All categories

3 years ago

Please help will give brainliest and 30 pts

Mathematics

1 answer:

Rus_ich [418]3 years ago

3 0

You’re not giving 30 points, you’re only giving 5

You might be interested in

Exponential growth and decay functions are written in standard form as

strojnjashka [21]

Answer:

Step-by-step explanation:

te falta el factor terminal sige con las letras delabesedario

5 0

2 years ago

If <img src="https://tex.z-dn.net/?f=log_8%20N%20%3D%20%5Cfrac%7Br%7D%7B2%7D%20%5C%3A%20%2C%20%5C%3A%20log_2%282N%29%3D%205%5C%3

Rudik [331]

Answer:

N = 2 $\frac{3r}{2}$

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

PLEASE HELP ASAP!!! CORRECT ANSWERS ONLY PLEASE!!!

Alex73 [517]

inequalities shown in

y greater than 0

6 0

3 years ago

Read 2 more answers

Solve the inequality 4x - 7<5

Alexandra [31]

Answer:

x ∈ (-oo ; 3)

Step-by-step explanation:

4x - 7 < 5

4x < 5 + 7

4x < 12

x < 12 : 4

x < 3

x ∈ (-oo ; 3)

8 0

3 years ago

Que is on pic.i can't able to type in text.

ad-work [718]

It's not difficult to compute the values of $A$ and $B$ directly:

$A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}$
$A=\tan^{-1}(\sin\theta)-\dfrac\pi4$

$B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt$
$B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}$
$B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2$

Let's assume $0$ , so that $|\csc\theta|=\csc\theta$ .

Now,

$\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}$
$\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)$
$\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)$

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in $A$ and $B$ .

3 0

2 years ago