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ArbitrLikvidat [17]
3 years ago
9

Here is a garden and a pond.

Mathematics
1 answer:
Anna11 [10]3 years ago
8 0

Answer:

<h2>Kayla needs 8 boxes of grass seed to cover the whole garden without including the pond.</h2>

Step-by-step explanation:

You can observe a representation of this problem in the image attached.

Notice that the grass seed is gonna be put only in the green zone, out of the pond. So, we need to find each area and subtract them.

A_{garden}= 14 \times 9 =126 \ m^{2}  \\A_{pond}=2.4 \times 3.8 = 9.12 \ m^{2}

The difference is

A_{seed}=126 - 9.12 =  116.88 \ m^{2}

Now, according to the problem, one box of grass seed covers 15 meters of the garden, let's divide this number with the seed area.

n=\frac{116.88}{15} \approx 7.792

Therefore, Kayla needs 8 boxes of grass seed to cover the whole garden without including the pond.

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a) Let's try to put x = 3 for the first equation and we must get an answer equal to 2.25.

y=7.72(3)-29.02=-5.86_{}

Since the value of <em>y</em> is not equal to 2.25 and the deviation is too large. this equation is not a good model,

b) We put x = 3 on the second equation and solve for <em>y</em>

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Since the value of <em>y</em> is not equal to 2.25 and the deviation is too large. this equation is not a good model,

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Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of <em>x</em>. If x = 5, we get

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which has a slight deviation on the given value of <em>y</em> on the table for <em>x</em> = 5. let's try for <em>x</em> = 7. We have

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and the answer has a small deviation compared to the actual value given. The other values of <em>x</em> can again be put on the equation and check their corresponding value of <em>y</em>, and the resulting values are as follows

\begin{gathered} y=0.4(8)^2+0.79(8)-4.93=26.99 \\ y=0.4(12)^2+0.79(12)-4.93=62.15 \\ y=0.4(14)^2+0.79(14)-4.93=84.53 \end{gathered}

And as you can see, the deviation of values from the table to calculated becomes smaller. Hence, this is the best model.

d) We put <em>x</em> = 3 on the third equation and solve for <em>y,</em>

y=4.19(1.02)^3=4.45_{}_{}

Again, the value that we get is not equal to 2.25, hence, this equation is not a good model. But since its value is close to 2.25, we try to other values of <em>x</em>. If x = 5, we get

y=4.19(1.02)^5=4.63

where the answer's deviation is too large compared to the value of <em>y</em> if x = 5 on the table given.

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