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sergeinik [125]

3 years ago

6

A square field has an area of 10 square miles. What is the approximate length, to the nearest thousandth, of one side of this fi

eld?
a.2.5 miles
b.3 miles
c.3.162 miles
d.3.612 miles

2 answers:

VladimirAG [237]3 years ago

6 0

Sea is the correct answer

blsea [12.9K]3 years ago

4 0

$A=10 [mi^{2}]\\
A=s*s=s^{2}\\
s= \sqrt{A} \\
s= \sqrt{10} =3.162$

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3 years ago

Express the fractions 1/2, 3/16, and 7/8 with an LCD

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You first need to find the LCD (lowest common denominator). You will need to find the smallest number that is a multiple of all numbers that is the denominator (2, 16, 8). Or, to say it another way, all the numbers in the denominator need to be a factor of this number.

You can find this by first checking if the largest number that is the denominator-- in this case 16-- is already the LCD, which means 16 is divisible by all the other numbers.

If this does not work, then multiply all the numbers together to get the LCD-- since you multiplied them together, you know that they will all be factors of the product.

However, you will be able to see that 16 is indeed the lowest common denominator:

2 × 8=16

8 × 2=16

16 × 1=16

So, after you find the LCD, multiply both the numerator and the denominator by the number that you would need to multiply the denominator to get the LCD (the whole point is that you want to get the denominator to be the LCD, but to do that you need to multiply both the top and bottom by the same number to keep the fraction the same).

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3 years ago

Find center,foci, and vertices of ellipse (x+3)^2/21+(y-5)^2/25=1

sasho [114]

I don't know if we can find the foci of this ellipse, but we can find the centre and the vertices. First of all, let us state the standard equation of an ellipse.

(If there is a way to solve for the foci of this ellipse, please let me know! I am learning this stuff currently.)

$\frac{(x-x_{1})^2}{a^2}+ \frac{(y-y_{1})^2}{b^2}=1$

Where $(x_{1},y_{1})$ is the centre of the ellipse. Just by looking at your equation right away, we can tell that the centre of the ellipse is:

$(-3,5)$

Now to find the vertices, we must first remember that the vertices of an ellipse are on the major axis.

The major axis in this case is that of the y-axis. In other words,

$b^2>a^2$

So we know that b=5 from your equation given. The vertices are 5 away from the centre, so we find that the vertices of your ellipse are:

$(-3,10)$
& $(-3,0)$

I really hope this helped you! (Partially because I spent a lot of time on this lol)

Sincerely,

~Cam943, Junior Moderator

6 0

4 years ago