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4 years ago

A group of friends ordered 3 burgers and 3 medium fries for a total of $17.55

Mathematics

2 answers:

S_A_V [24]4 years ago

7 0

This is gonna be a long answer but it will help...

Answer:

Cost price of the burger = $4.15

Step-by-step explanation:

Given data:

3 burgers were ordered

3 medium fries were ordered

Total order cost 17.55

Let the price of the burger be X

Let the price of the medium fries be Y

Equation for the total sale will be 3X + 3Y = 17.55 (equation 1)

from this statement 'The price of a burger is .75 more than two times the price of medium fries.'

X - 2Y = 0.75 (equation 2)

There are now two equations and two unknowns, a simultaneous equation is solved to find the cost of the burger.

Many methods can be used to solve these equations but in this case substitution will be used

From equation 2

X = 0.75 + 2Y (equation 3)

Substituting equation 3 into equation 1

3(0.75 + 2Y) + 3Y = 17.55

2.25 + 6Y + 3Y = 17.55

6Y + 3Y = 17.55-2.25

9Y = 15.3

Y = 1.7

Substituting Y into equation 3

X = 0.75 + 2Y

X = 0.75 + 2(1.7)

X = 0.75 + 3.4

X = 4.15

Therefore the cost price of the burger is $4.15

Pavel [41]4 years ago

5 0

2m+.75=1 burger cost

m= 1 medium fry

2m+.75+2m+.75+2m+.75+m+m+m=17.55

9m+2.25=17.55

-2.25 -2.25

9m=15.30

/9 /9

m=1.70

1.70*2=3.40

3.40+.75=4.15

1 burger costs $4.15

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The first digit of a string of 2002 digits is a 1. Any two-digit number formed by consecutive digits within this string is divis

Umnica [9.8K]

The largest possible last digit in the string of 2002 digits and number divisible by 19 or 31 is 9.

Given the first digit of a string of 2002 digits is 1 and the two digit number formed by consecutive digits within the string is divisible by 19 or 31.

We have to tell the last largest digit of such number.

Two digit numbers divisible by 19=19,38,57,76,95.

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Hence the largest possible last digit and number divisible by 19 or 31 in this string is 9.

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4 0

2 years ago

Complete the following statement of congruence: MON~= .

AlexFokin [52]

Answer:

C. RTS

Step-by-step explanation:

When two triangles are congruent, they will have congruent vertices, like each vertex has a partner.

Take a look at the shape of the triangle.

If you look at the triangle on the right, you can see it is almost like a right triangle.

"R" is at the right angle. "S" is on the short side and "T" is on the long side.

In the triangle on the left, "M" is at the right angle. "N" is on the short side and "O" is on the long side.

Thus, the pairs of congruent angles are:

M ≅ R

O ≅ T

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When you write the statement of congruence between triangles, order the letters by their congruent pair.

MON ≅ RTS

8 0

3 years ago

Which statement is true? A.27/19<11/30 B.17/31>19/14 C.16/26>30/31 D.35/30<22/12

Mrac [35]

A. $\frac{27}{19}$

To find greater than or smaller than relation, we multiply the terms like (numerator of L.H.S with denominator of R.H.S and put the value on the left side. Then multiply the denominator of L.H.S with numerator of R.H.S and put the value on right side. Now compare the digits.)

So, solving A, we get 810<209 ... This is false

B. $\frac{17}{31}>\frac{19}{14}$