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4 years ago

which numbers below can be placed in an empty cell so that the table continues to represent a fuction

Mathematics

1 answer:

serg [7]4 years ago

6 0

Where is the table? I need to see a table to give the answer

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The diameter of a circle is 6 centimeters. What is the circle's area?<br><br> Use 3.14 for pi

Lynna [10]

Step-by-step explanation:

diameter=

$2 \times\pi \times r = 6 \\ r = \frac{6}{2 \times 3.14} = 0.955 \\ area = \pi \times {r}^{2} = 3.14 \times {0.955}^{2} \\ = 2.87\: {cm}^{2}$

7 0

3 years ago

Read 2 more answers

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
$=\displaystyle\int(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\csc t\,\mathrm dt$

We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

etc.

3 0

3 years ago

A triangle of sides 15 cm 14 cm 13 cm prove

yarga [219]

Answer:

84 sq.m.

Step-by-step explanation:

Given : Sides of triangle are 13 cm,14 cm and 15 CM

To Find :find the area of the triangle

Solution:

We will use heron's formula :

Where

a = 13

b = 14

c = 15

Hence the area of the triangle is 84 sq.m.

Tip: use Heron's formula

6 0

3 years ago

Tiffany is solving an equation where both sides are quadratic expressions. She sets each quadratic equation equal to y and graph

Lapatulllka [165]

Answer:

Step-by-step explanation:

There are 3 possibilities for the graph as shown in the figure.

(a) having zero intersection

(b) having one intersection point (touching point)

So, the greatest possible number of intersections for these graphs is 2 as shown in figure (c).

3 0

3 years ago

Read 2 more answers

How do you do long division with 54 divided by 694

oksano4ka [1.4K]

694/54=12 with remainder 46 ( Answer is 12 R 47)

5 0

4 years ago