All categories

3 years ago

What’s the distance between (3,6) and (3,-4)

Mathematics

1 answer:

Firlakuza [10]3 years ago

8 0

Answer:

Step-by-step explanation:

Use the distance formula in order to find the distance.

Formula

$\sqrt{(x^2-x^1)^2+(y^2-y^1)^2$

Solve

3-3 = 0

0^2 = 0

-4 - 6 = -10

-10 ^ 2 = 100

$\sqrt{100}$ = 10

Answer: 10

You might be interested in

Kelly drove north for 9 miles and then east for 12 miles at an average rate of 42 miles per hour to arrive at the town of Prime.

mr_godi [17]

Answer:

Step-by-step explanation:

This is one of the more interesting motion problems I've seen. I like it! If Kelly is driving north (straight up) for 9 miles, then turns east (right) and drives for 12 miles, what we have there are 2 sides of a right triangle. The hypotenuse is created by Brenda's trip, which originated from the same starting point as Kelly and went straight to the destination, no turns. We need the distance formula to solve this problem, so that means we need to find the distance that Brenda drove. Using Pythagorean's Theorem:

$9^2+12^2=c^2$ and

$81+144=c^2$ and

$225=c^2$ so

c = 15.

Brenda drove 15 miles. Now we can fill in a table with the info:

d = r x t

Kelly 12+9 42 t

Brenda 15 45 t

Because they both left at the same time, t represents that same time, whatever that time is. That's our unknown.

If d = rt, then for Kelly:

21 = 42t

For Brenda

15 = 45t

Solve Kelly's equation for t to get

t = 1/2 hr or 30 minutes

Solve Brenda's equations for t to get

t = 1/3 hr or 20 minutes

That means that Brenda arrived at the destination 10 minutes sooner than Kelly.

4 0

4 years ago

One egg carton holds 12 eggs how many cartons are you able to fill with 60 eggs

PtichkaEL [24]

The answer should be 5 egg cartons

8 0

3 years ago

Read 2 more answers

Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7

Ilia_Sergeevich [38]

I'll go out on a limb and guess the system is

$\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x$

with initial condition $\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top$ . The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12$

The corresponding eigenvectors $\eta$ are such that

$\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}$

$\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}$

So the characteristic solution to the ODE system is

$\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

When $t=0$ , we have

$\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}$

from which it follows that $C_1=2$ and $C_2=5$ , making the particular solution to the IVP

$\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

$\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}$

5 0

4 years ago

Factor and solve (please explain) 10p^3-1960p=0 Will mark brainliest!!!

Sergeeva-Olga [200]

10p^3=1960p
/10. /10
P^3=196
P=5.8

5 0

4 years ago

Amanda has at most $40 to spend on flowers. She wants to buy a pair of red rose flowers for $18 dollars and spend the rest on li

Mkey [24]

Answer:

x ≤ 2

Step-by-step explanation:

2 red rose cost $ 18

x flowers cost $ 11x

"At most " means ≤.

11x + 18 ≤ 40

11x ≤ 40 - 18

11x ≤ 22

x ≤ 2

5 0

4 years ago