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Setler79 [48]
3 years ago
10

Jayden wants to lay sod on his front yard and on half of his back yard. His front yard has a length of 60 feet and a width of 80

feet. His back yard has a length of 20 feet and a width of 30 feet. How many square feet of sod does Jayden need to purchase?
Mathematics
2 answers:
finlep [7]3 years ago
4 0

Answer:

he needs 165 square feet to purchase

Step-by-step explanation:

its says half of the back yard witch is 20 length and 30 width if you divide 20÷2=10 and 30÷2=15 10+15=25 that's the backyard in the front if you do 80+60=140 feet than you do 140+25=165 feet now that's what you get for your answer

Arturiano [62]3 years ago
3 0

Answer:

5,100 sq.feet

Step-by-step explanation:

60x80= 4,800

and then 20x30= 600 you have to divide by 2 if he only wants 1/2 of the back yard. So 4,800+300= 5,100 sq. feet

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To the nearest hundredth, what is the value of m ?
cluponka [151]

Answer:

12.99

Step-by-step explanation:

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7 0
3 years ago
SOME ONE HELP ME I REALLY NEED A SOLUTION FOR THIS PROBLEM WITH A CLEAR EXPLANATION
Sergio [31]

I'm going to answer this question using logic.  Let x be Jenny's favorite number.  We are going to square root this number, \sqrt{x} and then multiply it by \sqrt{2}.

This product needs to be an integer.  How is that obtainable? To be an integer, we need to get rid of the nasty \sqrt{2} part.  The only way I can think of to get rid of it, is to multiply it by \sqrt{2}, because \sqrt{2} *\sqrt{2} = 2.  Thus we have some conditions we need to fulfill when choosing Jenny's favorite number.

When we take the square root of Jenny's favorite number, x, it must contain a perfect square and a 2 in its prime factorization.  For example, 8 works because 8 = 2 x 2 x 2, or 2² x 2.  

You notice 8 is made up of a perfect square multiplied by 2.  So when we take the square root of 8, we get:

\sqrt{8} = \sqrt{2^{2}*2 } = \sqrt{ 2^{2} }*\sqrt{2} = \sqrt{4} *\sqrt{2} = 2*\sqrt{2}

So 8 is the same thing as 2\sqrt{2}

So when we multiply this by \sqrt{2}, we will get an integer! So as long as Jenny's favorite number consists of a perfect square and two in its prime factorization, we will have an integer!

So possible choices are: 2,8,18.

Why does 18 work? Because \sqrt{18} = \sqrt{9}* \sqrt{2} = 3\sqrt{2}

When we multiply this by \sqrt{2}, we get 6, which is an integer.

b) Suppose instead of multiplying by \sqrt{2}, we divided by \sqrt{2}. Is the resulting quotient still an integer?

YES, because we can get rid of the \sqrt{2} part by dividing by \sqrt{2} as well.  This leaves only the "perfect square" part left in our square root, and obviously a perfect square is an integer when we square root it.

I hope that made sense! (⌐■_■)

3 0
3 years ago
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Alika [10]
When Stacy draws the first marble, her probability of drawing one of the 7 marbles from the 14 in the bag is 7/14. When she draws the second marble, her probability of drawing one of the 6 remaining white marbles from the remaining 13 in the bag is 6/13.

Her probability of drawing two white marbles in succession from the bag is ...
\dfrac{7}{14}\cdot\dfrac{6}{13}     matches the 2nd selection
4 0
3 years ago
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