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3 years ago

the square root of a number is between 8 and 9. which of the following can be the value of that number

Mathematics

1 answer:

kiruha [24]3 years ago

8 0

Where's the following??

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Which equation represents a line that passes through (4,1/3) and has a slope of 3/4?

Alexxandr [17]

Answer:

$\large\boxed{B.\ y-\dfrac{1}{3}=\dfrac{3}{4}(x-4)}$

Step-by-step explanation:

The point-slope form of an equation:

$y-y_1=m(x-x_1)$

m - slope

We have

$m=\dfrac{3}{4},\ \left(4,\ \dfrac{1}{3}\right)$

Substitute:

$y-\dfrac{1}{3}=\dfrac{3}{4}(x-4)$

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3 years ago

If 4 desk lamps cost $120 what is the average cost per lamp?

Rzqust [24]

120/4 = 30 So the average cost of a lamp is $30

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3 years ago

A rectangle’s length is two feet less than three times it’s width. If it’s width is 7 feet, then which of the following is its a

vaieri [72.5K]

The answer would be 19

Explanation is that 7x3 = 21 -2 = 19

7 0

3 years ago

Number 18 idk what to do help plz and fast

stepan [7]

It is right it is C 454.5

7 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

3 years ago