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3 years ago

What are all solutions to the differential equation dy/dx=sec^2*x ?

Mathematics

1 answer:

zysi [14]3 years ago

4 0

Answer:

y=tanx+C

Step-by-step explanation:

$\frac{dy}{dx} = {sec}^{2} \: x \\ \\ dy = {sec}^{2} \: x \: dx \\ integrating \: both \: sides \\ \\ \int \:1\: dy = \int {sec}^{2} \: x \: dx \\ \\ \huge \red{ \boxed{ \therefore \: y = tan \: x + c}} \\$

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Problem page a garden table and a bench cost $990 combined. the garden table costs $90 more than the bench. what is the cost of

iogann1982 [59]

$900, but not too sure if this is a question to trick you.

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3 years ago

In exponential growth functions, the base of the exponent must be greater than 1. How would the function change if the base of t

alexandr402 [8]

Answer:

How would the function change if the base of the exponent were between 0 and 1? If the base of the exponent were 1, the function would remain constant. The graph would be a horizontal line. If the base of the exponent were less than 1, but greater than 0, the function would be decreasing.

Step-by-step explanation:

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3 years ago

What is 244 divided 5

Grace [21]

48.8 is the answer to 244 divided 5

5 0

3 years ago

Read 2 more answers

The options are<br> 40<br> 80<br> 120<br> 140

Nataly [62]

According to given figure,

$\qquad \sf \dashrightarrow \: \angle ABC + 60° = 180°$

$\qquad \sf \dashrightarrow \: \angle ABC = 180 \degree - 60 \degree$

[ By linear pair ]

$\qquad \sf \dashrightarrow \: \angle ABC = 120 \degree$

now, we can see that :

$\qquad \sf \dashrightarrow \: \angle ABC + \angle BAC = \angle 1$

[ By Exterior angle property of Triangle ]

$\qquad \sf \dashrightarrow \: \angle 1 = 20 \degree + 120 \degree$

$\qquad \sf \dashrightarrow \: \angle 1 = 140 \degree$

4 0

2 years ago

A person invests $4000 at 2% interest compounded annually for 4 years and then invests the balance (the $4000 plus the interest

faltersainse [42]

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4000\\
r=rate\to 2\%\to \frac{2}{100}\to &0.02\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &4
\end{cases}
\\\\\\
A=4000\left(1+\frac{0.02}{1}\right)^{1\cdot 4}\implies A=4000(1.02)^4\implies A\approx 4329.73$

then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$4329.73\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &7
\end{cases}
\\\\\\
A=4329.73\left(1+\frac{0.08}{1}\right)^{1\cdot 7}\implies A=4329.73(1.08)^7\\\\\\ A\approx 7420.396$

add both amounts, and that's her investment for the 11 years.

7 0

3 years ago