All categories

3 years ago

Two customers went to a post office to buy postcards and large envelopes. Each postcard

Mathematics

1 answer:

astraxan [27]3 years ago

4 0

The right answer is Option A.

Step-by-step explanation:

Let,

x be the postcards

y be the large envelops

According to given statement;

14x+5y=12 Eqn 1

10x+15y=24.80 Eqn 2

Multiplying Eqn 1 by 3;

$3(14x+5y=12)\\42x+15y=36\ \ \ Eqn \ 3$

Subtracting Eqn 2 from Eqn 3;

$(42x+15y)-(10x+15y)=36-24.80\\42x+15y-10x-15y=11.20\\32x=11.20\\$

Dividing both sides by 32

$\frac{32x}{32}=\frac{11.20}{32}\\x=0.35$

Putting x=0.35 in Eqn 1;

$14(0.35)+5y=12\\4.90+5y=12\\5y=12-4.90\\5y=7.10$

Dividing both sides by 5

$\frac{5y}{5}=\frac{7.10}{5}\\y=1.42$

Therefore, one large envelope costs $1.42

The right answer is Option A.

Keywords: linear equations, subtraction

Learn more about linear equations at:

#LearnwithBrainly

You might be interested in

When there is a wedge between private costs and social costs, which of the following must be present?

Bogdan [553]

The negative externalities must be present.

4 0

2 years ago

Help me plz I really need help with ixl

qwelly [4]

Answer:

My apologies if this is incorrect, but perhaps it is p=4.

8 0

2 years ago

Read 2 more answers

HELP PLEASE Find 33/4 x 21/3. All answers must be in lowest terms

eimsori [14]

$\frac{231}{4}$ ✅

Step-by-step explanation:

$\frac{33}{4} \times \frac{21}{3} \\ = \frac{11 \times 21}{4} \\ = \frac{231}{4}$

Note:-

A fraction is said to be written in its lowest form if its numerator and denominator are prime, that is, they have no common factors other than 1.

$\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}$

6 0

2 years ago

The length of a rectangle is the sum of the width and 4. The area of the rectangle is 32

Sauron [17]

Answer:

The width of rectangle w = 4 units

Step-by-step explanation:

We are given:

Width of rectangle = w

Length of rectangle = w+ 4

Area of rectangle = 32 units

We need to find width of the rectangle

The formula used is: $Area\:of\:rectangle=Length\times Width$

Putting values and finding width of rectangle

$Area\:of\:rectangle=Length\times Width\\32=(w+4)\times w\\32=w^2+4w\\w^2+4w-32=0$

We need to solve this quadratic equation to find width (w)

We will use factorisation and break the middle term.

$w^2+4w-32=0\\w^2+8w-4w-32=0\\w(w+8)-4(w+8)=0\\(w-4)(w+8)=0\\w-4=0\:or\:w+8=0\\w=4\:or\:w=-8$

Since width of rectangle can't be negative, so rejecting w=-8

Therefore, the width of rectangle w = 4 units

4 0

2 years ago

Michelle is at the bakery buying donuts. She can buy 3 donuts for $2.25, 5 donuts for $3.25, or a dozen donuts for $7.68. Which

horsena [70]

Answer:

the dozen for 7.68 because it is 12 instead of 5 for 4.42 more and ur geting 7 more donuts

5 0

2 years ago