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3 years ago

PLZ ANSWER THIS QUICK! THANKS!

Mathematics

1 answer:

KengaRu [80]3 years ago

6 0

Answer:

a) 300m

b) 15 miles

Step-by-step explanation:

a) 1 : 30 000

= 1x 30 000

= 30 000cm

= 30 000cm/ 100 Representing in metres

= 300m

b) 1 : 6

= 1x 2.5 : 6x 2.5

= 2.5 : 15

= 15 miles

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Please show your work I will mark you Brainiest

Talja [164]

Answer:

1. 7

2. 5/8

3. 5

4. 20

Step-by-step explanation:

There is one burger that weighs 3/8 pounds, four that weigh 1/2 pound, and two that weigh 3/4 pounds. The difference between the heaviest and the lightest is 3/4-1/8=5/8. The total combined weight is 5 pounds. 5/ 1/4= 5*4=20. Hope this helps!

5 0

3 years ago

Read 2 more answers

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

3 years ago

A football team charges $30 per ticket and averages 20,000 people per game. Each person spend an average of $8 on concessions.Fo

olya-2409 [2.1K]

The ticket price that would maximize the total revenue would be $ 23.

Given that a football team charges $ 30 per ticket and averages 20,000 people per game, and each person spend an average of $ 8 on concessions, and for every drop of $ 1 in price, the attendance rises by 800 people, to determine what ticket price should the team charge to maximize total revenue, the following calculation must be performed: