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3 years ago

12

The perimeter of Raul's picture frame is 108 centimeters. The length of the picture frame is 18 centimeters. What is the width o

f the picture frame?

1 answer:

o-na [289]3 years ago

7 0

Perimeter of a rectangle= 2(length)+2(width)

Data:
Perimeter=108 cm
length=18 cm

therefore:
108 cm=2(18 cm)+2(width)
108 cm=36 cm+2 (width)
2 width=108 cm -36 cm=
2 width=72 cm
width=72cm/2
width=36 cm

Answer: the width of the picture frame would be 36 cm.

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Can someone help with an explanation? This is finding Surface Area.

Y_Kistochka [10]

The answer should be 95in^2

Explanation:

You’re essentially just finding the area of all the shapes on the triangle and adding them together.

Area of a triangle = 1/2•b•h
Area of a square = l•w or just s2

The area of the triangles given what we have in the diagram can be found with
1/2•5•7 = 17.5

There are 4 triangles so you multiply 17.5 by 4 which gives 70

The base of the triangle (the square) = 25

70+25=95

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2 years ago

What is the converse of the following conditional statement? If x = –10, then x2 = 100

tangare [24]

The converse would be If x² = 100 then x = -10

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3 years ago

How to write 6/8 as a decimal ?

Korvikt [17]

all have to do is divide the fraction to make it into a decimal ,

6/8= 0.75

6 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

If f(x) = 5x + 4, which of the following is the inverse of Ax)?

Levart [38]

Answer:

$f^{-1}$ (x) = $\frac{x-4}{5}$

Step-by-step explanation:

let y = f(x) and rearrange making x the subject, that is

y = 5x + 4 ( subtract 4 from both sides )

y - 4 = 5x ( divide both sides by 5 )

$\frac{y-4}{5}$ = x

change x to $f^{-1}$ (x) and y back to x, thus

$f^{-1}$ (x) = $\frac{x-4}{5}$

6 0

3 years ago