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3 years ago

The vertices of JKL are J(4,3), K(-3,4), L(-1,-1). Graph the image of JKL after a reflection in the line y=-x.

Mathematics

1 answer:

nalin [4]3 years ago

6 0

y=-x

J(-4,3)

K(3,-4)

L(1,1)

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Jake is arranging a poetry reading. There will be 6 poets. The reading should run 90 minutes. Each poet will read for the same l

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Answer:

Each poet reads for exactly 15 minutes.

Step-by-step explanation:

If it should run for 90 minutes, and there is 6 poets, therefore I'm pretty sure you divide 90 by 6, which is 15.

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3 years ago

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4 years ago

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Answer:

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3 years ago

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3 years ago

A farmer has $1500 available to build an E-shaped fence along a straight river so as to create two identical pastures. The mater

Dahasolnce [82]

Answer: The largest dimensions that are possible would be 50 foot by 125 foot.

Step-by-step explanation:

Since we have given that

Amount a farmer has available = $1500

Let x be the side perpendicular to the river.

Let y be the side parallel to the river.

Number of section perpendicular to the river = 3

cost of material for the side parallel to the river = $6 per foot

Cost of material for the side perpendicular to the river = $5 per foot

So, total cost becomes

$Cost=6y+5(3x)=1500\\\\Cost=6y+15x=1500\\\\y=\dfrac{1500-15x}{6}}$

Area would be

$A=x\times y\\\\A=x(\dfrac{1500-15x}{6})\\\\A=\dfrac{1}{6}(1500x-15x^2)\\\\A'=\dfrac{1}{6}(1500-30x)$

Now, put A' = 0 to get the critical points.

So, it becomes,

$\dfrac{1}{6}(1500-30x)=0\\\\1500-30x=0\\\\1500=30x\\\\x=\dfrac{1500}{30}\\\\x=50$

$A''=\dfrac{-30}{6}=-5$

so, at x= 50 it will give maximum dimensions.

$y=\dfrac{1500-15x}{6}=\dfrac{1500-15\times 50}{6}=\dfrac{750}{6}=125$

So, the largest dimensions that are possible would be 50 foot by 125 foot.

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3 years ago