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3 years ago

14

Jack's father let jack use a rectangular section of the back yard that had dimensions of 10 feet by 12 feet for his garden. he w

anted to put a fence around the outside of the garden. how long should the piece of fencing he needs to buy be?

1 answer:

Dafna1 [17]3 years ago

6 0

That would be 12*2 + 10*2 = 24 + 20 = 44 feet

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dlinn [17]

1. Ans: Option (B) $-6x^4$

Explanation:
Given: $2x^4 - 8x^4$
Take the common out:
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Hence: => $[tex]-6x^4$ [/tex] (Option B)

2. Ans: Option (D) $-3y^5$

Explanation:
Given: $6y^5 - 9y^5$
Take the common term(s) out:
=> $y^5(6-9)$

Hence: => $-3y^5$ (Option D)

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Explanation:
</span>Given: $6 - 12x + 13x^2 - 4x^2$

The standard form of polynomial function must have the highest powered value at the start, then the second highest and so on.

=> $13x^2 - 4x^2 - 12x + 6$
=> $9x^2 - 12x + 6$ (Option B)

4. Ans: Option (C) $7.3x^2 + 0.8x + 1.3$

Explanation:
In order to find the total number of C<span>ommon and Endler's guppies, you need to add both Common and Endler's guppies' polynomials, as follows:

</span>Common guppies: $3.1x^2 +6.0x + 0.3$ <span>
Endler's guppies: </span> $4.2x^2 - 5.2x + 1.0$
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Total number: $7.3x^2 +0.8x + 1.3$
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Hence the ans is Option(C) $7.3x^2 + 0.8x + 1.3$

5. Ans: Option (D) $2x^2(133 - 18 \pi )$

Explanation:
First let's find the total area of the yard:
Total Area of the Yard = 14x * 19x = $266x^2$

Now the area of the circular fountain:
Area of the Circular Fountain = $\pi r^2$
Since, r=6x
Therefore,
Area of the Circular Fountain = $\pi (6x)^2 = 36 \pi x^2$

Now the final Area of the yard would be:
Final area of the Yard = Total Area of the Yard - Area of the Circular Fountain
Final area of the Yard = $266x^2$ - $36 \pi x^2$
=> Final area of the Yard = $2x^2(133 - 18 \pi)$ (Option D)

6. Ans: Option (A) $56x^2$

Explanation:
First let's find the total area of the lot:
Total Area of the lot= 6x * 10x = $60x^2$

Now the area of the stadium:
Area of the stadium = 1x * 4x = $4x^2$

Now the final Area of the lot would be:
Final area of the lot= Total Area of the lot - Area of the stadium
Final area of the lot= $60x^2$ - $4x^2$
=> Final area of the lot = $56x^2$ (Option A)

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