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2 years ago

Which function has the greater maximum value, f(x) or g(x)?

Mathematics

1 answer:

Stolb23 [73]2 years ago

6 0

Answer:

Step-by-step explanation:

fx is the nonsimplest form gx is the simplest i hope this helps

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Hannah receives a weekly allowance. Each week, she puts $10 in her savings account and

vesna_86 [32]

Answer: $18 a week

Step-by-step explanation: hope it helps

8 0

2 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

When cross canceling fractions, can you multiply vertically?

mr Goodwill [35]

I think the answer is no

3 0

3 years ago

How much wrapping paper did he use ,in square inches

Paul [167]

Answer:

96.76 square inches

Step-by-step explanation:

5.2x5.2=27.04

6.7x5.2/2=17.42

17.42x4=69.68

69.68+27.04=96.76

Hope this helps <3 Comment if you want more thanks and be sure to give brainliest (4 left) <3

6 0

2 years ago

You only have to get this second answer right (30 points)

DanielleElmas [232]

Answer:

6 nights with 4 logs left over

5 0

2 years ago

Read 2 more answers