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3 years ago

Ben drives 405 miles on 15 gallons. How many miles does Ben drive on one gallon? HELP ASAP

Mathematics

2 answers:

Lera25 [3.4K]3 years ago

8 0

Iauida girgel 39 for mayos

AfilCa [17]3 years ago

8 0

This is simple: divide 405 miles by 15 gallons. The answer should be 27 miles per gallon.

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Francis is at the state fair. A fair concession stand is selling funnel cakes for $3.50 and deep-fried Oreos for $2.00. Write an

VladimirAG [237]

The equation that models the number of funnel cakes and Oreos he can buy is 3.50x + 2.0y = 42

Data given;

Cost of cakes = $3.50

Cost of Oreos = $2.00
The total amount spent = $42.00

<h3>What is the Equation</h3>

To solve this problem, we just need to write out an equation to show how he can spend $42.00 in the fair on Oreos and Cakes.

Let x represent the cakes

Let y represent the Oreos

The equation is thus;

$3.50x + 2.0y = 42.00$

The equation that shows the number of Cakes and Oreos can by is

3.50x + 2.0y = 42

Learn more about equation here;

brainly.com/question/13729904

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2 years ago

Which number is greatest?

raketka [301]

A) =27.2
b) = 30.1
c) =31.2
d) =36.6

answer = D

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3 years ago

Read 2 more answers

What decimal is equal to 31/100

BlackZzzverrR [31]

Answer:

0.31

Step-by-step explanation:

So if you divide 31 by 100 all you have to do is move the decimal from right to left the number of zeros you have. Here we have 2 zeros so we move from right to left twice.

0.31

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3 years ago

PRACTICE 1

kari74 [83]

Answer:

ones

Step-by-step explanation:

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2 years ago

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h a

kkurt [141]

Answer:

$b=h=\sqrt{6}$ m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box= $2b^2+4bh$

$2b^2+4bh=36$

$b^2+2bh=18$

$2bh=18-b^2$

$h=\frac{18-b^2}{2b}$

Volume of box, V= $b^2h$

Substitute the values

$V=b^2\times \frac{18-b^2}{2b}$

$V=\frac{1}{2}(18b-b^3)$

Differentiate w. r.t b

$\frac{dV}{db}=\frac{1}{2}(18-3b^2)$

$\frac{dV}{db}=0$

$\frac{1}{2}(18-3b^2)=0$

$\implies 18-3b^2=0$

$\implies 3b^2=18$

$b^2=6$

$b=\pm \sqrt{6}$

$b=\sqrt{6}$

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

$\frac{d^2V}{db^2}=-3b$

At $b=\sqrt{6}$

$\frac{d^2V}{db^2}=-3\sqrt{6}$

Hence, the volume of box is maximum at $b=\sqrt{6}$ .

$h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}$

$h=\frac{18-6}{2\sqrt{6}}$

$h=\frac{12}{2\sqrt{6}}$

$h=\sqrt{6}$

$b=h=\sqrt{6}$ m

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3 years ago