer:
zaharov [31]3 years ago
5 0
What grade are u in so i can try to help u
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I'm have some issues with the problem shown in the screenshot and would love some help.
KiRa [710]

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;

\frac{d V }{dx}  =  \frac{d( 4 \cdot \:  {x}^{3}  - 106 \cdot \:  {x}^{2}  + 630\cdot \:  {x} )  }{dx} = 0

Which gives;

\frac{d V }{dx}  =12\cdot \:  {x}^{2}  -  212\cdot \:  {x} + 630 = 0

6•x² - 106•x + 315 = 0

x  =  \frac{ - 6 \pm \sqrt{106 ^2 - 4 \times 6 \times 315} }{2 \times 6}

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.

  • Height = x ≈ 4.55 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

#SPJ1

7 0
1 year ago
Which one of the following decimals in the largest?
Natalija [7]
B .35 line the numbers up by the decimals then go from top to bottom starting from each individual column.
7 0
2 years ago
A student is given that point P(a, b) lies on the terminal ray of angle Theta, which is between StartFraction 3 pi Over 2 EndFra
Harman [31]

Answer:

<em>A.</em>

<em>The student made an error in step 3 because a is positive in Quadrant IV; therefore, </em>

<em />cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

Step-by-step explanation:

Given

P\ (a,b)

r = \± \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{-a}{\sqrt{a^2 + b^2}} = -\frac{\sqrt{a^2 + b^2}}{a^2 + b^2}

Required

Where and which error did the student make

Given that the angle is in the 4th quadrant;

The value of r is positive, a is positive but b is negative;

Hence;

r = \sqrt{(a)^2 + (b)^2}

Since a belongs to the x axis and b belongs to the y axis;

cos\theta is calculated as thus

cos\theta = \frac{a}{r}

Substitute r = \sqrt{(a)^2 + (b)^2}

cos\theta = \frac{a}{\sqrt{(a)^2 + (b)^2}}

cos\theta = \frac{a}{\sqrt{a^2 + b^2}}

Rationalize the denominator

cos\theta = \frac{a}{\sqrt{a^2 + b^2}} * \frac{\sqrt{a^2 + b^2}}{\sqrt{a^2 + b^2}}

cos\theta = \frac{a\sqrt{a^2 + b^2}}{a^2 + b^2}

So, from the list of given options;

<em>The student's mistake is that a is positive in quadrant iv and his error is in step 3</em>

3 0
3 years ago
Please help 10 Points​
QveST [7]

Answer:

c

Step-by-step explanation:

8 0
3 years ago
I’ll give brainliest to whoever answers thiss
andrey2020 [161]

Answer:

21.25

Step-by-step explanation:

diagonal is 21.25

the area is 216.75

the perimeter is 59.5

6 0
2 years ago
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