All categories

3 years ago

1) What is the square root of 81

Mathematics

1 answer:

vekshin13 years ago

3 0

It’s 9. Square root is like the common factor of that number doubled by itself ... & 9x9 equals 81

You might be interested in

Solve for x in each of the following diagrams.

Vikentia [17]

In the left one x equals 12.9 and on the right one x equals 11

5 0

3 years ago

Select the correct answer.

ioda

Answer:

B. -5.20.

Step-by-step explanation:

We can see that Janet made more in week 4 then in week 2, so that means the answer will have to be a negative. To find the exact answer, we take away 21.74 from 16.54. This gets us -5.2, which is equivalent to -$5.20. I hope this helps!

6 0

3 years ago

Read 2 more answers

Does anyone know the answer to this question

Ivanshal [37]

$265.32

r = R/100
r = 4/100
r = 0.04 rate per year,
A = P(1 + r/n)nt
A = 1,000.00(1 + 0.04/1)(1)(6)
A = 1,000.00(1 + 0.04)(6)
A = $1,265.32
1,265.32-1,000=265.32

7 0

3 years ago

The center of an ellipse is located at (3, 2) One focus is located at (6, 2) and its associated directrix is represented by the

tangare [24]

Answer:

Step-by-step explanation:

This is really hard to try and explain over the internet here, but I'll do my best, assuming you have some experience with ellipses.

We are given the center of (3, 2), and the focus of (6, 2). The focus is 3 units to the right of the center; the other focus is 3 units to the left of the center at (0, 2). The focus has coordinates of (-ae, 2) and (ae, 2) from left to right and the distance between them then is 2ae. There are 6 units between the 2 foci, so 2ae = 6.

The directrix is 8 1/3 units from the center to the right AND the left. The directrices have distances of 2a/e between them. There are 16 2/3 units between the 2 directrices, so

$\frac{2a}{e}=\frac{50}{3}$

We can solve for the eccentricity of this ellipse given the fact that the eccentricity is the ratio of the distance between the foci to the distance between the directrices. Therefore,

$\frac{2ae}{2\frac{a}{e} }=\frac{6}{\frac{50}{3} }$

The 2's andd the a's cancel, leaving you with

$e^2=\frac{18}{50}=\frac{9}{25}$

That gives us then that

$e=\frac{3}{5}$

Going back to the identity for the distance between the foci, 2ae = 6, we fill in e and solve for a:

$2a(\frac{3}{5})=6$ and

$\frac{6a}{5}=6$

which gives us that a = 5

Now use the identity $b^2=a^2(1-e^2)$ to solve for b:

$b^2=25(1-\frac{9}{25})$ and

$b^2=25(\frac{16}{25})$ and

b = 4

Because this is a horizontally stretched ellipse, it is of the form

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

we fill in as

$\frac{(x-3)^2}{25}+\frac{(y-2)^2}{16}=1$

Again, since you are expected to be able to solve for the equation of an ellipse at this advanced level, I am assuming that the equations I gave above as a means to solve for the different characteristics of an ellipse are familiar to you. This is definitely NOT beginning conics!

3 0

3 years ago

Read 2 more answers

Rotate the parallelogram 90° clockwise about point q

oee [108]

The rotation is sketched in the photo, with the shaded parollellagram being the origional

8 0

3 years ago

Read 2 more answers

1) What is the square root of 81​

1) What is the square root of 81