Determine whether (2, - 1) and (-4, 2) satisfy the inequality 2x

Point x is located at (-2,-6) and point z is located at (0,5). Find the y value for the point y that is located 1/5 the disteanc

MA_775_DIABLO [31]

Answer:

-3.5

Step-by-step explanation:

We have to find the distance between x and z along the y-axis.

This will be: 5 - -6 = 11 units

1/5 th of 11 = 1/5 × 11

= 11/5

= 2 1/5 = 2.5

Now add 2.5 to -6

2.5 + -6 = -3.5

The value of y = -3.5

3 0

3 years ago

Two cars are traveling north along a highway. The first drives at 40 mph, and the second, which leaves 3 hours later, travels at

belka [17]

Answer:

Step-by-step explanation:

We will make a table and fill it in according to the information provided. What this question is asking us to find, in the end, is how long did it take the cars to travel the same distance. In other words, how long, t, til car 1's distance = car 2's distance. The table looks like this:

d = r * t

car1

car2

We can fill in the rates right away:

d = r * t

car1 40

car2 60

Now it tells us that car 2 leaves 3 hours after car 1, so logically that means that car 1 has been driving 3 hours longer than car 2:

d = r * t

car1 40 t + 3

car2 60 t

Because distance = rate * time, the distances fill in like this:

d = r * t

car1 40(t + 3) = 40 t+3

car2 60t = 60 t

Going back to the interpretation of the original question, I am looking to solve for t when the distance of car 1 = the distance of car 2. Therefore,

40(t+3) = 60t and

40t + 120 = 60t and

120 = 20t so

t = 6 hours.

8 0

3 years ago

Help me plzzzzzzz! I need this today!

makkiz [27]

Answer:

it looks like y int is about 2550

Step-by-step explanation:

3 0

2 years ago

Square root 2x-1+2=5 what is x equal to?

KATRIN_1 [288]

Because your question isn't specific or formatted exactly, I cannot guarantee that my answer is what you expect.

√2x - 1 + 2 = 5
√(2x + 1)² = 5²
2x + 1 = 25
2x = 24
/2 /2
x = 12

Therefore x = 12.
Proof:

√2x - 1 + 2 = 5
√2(12) - 1 + 2 = 5
  √24 - 1 + 2 = 5
  √25 = 5
5 = 5

8 0

3 years ago

In a previous exercise we formulated a model for learning in the form of the differential equation dp dt = k(m − p) where p(t) m

DaniilM [7]

I assume you mean

$\dfrac{dP}{dt} = k(M-P)$

ANSWER
An expression for P(t) is

$P = M - Me^{-kt}$

EXPLANATION
This is a separable differential equation. Treat M and k as constants. Then we can divide both sides by M - P to get the P term with the differential dP and multiply both sides by dt to separate dt from the P terms

$\begin{aligned} \dfrac{dP}{dt} &= k(M-P) \\ \dfrac{dP}{M-P} &= k\, dt
\end{aligned}$

Integrate both sides of the equation.

$\begin{aligned}
\int \dfrac{dP}{M-P} &= \int k\, dt \\
-\ln|M-P| &= kt + C \\
\ln|M-P| &= -kt - C\end{aligned}$

Note that for the left-hand side, u-substitution gives us

$u = M - P \implies du = -1dP \implies dP = -du$

hence why $\int \frac{dP}{M-P} \ne \ln|M - P|$

Now we use the definition of the logarithm to convert into exponential form.

The definition is

$\ln(a) = b \iff \log_e(a) = b \iff e^b = a$

so applying it here, we get

$\begin{aligned} \ln|M-P| &= -kt - C \\ |M - P| &= e^{-kt - C} \\ 
M - P &= \pm e^{-kt - C} 
 \end{aligned}$

Exponent properties can be used to address the constant C. We use $x^{a} \cdot x^{b} = x^{a+b}$ here:

$\begin{aligned}
 M - P &= \pm e^{-kt - C} \\
M - P &= \pm e^{- C - kt} \\ 
M - P &= \pm e^{- C + (- kt)} \\ 
M - P &= \pm e^{- C} \cdot e^{- kt} \\ 
M - P &= Ke^{- kt} && (\text{\footnotesize Let $K = \pm e^{-kt}$ }) \\ 
M &= Ke^{- kt} + P\\
P &= M - Ke^{- kt}
\end{aligned}$

If we assume that P(0) = 0, then set t = 0 and P = 0

$\begin{aligned} 
0 &= M - Ke^{- k\cdot 0} \\
0 &= M - K \cdot 1 \\
M &= K
 \end{aligned}
$

Substituting into our original equation, we get our final answer of

$P = M - Me^{-kt}$

6 0

3 years ago

Read 2 more answers

Determine whether (2, - 1) and (-4, 2) satisfy the inequality 2x – 3y > 4.