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3 years ago

Complete each statement from the information given and state the reasons and the triangle criterion you used.

Mathematics

1 answer:

TiliK225 [7]3 years ago

5 0

Answer:

Triangle GAS = Triangle IOL by rule AAS

Step-by-step explanation:

The points must be ordered correctly so they correspond within the triangles. Point G = Point I, Point A = Point O, Point S = Point L.

A rule for determining Triangle Congruency is AAS. If two angles and a side of a triangle are equal to one of another triangle, the triangles are congruent.

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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

strojnjashka [21]

Answer:

2500 pence (C = 2500)

Step-by-step explanation:

$C = 10t + 20m$

From the given information, t is the number of texts so essentially you will plug in the number of texts provided (150) in the place of the variable t.

$C = 10(150) + 20m$

Apply the same thing to minutes, plug in the number of minutes (50) in the place of the variable m.

$C = 10(150) + 20(50)$

Now multiply the substitutes with the co-efficient.

$C = 1500 + 1000$ (10*150=1500 & 20*50=1000)

Add the addends

C = 2500

7 0

2 years ago

What is the slope of the line through (-9,6) and (-3, 9)?

Artyom0805 [142]

To find slope, y2-y1/x2-x1
9-6/-3-(-9)=
3/6= 1/2
Slope = 1/2.
Y=1/2x.

3 0

3 years ago

Find the indicated limit, if it exists.

Paha777 [63]

The limit is the y value that the graph approaches from both sides as x approaches the target limit

it they approach different values from the left and right sides, the limit does not exist

we want to find the limit as x approaches 5

we are given that f(x)=5-x for x<5 and f(5)=8 and f(x)=x+3 for x>5

evaluate them all for f(5)
f(x)=5-x
f(5)=5-5
f(5)=0

f(5)=8

f(x)=x+3
f(5)=5+3
f(5)=8

it approaches 0 and 8 from both sides
the limit doesn't exist
D

8 0

3 years ago

Which relation is displayed in the graph?

uranmaximum [27]

Look at the picture.

Your answer is

$\{(-3;-1);\ (-2;\ 2);\ (1;\ 0);\ (3;\ 1);\ (4;-2)\}$

4 0

3 years ago

Read 2 more answers