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4 years ago

I have a quiz on this tomorrow and i'm completely lost someone please help

Mathematics

1 answer:

Minchanka [31]4 years ago

5 0

All the points on those lines are a part of the domains and the ranges. The domain points are the "x" values of the points on those lines. The range points are the "y" values of the points on those lines.

The first square in the top left hand corner:
Domain: -3, -1, 0, 1, 2
Range: 0, -1, 1, 0, -1

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A rectangular flower bed that is 5 ft long and 4 feet wide is surrounded by a square lawn that is

Trava [24]

Answer:

124 ft2

Step-by-step explanation:

The yard is a 12ft square so the length and width would both be 12ft. You multiply the length and width of the yard to get the area of the yard so 12x12=144. You have the length and width of the flower bed so 5x4=20. Once you have the area of both you can subtract the area the flower bed takes up of the lawn. 144-20=124 ft2.

5 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

Find a normal vector n to the plane z−5(x−2)=2(8−y)

Hitman42 [59]

A normal vector is the set of coefficients of x, y, and z when the equation is written in standard (or general) form.

Subtracting the left side, we have

... 2(8 -y) -z +5(x -2) = 0

... 16 -2y -z +5x -10 = 0

... 5x -2y -z +6 = 0 . . . . . general form

A normal vector is (5, -2, -1).

7 0

3 years ago

ANSWER ALL FOUR FOR BRAINLIEST

sergij07 [2.7K]

Answer: (in fractions)

a. 1/4096

b. 1/15625

c. 1/16807

d. 1/432

7 0

3 years ago

11 Select the decimal that is equivalent to 36 Choose 1 answers 0.28 0.23 0.37 0.37 None of the above

rusak2 [61]

None of those are near 36 at all, so none of the above would be your answer.

6 0

4 years ago