All categories

3 years ago

PLEEEEEEEEEEEAAAAAAAAAASSSSSSSEEEE HELP I WIL GIVE BRAINLIEST

Mathematics

1 answer:

Mkey [24]3 years ago

6 0

Answer: I think the answer is B

Step-by-step explanation: Cause in each year the numbers increase.

You might be interested in

Who else goes to Don Callejon school?<br> (put an equation so moderators don't delete)

Rom4ik [11]

3 x 5 = 0 I don’t :)

3 0

3 years ago

Read 2 more answers

Lin and her brother each created a scale drawing of the backyard, but at different scales. Lin used a scale of 1 inch to 1 foot.

Gnom [1K]

Idkidkodkidkidkidkidkidkidkidk

7 0

3 years ago

Seven people are in an elevator which stops at ten floors. In how many ways can they get off the elevator?

zlopas [31]

The number of ways people can get off the elevator is 604800 ways.

In this question,

Number of people, n = 7

Number of floors, r = 10

The first person can leave elevator in one of 10 ways. Then second person has to choose from one of remaining 9 floors. Then third person in 8, fourth in 7 ways and so on.

Number of ways people can get off the elevator can be calculated as

⇒ $nP_r=\frac{n!}{(n-r)!}$

⇒ $10P_{7}=\frac{10!}{(10-7)!}$

⇒ $10P_{7}=\frac{10!}{(3)!}$

⇒ $10P_{7}=\frac{(10)(9)(8)(7)(6)(5)(4)3!}{(3)!}$

⇒ $10P_{7}=(10)(9)(8)(7)(6)(5)(4)$

⇒ 604800 ways.

Hence we can conclude that the number of ways people can get off the elevator is 604800 ways.

Learn more about number of ways here

brainly.com/question/27992615

#SPJ4

4 0

2 years ago

D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x

MA_775_DIABLO [31]

The solution depends on the value of $k$ . To make things simple, assume $k>0$ . The homogeneous part of the equation is

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0$

and has characteristic equation

$r^2-16k=0\implies r=\pm4\sqrt k$

which admits the characteristic solution $y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}$ .

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be $y_p=ae^{4x}+be^x$ . Then

$\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x$

So you have

$16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x$
$(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x$

This means

$16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}$
$b(1-16k)=30\implies b=\dfrac{30}{1-16k}$

and so the general solution would be

$y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x$

8 0

3 years ago

Suppose that J and K are points on the number line.

V125BC [204]

K could either be a 16 or -24

5 0

3 years ago