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olya-2409 [2.1K]
3 years ago
15

help,for,points and brainliest math easy math hi a b c d e f g h i j k l m n o p q r s t u v w s y and z

Mathematics
1 answer:
Marrrta [24]3 years ago
3 0
The answers are
D. 21
E. 30
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the surface of a rectangle is 352mm squared. two of the dimensions are 4 mm 8mm. find the measure of the other dimension​
Nostrana [21]

Answer:

pihrfghb4yysrjrhna4r\h

Step-by-step explanation:

4 0
3 years ago
The constraints of a problem are listed below. What are the vertices of the feasible region?
Luda [366]

Answer: option 1.

Explanation:

feasible region is that region which is formed by the lines of constraints.

feasible region is shaded in the attached graph

inequalities becomes equalities to draw the graph

and lines will head towards the origin if constraint satisfied by putting x= 0, y=0

and on the contrary lines will move away from origin when condition of constraint does not satisfied.

3 0
3 years ago
Read 2 more answers
Find the average value of f over region
yan [13]
The area of D is given by:

\int\limits \int\limits {1} \, dA = \int\limits_0^7 \int\limits_0^{x^2} {1} \, dydx  \\  \\ = \int\limits^7_0 {x^2} \, dx =\left. \frac{x^3}{3} \right|_0^7= \frac{343}{3}

The average value of f over D is given by:

\frac{1}{ \frac{343}{3} }  \int\limits^7_0  \int\limits^{x^2}_0 {4x\sin(y)} \, dydx  = -\frac{3}{343}  \int\limits^7_0 {4x\cos(x^2)} \, dx  \\  \\ =-\frac{3}{343} \int\limits^{49}_0 {2\cos(t)} \, dt=-\frac{6}{343} \left[\sin(t)\right]_0^{49} \, dt=-\frac{6}{343}\sin49
3 0
3 years ago
What value of x makes this equation true? x/3 −6=10
Setler [38]

Answer:

48

Step-by-step explanation:

x/3 - 6 = 10

      +6   +6

x/3=16

16*3=48

5 0
3 years ago
You are constructing a cardboard box from a piece of cardboard with the dimensions 2 m by 4 m. You then cut equal-size squares f
postnew [5]

Answer:

Dimensions:

l =3,15 m

w=1,15 m

x= 0,42 m   (height)

V(max) = 1,52 m³

Step-by-step explanation:

The cardboard is L = 4 m   and   W = 2 m

Let call x the length of the square to cut in each corner, then, volume of open box is:

For the side L       is  L - 2*x               l =  4 - 2*x

For the side W      is  W - 2*x             w=  2 - 2*x

The height              is  x

Volume of the open box, as function of x is:

V(x) = ( 4 -2x) * ( 2 - 2x) *x    ⇒  V(x) = ( 8 - 8x -4x + 4x²) *x

V(x) = ( 8 -12x + 4x² )*x     V(x) = 8x - 12x² + 4x³

V(x) = 8x - 12x² + 4x³

Taking derivatives on both sides of the equation

V´(x) = 8 - 24x + 12x²

V´(x) = 0       8  - 24x + 12x² = 0     reordering    12x² - 24x + 8  = 0

or    3x² - 6x + 2 = 0

A second degree equation. Solving for x

x₁,₂  = ( 6 ± √36 - 24 ) /6

x₁,₂  = ( 6 ± 3.46) /6

x₁  =  6 + 3,46 /6     x₁  = 1.58  we dismiss such solution because 1,58 * 2 = 3,15 and is bigger than 2 one of the side of the cardboard

x₂  =( 6 - 3,46 ) / 6

x₂  = 0,42 m

Dimensions of the open box

l  = 4 - 2*x     l  = 4 - 0,85      l  =  3,15 m

w = 2 -2*x    w  = 2 - 0,85     w = 1,15 m

x = 0,42 m

V(max) =3,15*1,15*0,42

V(max) = 1,52 m³

8 0
3 years ago
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