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3 years ago

5

A truangular bandana has an area of 42 square inches. The height of the triangle is 5 1/4 inches. Enter and solve an equation ti

find the length of the base of the triangle.

1 answer:

nasty-shy [4]3 years ago

4 0

Answer: The length of the base of the triangle is 16 inches.

Step-by-step explanation:

Hi, to answer this question we have to apply the next formula

Area of a triangle (A) = base x height x 1/2

Replacing with the values given:

42 = b (5 1/4) 1/2

Solving for b (lenght of the base)

42 / (5 1/4) 1/2 = b

16 in = b

The length of the base of the triangle is 16 inches.

Feel free to ask for more if needed or if you did not understand something.

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Find all solutions to each quadratic equation

romanna [79]

I hope this helps you

2x^2-4x+7=0 a=2 b=-4 c= 7

disctirminant =b^2-4ac

disctirminant =(-4)^2-4.2.7

disctirminant = 16-56= -40

x=-b+square root of disctirminant ÷2a

x=4+2square root of (-10)/4

x=2+square root of (-10)/2

x'=4 -2 square root of (-10)/4

x'=2 -square root of (-10)/2

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3 years ago

I REALLY NEED HELP, PLEASE HELP !! WORTH 30 POINTS !!

valina [46]

Answer:

2/9

Step-by-step explanation:

y=6(3)^x

Let x =-3

y = 6 * 3^-3

Remember negative exponents move it from the numerator to the denominator

y = 6 *1/3^3

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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

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

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

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

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