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nikdorinn [45]
3 years ago
7

1. How many feet of fencing are needed to completely enclose a square plot of land with sides of length 8 feet?

Mathematics
1 answer:
max2010maxim [7]3 years ago
6 0

1. 28

2. 2 7/12 or approximately 2.58

3. I believe 580 but I could be wrong.

4. 213.75

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On their farm, Adam's family maintains a storage that can hold 9.8 cubic yards (yd) of grain. Use the fact that 1 yard is approx
Inga [223]

ANSWER: V=19.9*(0.9144m)^3=15,2146416738816 m^3

1 yard =0.9144 m  

We will take the 18.0 cubic yards, cross multiply with 0.9144 and then divide the result by one to get  

13.76m3

Sure hope this helps

6 0
2 years ago
What is the value of x?​
guajiro [1.7K]

Answer:

x ≈ 8.66

Step-by-step explanation:

Using the sine ratio in the right triangle

sin60° = \frac{opposite}{hypotenuse} = \frac{x}{10}

Multiply both sides by 10

10 × sin60° = x, thus

x ≈ 8.66 ( to 2 dec. places )

4 0
3 years ago
Is it possible to split a jackpot of $2,783,000 equally among 9 winners? Explain your answer.​
Katena32 [7]

Answer:

No it isnt possible

Step-by-step explanation:

it is not because 2,783,000 is not divisible by 9

4 0
2 years ago
2+2-2-3-87+20090 then split in half
Svetllana [295]

20002 because all of it added together and then divided by two gives you 20002.

6 0
3 years ago
Show work please.<br><br> solve system of equations using matrices.
nadya68 [22]

Answer:

(t, t -1, t)

Step-by-step explanation:

You have three unknowns but only 2 equations, so you can't really SOLVE this...you can get a solution with a variable still in it (I forget what this is called.  I think it refers to infinite many solutions).  Here's how it works:

Set up your matrix:

\left[\begin{array}{ccc}1&-2&1\\2&-1&-1\\\end{array}\right] \left[\begin{array}{ccc}2\\1\\\end{array}\right]

You want to change the number in position 21 (the 2 in the scond row) to a 0 so you have y and z left.  Do this by multiplying the top row by -2 then adding it to the second row to get that 2 to become a 0.  Multiplying in a -2 to the top row gives you:

\left[\begin{array}{ccc}-2&4&-2\\2&-1&-1\\\end{array}\right]\left[\begin{array}{ccc}-4\\1\\\end{array}\right]

Then add, keeping the first row the same and changing the second to reflect the addition:

\left[\begin{array}{ccc}-2&4&-2\\0&3&-3\\\end{array}\right] \left[\begin{array}{ccc}-4\\-3\\\end{array}\right]

The second equation is this now:

3y - 3z = -3.  Solving for y gives you y = z - 1.  Let's let z = t (some random real number that will make the system true.  Any number will work.  I'll show you at the end.  Just bear with me...)

lf z = t, and if y = z - 1, then y = t - 1.  So far we have that y = t - 1 and z = t.  Now we solve for x:

From the first equation in the original system,

x - 2y + z = 2.  Subbing in t - 1 for y and t for z:

x - 2(t - 1) + t = 2.  Simplify to get

x - 2t + 2 + t = 2  and  x - t = 0, and x = t.  So the solution set is (t, t - 1, t).  Picking a random value for t of, let's say 2, sub that in and make sure it works.  If:

x - 2y + z = 2, then t - 2(t - 1) + t = 2 becomes t - 2t + 2 + t = 2, and with t = 2, 2 - 2(2) + 2 + 2 = 2.    Check it:  2 - 4 + 4 = 2 and 2 = 2.  You could pick any value for t and it will work.

6 0
3 years ago
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