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3 years ago

8

On an architects plan for a house, the scale reads 1/2 in. = 5 ft. How long is the actual length of a room that's 2 in. Long on

the drawing?

1 answer:

Nady [450]3 years ago

6 0

1/2:5

1:10

2:20

The actual length of the room is 20 feet.

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What are the answers to this someone please help.

statuscvo [17]

1234567890233572o7o5y345uo22u.3936584643728284/43794729237932972478864966666666666666666666666666666666663870434573428734907 goood luck!

6 0

3 years ago

Identify the coordinates of the point (3,-2), translated 5 units left and 6

kap26 [50]

Answer:

(- 2, 4 )

Step-by-step explanation:

A translation of 5 units left means subtract 5 from the original x- coordinate.

A translation of 6 units up means add 6 to the original y- coordinate.

Thus

(3, - 2 ) → (3- 5, - 2 + 6 ) → (- 2, 4 )

3 0

4 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

How can you use unit rates to solve pricing problems?

melomori [17]

Answer:

a ratio is a comparison of two numbers or measurements

the objects being compared are called the terms of the ratio.

if the tow terms are different units this is called a rate.

example:

David covers 600 miles in 8 hours and John covers 380 miles in 5 hours. Who is driving faster ?

solution:

Distance covered by David in 1 hour is

= 600 / 8

= 75 miles per hour

Distance covered by John in 1 hour is

= 380 / 5

= 76 miles per hour

When we compare the above unit rates (Distance covered in 1 hour), John is driving faster than David.

5 0

3 years ago

Read 2 more answers

What’s the volume of the cylinder

FrozenT [24]

Answer:

188.496 mm^3 (or 188.495559 mm^3 if you want to be specific)

Step-by-step explanation:

The radius would be the distance from the middle of the circle on top, we are given that the diameter of the circle is 4mm, the radius is half of that so you would get 2 as the radius (or R). H is the height of the cylinder which is 15 mm, so plugging it in (pi * (2^2) * 15) would get you 188.496 mm^3 (rounded of course).

3 0

3 years ago