All categories

2 years ago

Https://gyazo.com/632895ced27ab21bbe662b4d054b7b7d

Mathematics

1 answer:

Gekata [30.6K]2 years ago

3 0

What??..........rubeuhruhfeiuhwkuhw

You might be interested in

Add polynomial expressions

lisov135 [29]

$12 {g}^{2} + 4g - 1$

7 0

2 years ago

Gary is making a casserole but only needs 1/3 of the original serving size. If the number of servings Gary needs is 4, how many

Deffense [45]

Answer:

What equation will model the situation?

✔ x/3 = 4

How many servings did the original recipe make?

✔ 12

Step-by-step explanation:

Hope this helps!

7 0

2 years ago

Read 2 more answers

"5 more than 3 times a number" translates to an equation is it true or false ?

lozanna [386]

Answer:

3n + 5

Step-by-step explanation:

what do u mean by is it true or false?

5 0

2 years ago

Find the solution of the differential equation dy/dt = ky, k a constant, that satisfies the given conditions. y(0) = 50, y(5) =

irga5000 [103]

Answer: The required solution is $y=50e^{0.1386t}.$

Step-by-step explanation:

We are given to solve the following differential equation :

$\dfrac{dy}{dt}=ky~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.

From equation (i), we have

$\dfrac{dy}{y}=kdt.$

Integrating both sides, we get

$\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]$

Also, the conditions are

$y(0)=50\\\\\Rightarrow ae^0=50\\\\\Rightarrow a=50$

and

$y(5)=100\\\\\Rightarrow 50e^{5k}=100\\\\\Rightarrow e^{5k}=2\\\\\Rightarrow 5k=\log_e2\\\\\Rightarrow 5k=0.6931\\\\\Rightarrow k=0.1386.$

Thus, the required solution is $y=50e^{0.1386t}.$

8 0

3 years ago

Read 2 more answers

Round to the nearest thousand . Use the number line to model your thinking

ivolga24 [154]

Answer:

Step-by-step explanation:

the line grath is your answer

YOU WELCOME

8 0

3 years ago