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abruzzese [7]
3 years ago
14

Bill spent 70 on fertilizer and weed killer for his lawn .Each pound of fertilizer cost 50 cents and each once of weed killer co

st 3.50 . Write an equation that correctly represent this relationship if x represents the number of pounds of fertilizer and y the number of ouncescof weed killer
Mathematics
1 answer:
meriva3 years ago
5 0

An equation that correctly represent this relationship if x represents the number of pounds of fertilizer and y the number of ounces of weed killer is  0.5x + 3.5y = 70

<u>Solution:</u>

Given that, Bill spent 70 on fertilizer and weed killer for his lawn

Each pound of fertilizer cost 50 cents and each once of weed killer cost 3.50

We have to write an equation, if x represents the number of pounds of fertilizer and y the number of ounces of weed killer.

Now,<em> total amount = amount for fertilizers + amount for weed killers.</em>

Total amount = number of pounds of fertilizers x cost per pound + number of ounces of weed killer x cost per ounce.

\begin{array}{l}{\rightarrow 70=x \times 0.50+y \times 3.50} \\\\ {\rightarrow 0.5 x+3.5 y=70}\end{array}

Hence, the equation is 0.5x + 3.5y = 70

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A person stands 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 meters
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Answer:

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

Step-by-step explanation:

Given that,

A person stand 10 meters east of an intersection and watches a car driving towards the intersection from the north at 13 m/s.

From Pythagorean Theorem,

(The distance between car and person)²= (The distance of the car from intersection)²+ (The distance of the person from intersection)²+

Assume that the distance of the car from the intersection and from the person be x and y at any time t respectively.

∴y²= x²+10²

\Rightarrow y=\sqrt{x^2+100}

Differentiating with respect to t

\frac{dy}{dt}=\frac{1}{2\sqrt{x^2+100}}. 2x\frac{dx}{dt}

\Rightarrow \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}. \frac{dx}{dt}

Since the car driving towards the intersection at 13 m/s.

so,\frac{dx}{dt}=-13

\therefore \frac{dy}{dt}=\frac{x}{\sqrt{x^2+100}}.(-13)

Now

\therefore \frac{dy}{dt}|_{x=24}=\frac{24}{\sqrt{24^2+100}}.(-13)

               =\frac{24\times (-13)}{\sqrt{676}}

               =\frac{24\times (-13)}{26}

               = -12 m/s

Negative sign denotes the distance between the car and the person decrease.

Therefore the rate change of distance between the car and the person at the instant, the car is 24 m from the intersection is 12 m/s.

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Step-by-step explanation:

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