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NemiM [27]
3 years ago
5

What is the value of y? 1 2 7 8

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
4 0

Answer:

7

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:

V=a^3

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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Which translation will change figure ABCD to figure A'B'C'D'
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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
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