Can somebody help me on this because my brain can’t do anything right now

Which is an equation of an asymptote of the hyperbola?

Vikentia [17]

Follow the method of rise over run and a is your answer

3 0

3 years ago

N-24/8 = N/6 N equals to what?

Fed [463]

Answer:

We have 6( n - 24 ) = 8N;

Then 6n - 144 = 8N;

-144 = 8N - 6N;

-144 = 2N;

N = (-144)÷2;

N = - 72;

Step-by-step explanation:

6 0

2 years ago

Find the solution to the equation below.

Andreyy89

Answer:

B and D

Step-by-step explanation:

The quadratic formula is $x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}$ . Here a=20, b=-26, and c=8.

Substitute and you'll have:

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a} =\frac{26+/-\sqrt{26^2-4(20)(8)} }{2(20)}=\frac{26+/-\sqrt{676-640} }{40)}$

$\frac{26+/-\sqrt{36} }{40}=\frac{26+/-6 }{40}$

This means the solutions are 26 - 6/ 40 and 26 + 6/ 40.

It simplifies to 1/2 and 4/5.

3 0

3 years ago

Please help!!!! My assignment is almost due!!!

NeX [460]

Answer:

15

Step-by-step explanation:

the triangles are similar so will be the same proprotion,

I have attached a picture showing how I solved this. the second traingle is 3 times as large (15 by 12) compared to the first triangle which is 4 by 5.

3 0

2 years ago

Read 2 more answers

A leprechaun places a magic penny under a girls pillow. The next night there are 2 magic pennies under her pillow. The following

kiruha [24]

Answer:

After 47 days she will have more than 90 trillion pennies.

Step-by-step explanation:

At the beginning there was 1 penny. At the second day the amount of pennies under the pillow became 2.

The amount of pennies doubled each day. So the series is,

$1,2,4,8,16,32,.....$

This series is in geometric progression.

As the pennies from each of the previous days are not being stored away until more pennies magically appear so the sum of series will be,

$S_n=\dfrac{a(r^n-1)}{r-1}$

where,

a = initial term = 1,

r = common ratio = 2,

As we have find the number of days that would elapse before she has a total of more than 90 trillion, so

$\Rightarrow 90\times 10^{12}\le \dfrac{1(2^n-1)}{2-1}$

$\Rightarrow 90\times 10^{12}\le \dfrac{2^n-1}{1}$

$\Rightarrow 90\times 10^{12}\le 2^n-1$

$\Rightarrow 2^n\ge 90\times 10^{12}+1$

$\Rightarrow \log 2^n\ge \log (90\times 10^{12}+1)$

$\Rightarrow n\times \log 2\ge \log (90\times 10^{12}+1)$

$\Rightarrow n \ge \dfrac{\log (90\times 10^{12}+1)}{\log 2}$

$\Rightarrow n \ge 46.4$

$\Rightarrow n\approx 47$

8 0

3 years ago