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aalyn [17]
3 years ago
8

Alisa hopes to play beach volleyball in the Olympics someday. She has convinced her parents to allow her to set up a beach volle

yball court in their backyard. She goes to the hardware store to shop for sand and sees the following signs on pallets containing bags of sand.
Sand A- 1 bag is 60 lbs
3 bags are for $120
Sand B- 1 bag is 50 lbs
2 bags are for $150

1- What is the rate that Brand A is selling for? (Price for 1 pound) or (pounds per 1 dollar)
2- Which brand is offering the better value? (Explain)
Mathematics
2 answers:
VladimirAG [237]3 years ago
5 0
The unit rate of brand A is,
1 pound : $2
The unit rate of brand b is,
1 pound : $3
Given that information, along with the fact that brand A is selling the sand at a price of $120 for 3 bags compared to brand B with 2 bags for $150, brand A is the better buy.
Naily [24]3 years ago
4 0
60x3\120 = $1.50 per pound
100/150=.67
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Sudhir bought an almirah for RS 13600 and spent RS 400 on its transportation . He sold it for RS 16800 . find the his gain perce
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Answer:

His gain percent is 20%

Step-by-step explanation:

In this question, we want to find the percentage gain of the business.

We need the following parameters;

The cost price which is the sum of the amount she bought the goods + the amount spent on transport

Using the values we have in the question;

The cost price will be;

13,600 + 400 = Rs 14,000

The selling price = Rs 16,800

Mathematically the percentage gain will be;

% gain = (selling price- cost price)/selling price * 100%

% gain = (16,800-14,000)/14,000 * 100/1

% gain = 2800/14,000 * 100/1 = 1/5 * 100/1 = 20%

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3 years ago
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2
Mekhanik [1.2K]

Answer:

x=1

y=s

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Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

y = e^{-2t} *sin 4t\\0 = e^{-2t} *sin 4t\\\\0 = e^{-2t} *(\frac{e^{4t}-e^{-4t}}{2j} )\\0 = e^{-2t} *(e^{4t}-e^{-4t} )\\0 = e^{-2t} *e^{4t}*(1-e^{-8t} )\\\\0 = e^{2t}*(1-e^{-8t} )\\\\\\Either   \\0 = e^{2t}\\t = -inf\\or\\0 = (1-e^{-8t} )\\(e^{-8t} ) = 1\\ -8t = ln(1) =0\\t=0\\\\

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t  = e^−2(0)  = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

\frac{dx}{dt}=-2te^{-2t} cos4t + e^{-2t}(-4sin4t)=-2(0)e^{-2(0)} cos4(0) + e^{-2(0)}(-4sin4(0))=0\\\\ \frac{dy}{dt} =-2te^{-2t} sin4t + e^{-2t}(4cos4t)=-2(0)e^{-2(0)} sin4(0)+ e^{-2(0)(4cos4(0)) }= 1\\\\\frac{dz}{dt}  = -2te^{-2t}  = -2(0)e^{-2(0)}  = 0

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

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miskamm [114]

Answer:

Step-by-step explanation:

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