You and your friend go to a taco truck for lunch. you order 3 soft tacos and 3 burritos and spend $11.25. your friend orders 4 soft tacos and 2 burritos and spends $10.00.

A soft taco costs $1.25 and a burrito costs $2.50

Step-by-step explanation:

To start to find the cost of each, you need to set up a system of equations. Start by using x for the cost of a soft taco. Then use y for the cost of a burrito. First you can compile an equation for what you paid.

3x + 3y = 11.25

Now you can do the same with your friend's purchase.

4x + 2y = 10.00

Now to solve we can use a number of different methods. We'll use elimination this time. To use that, multiply the first equation by -2 and the second equation by 3. Then line them up like this:

-6x - 6y = -22.50

12x + 6y = 30.00

Now we add them straight through to cancel the y's and solve for x.

6x = 7.50

x = 1.25

Now we know the cost of a soft taco is $1.25 and we can find the burrito cost by using the taco cost in either equation.

3x + 3y = 11.25

3(1.25) + 3y = 11.25

3.75 + 3y = 11.25

3y = 7.50

y = 2.50

So we know that a burrito costs $2.50

d

step-by-step explanation:

d

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)