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3 years ago

What is the value of y? 1 2 7 8

Mathematics

1 answer:

SVETLANKA909090 [29]3 years ago

4 0

Answer:

Step-by-step explanation:

$8y+18=12y-10\\ 12y-8y=10+18\\ 4y=28\\ y=\frac{28}{4} \\ y=7$

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The volume of a cubical box is 474.552m3.Find the length of each side of the box

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The formula of a volume of a cube:

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We have $V=474.552\ m^3$

Substitute:

$a^3=474.552\to a=\sqrt[3]{474.552}\\\\a=7.8\ m$

Answer: The length of each side of the box is equal 7.8 m.

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2 years ago

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3 years ago

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GrogVix [38]

Answer:

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2 years ago

A 160-inch strip of metal 20 inches wide is to be made into a small open trough by bending up two sides on the long side, at rig

babymother [125]

Answer:

The value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.

Step-by-step explanation:

The dimensions of given metal strip are

Length = 160 inch

Width = 20 inch

Let the side bend x inch from each sides to make a open box.

Dimensions of the box are

Length = 160-2x inch

Breadth = 20-2x inch

Height = x inch

The volume of a cuboid is

$V=length\times breadth \times height$

Volume of box is

$V(x)=(160-2x)\times (20-2x)\times x$

$V(x)=(160-2x)(20-2x)x$

$V(x)=4 x^3 - 360 x^2 + 3200 x$

Differentiate with respect to x.

$V'(x)=12x^2 - 720 x + 3200$

Equate V'(x)=0, to find the critical points.

$0=12x^2 - 720 x + 3200$

Using quadratic formula,

$x=30\pm 10\sqrt{\frac{\left(19\right)}{3}}$

The critical values are

$x_1=30+10\sqrt{\frac{\left(19\right)}{3}}\approx 55.166$

$x_2=30-10\sqrt{\frac{\left(19\right)}{3}}\approx 4.834$

Differentiate V'(x) with respect to x.

$V'(x)=24x - 720$

The value of double derivative at critical points are

$V'(55.166)=24(55.166) - 720=603.984$

$V'(4.834)=24(4.834) - 720=-603.984$

Since the value of double derivative at x=4.834 is negative, therefore the trough have a maximum volume at x=4.834 inches.

8 0

3 years ago

Read 2 more answers