The quantity of meat and cheese sold by the deli according to the description accounts in the task content are; 4.5 and 2.5 pounds respectively.
<h3>What is the quantity of meat and cheese sold by the deli?</h3>
The quantity of meat sold by the deli as represented by the variable X in the task content can be calculated by solving the systems of equations.
The quantity of cheese sold by the deli as represented by the variable y in the task content can be calculated by solving the systems of equations.
Consequently, solving the system of equations by means of substitution, we have;
y = (30.50-4x)/5
Hence, we have;
11x + 14((30.50-4x)/5) = 84.50
x = 4.5 pounds of meat and
y = 2.5 pounds of cheese.
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The answer should be A and D
Answer:
x=3
Step-by-step explanation:
I am assuming that your equation is:
![\sqrt{8x + 12} + 4 = 10](https://tex.z-dn.net/?f=%20%5Csqrt%7B8x%20%2B%2012%7D%20%20%2B%204%20%3D%2010)
Subtract 4 from both sides to get:
![\sqrt{8x + 12} = 6](https://tex.z-dn.net/?f=%20%5Csqrt%7B8x%20%2B%2012%7D%20%20%3D%206)
Square both sides to get:
![(\sqrt{8x + 12})^{2} = {6}^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7B8x%20%2B%2012%7D%29%5E%7B2%7D%20%20%3D%20%20%7B6%7D%5E%7B2%7D%20)
![8x + 12 =3 6](https://tex.z-dn.net/?f=8x%20%2B%2012%20%3D3%206)
![8x =24](https://tex.z-dn.net/?f=8x%20%3D24)
Divide both sides by 8 to get:
![x = \frac{24}{8} = 3](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B24%7D%7B8%7D%20%20%3D%203)
Answer:
4(5+2y)=3(3y+7) and -(y+1)
Step-by-step explanation:
You combine like terms
20-21 =-1
8y -9y = -1y
so your answer is anything that is the same as -1 - y
2(10+4y-7y-19) =2( -3y-9) = -6y-18 NO
2(10 +4y) - 3(3y-7) = 20 +8y - 9y +21 = -y +41 NO
4(5+2y-5y-17)= 4(-3y -12) = -12y -48 NO
4(5+2y) - 3(3y+7) = 20 +8y - 9y -21 = -1 = y YES
-(y+1)= -y-1 YES
-(1-y) = -1+y NO
You will need 3quartz
step by step:
98+52=150
200÷4=50
150÷50=3
so the answer is 3