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3 years ago

What is 3/4*16/21 Simplify fractors

1 answer:

4 0

$\bf \cfrac{3}{4}\cdot \cfrac{16}{21}\implies \cfrac{~~\begin{matrix} 3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\cdot \cfrac{~~\begin{matrix} 4 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\cdot 4}{~~\begin{matrix} 3 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\cdot 7}\implies \cfrac{4}{7}$

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What is the equation of the following linear funchon?

juin [17]

Answer:

3x+8

Step-by-step explanation:

linear function formula ax+b

when x=3 3a+b=17 and when x=6 6a+b=26

3a=9 a=3,

when x=3 3.3+b=17 9+b=17 b=8

3x+8

3 0

2 years ago

Find the value of Y .....

vekshin1

Answer:

$y = 6 \sqrt{3} \: units$

Step-by-step explanation:

In order to find the value of y, first we need to find the length of the perpendicular dropped from one of the vertices of the triangle to its opposite side.

By geometric mean theorem:

Length of the perpendicular

$=\sqrt{9\times 3}$

$=\sqrt{27}\: units$

Next, by Pythagoras theorem:

${y}^{2} = {9}^{2} + {( \sqrt{27} )}^{2} \\ \\ {y}^{2} = 81 + 27 \\ \\ {y}^{2} = 108 \\ \\ y = \sqrt{108} \\ \\ y = \sqrt{36 \times 3} \\ \\ y = \sqrt{ {6}^{2} \times 3} \\ \\ y = 6 \sqrt{3} \: units$

4 0

3 years ago

A function f is _____ on an open interval I if, for any choice of x1 and x2 in I, with 1 less than

Zepler [3.9K]

Answer:

If $f(x_1)\leq f(x_2)$ whenever $x_1\leq x_2$ f is increasing on I.

If $f(x_1)\geq f(x_2)$ whenever $x_1\leq x_2$ f is decreasing on I.

Step-by-step explanation:

These are definitions for real-valued functions f:I→R. To help you remember the definitions, you can interpret them in the following way:

When you choose any two numbers $x_1\leq x_2$ on I and compare their image under f, the following can happen.

$f(x_1)\leq f(x_2)$ . Because x2 is bigger than x1, you can think of f also becoming bigger, that is, f is increasing. The bigger the number x2, the bigger f becomes.

$f(x_1)\geq f(x_2)$ . The bigger the number x2, the smaller f becomes so f is "going down", that is, f is decreasing.

Note that this must hold for ALL choices of x1, x2. There exist many functions that are neither increasing nor decreasing, but usually some definition applies for continuous functions on a small enough interval I.

3 0

3 years ago

What is w-4>9 written on a number line?

Aleksandr-060686 [28]

There would be an open circle on positive 12 with the arrow moving to the right.

7 0

3 years ago

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