Answer:
Step-by-step explanation:
Hi, to answer this question we have to write a system of equations:
3 T + 4 B =425
T + B = 125
Where T is the number of tacos sold, and B is the number of burritos sold.
Multiplying the second equation by 3, and subtracting it to the first equation:
3 T + 4 B =425
3T + 3B = 375
___________
B =50
Replacing B in any equation:
T + B = 125
T +50 =125
T =125-50
T = 75
Feel free to ask for more if needed or if you did not understand something.
The answer is A. 42
Solution:
Let x= ones digit, y=tens digit
1st condition (original number) : 7(x+y)=10y + x
2nd condition (new number by reversing the digits): 18+x+y=10x+y
simplifying:
1st condition: 6x=3y
2nd condition: x=2
substituting x=2 to 6x=3y
<span>y=4</span>
B. Triangle B and Triangle D
Triangle A has all equal sides and Triangle C is a right triangle.
Answer: M = 2
Step-by-step explanation:
Given equation: m+3=5
You need to isolate m so you can find its value. Subtract 3 from the left side, and do the same to the right side. Subtracting 3 from 5 as well makes the number become 2 on the right side of the equation. Therefore, m=2.
A rule you should always remember is when you have a value on each side of the equal sign, whatever form of addition, subtraction, multiplication, or division should be done to the other side as well.
well, first off let's check those two points, we know it's centerd at (-26 , 120) and we also know it passes through (0 , 0), so the distance between those two points is its radius
![~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{-26}~,~\stackrel{y_2}{120})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ \stackrel{radius}{r}=\sqrt{(~~-26 - 0~~)^2 + (~~120 - 0~~)^2} \implies r=\sqrt{(-26)^2 + (120 )^2} \\\\\\ r=\sqrt{( -26 )^2 + ( 120 )^2} \implies r=\sqrt{ 676 + 14400 } \implies r=\sqrt{ 15076 } \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B0%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-26%7D~%2C~%5Cstackrel%7By_2%7D%7B120%7D%29%5Cqquad%20%5Cqquad%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7Bradius%7D%7Br%7D%3D%5Csqrt%7B%28~~-26%20-%200~~%29%5E2%20%2B%20%28~~120%20-%200~~%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%28-26%29%5E2%20%2B%20%28120%20%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20r%3D%5Csqrt%7B%28%20-26%20%29%5E2%20%2B%20%28%20120%20%29%5E2%7D%20%5Cimplies%20r%3D%5Csqrt%7B%20676%20%2B%2014400%20%7D%20%5Cimplies%20r%3D%5Csqrt%7B%2015076%20%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\textit{equation of a circle}\\\\ (x- h)^2+(y- k)^2= r^2 \hspace{5em}\stackrel{center}{(\underset{-26}{h}~~,~~\underset{120}{k})}\qquad \stackrel{radius}{\underset{\sqrt{15076}}{r}} \\\\[-0.35em] ~\dotfill\\\\ ( ~~ x - (-26) ~~ )^2 ~~ + ~~ ( ~~ y-120 ~~ )^2~~ = ~~(\sqrt{15076})^2 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill (x+26)^2+(y-120)^2 = 15076~\hfill](https://tex.z-dn.net/?f=%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%28x-%20h%29%5E2%2B%28y-%20k%29%5E2%3D%20r%5E2%20%5Chspace%7B5em%7D%5Cstackrel%7Bcenter%7D%7B%28%5Cunderset%7B-26%7D%7Bh%7D~~%2C~~%5Cunderset%7B120%7D%7Bk%7D%29%7D%5Cqquad%20%5Cstackrel%7Bradius%7D%7B%5Cunderset%7B%5Csqrt%7B15076%7D%7D%7Br%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%20~~%20x%20-%20%28-26%29%20~~%20%29%5E2%20~~%20%2B%20~~%20%28%20~~%20y-120%20~~%20%29%5E2~~%20%3D%20~~%28%5Csqrt%7B15076%7D%29%5E2%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20%28x%2B26%29%5E2%2B%28y-120%29%5E2%20%3D%2015076~%5Chfill)
