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Sonja [21]
3 years ago
8

Michael bought four tacos and a burrito and paid $9.50. Noah bought two tacos and three burritos and paid $11. Based on the info

rmation, what is the cost of one taco?
$1.00
$1.50
$1.75
$2.50
Mathematics
2 answers:
natulia [17]3 years ago
6 0
2t+3b=11 (1)*2
4t+b=9.5  (2)
4t+6b=22  (3)
(3)-(2)
5b=12.5
<u>b=2.5</u>
4t+2.5=9.5  (2)
4t=7
<u>taco=$1.75</u>
Afina-wow [57]3 years ago
3 0
In order to solve this equation you could use guess and check with multiple choice, or you can set up a system of equations.
First we must define our variables:
x=price of taco
y=price of burrito
Michael bought 4 tacos (4x) and (+) and Burrito (y) for $9.50 (=9.5)
4x+y=9.50
Noah bought two tacos (2x) and (+) three burritos (3y) for (=) $11 (11)
2x+3y=11
To solve this system of equations you can easily see substitution as the easiest method. For substitution we must isolate one variable:
4x+y=9.50
y=-4x+9.5
We can now replace the y in the other equation:
2x+3(-4x+9.5)=11
Then solve this equation to get the price of a taco (x)
2x-12x+28.5=11
-10x=-17.5
x=1.75
The price of a taco would be $1.75
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The actual height of the Gateway Arch in St. Louis is 630 feet. A scale model of the arch is 9 inches tall. Which statement(s) a
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The length of a 200 square foot rectangular vegetable garden is 4feet less than twice the width. Find the length and width of th
inna [77]

Answer:

Length = 18.099 ft

Width = 11.049 ft

Step-by-step explanation:

let the length of the field be x ft

and the width be y ft

as per the condition given in problem

x=2y-4   -----------(A)

Also the area is given as 200 sqft

Hence

xy=200

Hence from A we get

y(2y-4)=200

taking 2 as GCF out

2y(y-2)=200

Dividing both sides by 2 we get

y(y-2)=100

y^2-2y=100

subtracting 100 from both sides

y^2-2y-100=0

Now we solve the above equation with the help of Quadratic formula which is given in the image attached with this for any equation in form

ax^2+bx+c=0

Here in our case

a=1

b=-1

c=-100

Putting those values in the formula and solving them for y

y=\frac{-(-2)+\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

Solving first

y=\frac{2+\sqrt{4+400}{2}

y=\frac{2+\sqrt{404}{2}

y=\frac{2+20.099}{2}

y=\frac{22.099}{2}

y=11.049

Solving second one

y=\frac{-(-2)-\sqrt{(-2)^2-4 \times (-1) \times 100}}{2 \time 1}

y=\frac{2-\sqrt{4+400}{2}

y=\frac{2-\sqrt{404}{2}

y=\frac{2-20.099}{2}

y=\frac{-18.99}{2}

y=-9.045

Which is wrong as the width can not be in negative

Our width of the field is

y=11.099

Hence the length will be

x=2y-4

x=2(11.049)-4

x=22.099-4

x=18.099

Hence our length x and width y :

Length = 18.099 ft

Width = 11.049 ft

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3 years ago
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