Answer:
The third equation: 
Step-by-step explanation:
The two points on the line are 
and 
.
Slope of the line passing through two points 
and 
is given as:
Here, and are and .
Therefore, slope is equal to, 
Now, equation of a straight line with slope m and points and is given as:
Now, if we use the 2nd form, then 
.
So, the equation is given as :
Answer:
answer is 4
Step-by-step explanation:
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To use the product of powers rule, the powers must have the same base. In this expression, the bases are 3 and 2, which are different, so the rule does not apply.
Answer:
2x + 12y + 43 ≥ 40
17x + 12y + 8z ≥ 20
14x + 6y + z ≤ 50
x ≥ 0
y ≥ 0
z ≥ 0
Step-by-step explanation:
given:
Cost Eggs = $2
Cost of edema = $5
cost of elbow Macaroni = $3
Lets eggs = x,
edamame = y
elbow macaroni = z
TC = 2x+5y+3z
Therefore;
2x + 12y + 43 ≥ 40
17x + 12y + 8z ≥ 20
14x + 6y + z ≤ 50
x ≥ 0
y ≥ 0
z ≥ 0
the first objective is to make sure the total cost is subject to the required nutritional requirements.
So the total cost function (TC) is denoted by the number of servings multiplied for each costs. Eggs cost $2, edamame $5, and macaroni $3.
The problem subjects that each meal contains at least 40g of carbohydrates (this is the condition).
to get this we need to add what each meal component adds to the total, eggs add 2g of carbs, edamame 12g, and macaroni 43g.
Same should be done for protein, we require at least 20 grams of protein, Eggs add 17g, edamame adds 12g, and macaroni adds 8g.
and lastly we don't want more than 50 grams of fat, Eggs add 14g, edamame add 6g and macaroni 1g.
check the picture below on the top side.
we know that x = 4 = b, therefore, using the 30-60-90 rule, h = 4√3, and DC = 4+8+4 = 16.
![\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=8\\ b=\stackrel{DC}{16}\\ h=4\sqrt{3} \end{cases}\implies A=\cfrac{4\sqrt{3}(8+16)}{2} \\\\\\ A=2\sqrt{3}(24)\implies \boxed{A=48\sqrt{3}}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%0AA%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%0A%5Cbegin%7Bcases%7D%0Aa%2Cb%3D%5Cstackrel%7Bbases%7D%7Bparallel~sides%7D%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Aa%3D8%5C%5C%0Ab%3D%5Cstackrel%7BDC%7D%7B16%7D%5C%5C%0Ah%3D4%5Csqrt%7B3%7D%0A%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B4%5Csqrt%7B3%7D%288%2B16%29%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D2%5Csqrt%7B3%7D%2824%29%5Cimplies%20%5Cboxed%7BA%3D48%5Csqrt%7B3%7D%7D)
now, check the picture below on the bottom side.
since we know x = 9, then b = 9, therefore DC = 9+6+9 = 24, and h = b = 9.
![\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=6\\ b=\stackrel{DC}{24}\\ h=9 \end{cases}\implies A=\cfrac{9(6+24)}{2} \\\\\\ A=\cfrac{9(30)}{2}\implies \boxed{A=135}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%0AA%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%0A%5Cbegin%7Bcases%7D%0Aa%2Cb%3D%5Cstackrel%7Bbases%7D%7Bparallel~sides%7D%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Aa%3D6%5C%5C%0Ab%3D%5Cstackrel%7BDC%7D%7B24%7D%5C%5C%0Ah%3D9%0A%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B9%286%2B24%29%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D%5Ccfrac%7B9%2830%29%7D%7B2%7D%5Cimplies%20%5Cboxed%7BA%3D135%7D)
