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3 years ago

6

Jason is drawing a blueprint of his bedroom for his new house. Using a scale of 1 inch = 3.5 feet, he drew his bedroom 3 inches

wide and 3.5 inches long. What will be the approximate area of his actual new bedroom?

1 answer:

svetoff [14.1K]3 years ago

4 0

The approximate area will be 128.63 ( or 128.625 ) feet squared

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Someone help pls I forgot how to do it

yawa3891 [41]

(2,4)

Step-by-step explanation:

I'm not sure in the slightest but with the formula I learned I think this is correct

3 0

2 years ago

Find all possible values of α+

const2013 [10]

Answer:

$\rm\displaystyle 0,\pm\pi$

Step-by-step explanation:

please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation

===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

cancel tanγ from both sides which yields:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )$

factor out tanγ:

$\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)$

divide both sides by tanαtanβ-1 and that yields:

$\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}$

multiply both numerator and denominator by-1 which yields:

$\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)$

recall angle sum indentity of tan:

$\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )$

let α+β be t and transform:

$\rm\displaystyle \tan( \gamma ) = - \tan( t)$

remember that tan(t)=tan(t±kπ) so

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )$

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )$

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )$

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

$\rm\displaystyle \gamma = - \alpha - \beta \pm \pi$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }$

<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

$\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )$

likewise by Opposite Angle Identity we obtain:

$\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )$

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

$\rm\displaystyle \gamma = - \alpha - \beta \pm 0$

isolate -α-β to left hand side and change its sign:

$\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }$

and we're done!

8 0

3 years ago

Read 2 more answers

A box is to be constructed from a sheet of cardboard that is 10 cm by 60 cm by cutting out squares of length x by x from each co

Stells [14]

Volume of a rectangular box = length x width x height<span>

From the problem statement,
length = 60 - 2x
width = 10 - 2x
height = x</span>

<span>
where x is the height of the box or the side of the equal squares from each corner and turning up the sides

V = (60-2x) (10-2x) (x)
V = (60 - 2x) (10x - 2x^2)
V = 600x - 120x^2 -20x^2 + 4x^3
V = 4x^3 - 100x^2 + 600x

To maximize the volume, we differentiate the expression of the volume and equate it to zero.

V = </span>4x^3 - 100x^2 + 600x<span>
dV/dx = 12x^2 - 200x + 600
12x^2 - 200x + 600 = 0</span>

<span>x^2 - 50/3x + 50 = 0

Solving for x,
x1 = 12.74 ; Volume = -315.56 (cannot be negative)
x2 = 3.92 ; Volume = 1056.31

So, the answer would be that the maximum volume would be 1056.31 cm^3.
</span><span>
</span>

6 0

3 years ago

A potential energy function is given by U = A x3 + B x2 − C x + D . For positive parameters A, B, C, D this potential has two eq

Ivenika [448]

Answer:

7.8655

Step-by-step explanation:

Given the functionl

$U=Ax^3+Bx^2-cx+D$

Where,

$A=1.45J/m^3\\B=2.85J/m^2\\c= 1.7J/m\\D=0.6J$

We have, replacing,

$U=1.45x^3+2.85x^2-1.7x+0.6$

We need to derivate and verify for stable equilibrum that,

$\frac{d^2 U}{dx^2}>0$

Then,

$U'(x) = 4.32x^2+5.7x-1.7$

Roots in,

$x_1=-1.57008\\x_2=0.250636$

We obtain U"(x), then we have

$\frac{d^2U}{dx}=8.64x+5.7$

For $x=-1.57, U"(x)=-7.86549$ Unestable

For $x=0.250636 U"(x)=7.8655>0$ Stable

7 0

3 years ago

Measure the distance from point A to points B, C, and D. Round to the nearest eighth of an inch. Reduce if possible.

Vlad1618 [11]

Answer:

if its on a coordinate plane the formula would be:

√(2x-1x)²+(2y-1y)²

Step-by-step explanation:

so if the coordinates were

A= (-3,2)

B= (4,2)

C= (4,-1)

D= (-3,-1)

for A to B it would be

√(4 - (-3)² + (2-2)²

√(7)² + (0)²

√(7·7) + (0·0)

√49+0

√49

or

7

I hope the explanation helped

7 0

2 years ago