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2 years ago

Please help I will mark Brainly

Mathematics

1 answer:

Sloan [31]2 years ago

7 0

Answer:

The answer to number 1 is 10. The answer to number 2 is 4-6, inclusive.

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Help with number 17 plz!

love history [14]

Answer:

740 meters of fence
22,000 square meters

Step-by-step explanation:

Two of the sides total 150 +220 = 370 meters. The other two sides are the same lengths, so the sum of lengths of the four sides is ...

2·370 m = 740 m . . . . . length of fence needed

The area of a parallelogram is the product of base and height.

A = bh = (220 m)(100 m) = 22,000 m^2 . . . . area the fence encloses

5 0

3 years ago

I need an equation that has variables on both sides were x = 17

Andrei [34K]

3 (3 + x) = 5x - 25

9 + 3x = 5x - 25

Add 25 to each side

34 + 3x = 5x

Subtract 3x from each side

34 = 2x

Divide each side by 2

17 = x

5 0

3 years ago

2. JI. has coordinates J(-6, 1) and Z(-4, 3). Find the coordinates of the midpoint.

AleksandrR [38]

Answer:

(- 5, 2 )

Step-by-step explanation:

Given endpoints (x₁, y₁ ) and (x₂, y₂ ) then the midpoint is

( $\frac{x_{1}+x_{2} }{2}$ , $\frac{y_{1}+y_{2} }{2}$ )

Here (x₁, y₁ ) = (- 6, 1) and (x₂, y₂ ) = (- 4, 3) thus

midpoint = $\frac{-6-4}{2}$ , $\frac{1+3}{2}$ ) = ( $\frac{-10}{2}$ , $\frac{4}{2}$ ) = (- 5, 2 )

7 0

2 years ago

Maddie does not have any pretzels.

Firdavs [7]

Answer:

(p - 8) / 2 = Maddie's pretzels

7 0

2 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

2 years ago