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Otrada [13]
3 years ago
6

Roberto will save 1/6 of his alowence each day. If he gets $2.00 a day, about how much money will he save each day? Round your a

nswer to the nearest penny.
Mathematics
1 answer:
Tresset [83]3 years ago
5 0
2.00/6 = $0.33 rounded to the nearest penny, it actually is $0.333333333333 (infinite)

I hope this helped you out :)
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Determine the Unit Rate.
Mice21 [21]

Answer:

$1.5 each

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
of a parabolic arch and is to have a span of 100 feet. The height of the arch a distance of 40 feet from the center is to be 10
JulsSmile [24]

Answer:

y = -0.11x^2 + 1.111x

y = 28 ft  .... Height at center

Step-by-step explanation:

Given:-

- The span of the arc is = 100 ft

- The height of the arch is 40 ft at 10 ft from center.

Find:-

- The equation of parabolic arch and the height of the arch at center.

Solution:-

- We will take the height y as a function of width x of the parabolic arch. The general equation of the arch is such that it passes through origin. The equation is given in the form as:

                               y = f(x) = ax^2 + bx

Where,

           a, b, and c are constants to be determined.

- We will use the condition i.e the span of entire arch is 100 ft. So we could say that y = 0 for x = 100 ft. Then we have:

                               0 = f(100) = a(100)^2 + b(100)   ..... 1

- Using second condition i.e y = 10 ft at 40 ft from center. Since, due to symmetry we know that center lies at x = 50 ft. Then y = 10 ft at x = 10 ft. The condition can be expressed in the form:

                               10 = f(10) = a(10)^2 + b(10)  ..... 2

- Solving the 2 Equations simultaneously, we have:

                               0 = a(100)^2 + b(100)

                               10*10 = a*10*(10)^2 + b(10)*10

                               100 = a(10)^3 + b(100)

- Subtract both equations:

                               100 = a*(10^3 - 100^2)

                               a = 100 / ( 1000 - 10000)

                               a = -0.11

- Then using a = -0.11 evaluate b:

                               -1.11 + 10b = 10

                                b = 11.11 / 10 = 1.111

- The equation of the parabola is:

                                y = -0.11x^2 + 1.111x                                

-The height of the arch at center where x = 50 ft.

                                y = -0.11(50)^2 + 1.111(50)

                                y = -27.5 + 55.5

                                y = 28 ft                                                        

- The height of the parabolic arch at center is given as y = 28 ft.

4 0
4 years ago
What is the name of the relationship between ∠3 ​ and ∠7?
lara31 [8.8K]

answer is B) Corresponding angles

6 0
2 years ago
Read 2 more answers
What is the equation of the given line in the point slope -form ?
ladessa [460]

Answer:

see below

Step-by-step explanation:

We are given two points (3,4) and (0,1)

so we can find the slope

m = ( y2-y1)/(x2-x1)

    = (1-4)/(0-3)

  = -3/-3

  = 1

The point slope form is

( y-y1) = m(x-x1)

y - (4) = 1 (x-3)

y - (4) =  (x-3)

or using the other point

y - 1 = 1( x-0)

y - 1 = x

8 0
3 years ago
Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ<2π0≤θ<2π (a)(9,0
DochEvi [55]

Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

x=r  \cos \theta, y= r \sin \theta

r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )

a) converts (9, 0) to polar coordinates  ( r , θ)

r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9

\theta= \tan^-^1 (\frac{0}{9} )\\\\=0

b) Convert (18,\frac{18}{\sqrt{3} } ) to polar coordinates ( r, θ)

r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}

\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}

c)  converts (-5, 5) to polar coordinates  ( r , θ)

r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}

\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}

d)  converts (-1, √3) to polar coordinates  ( r , θ)

r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2

\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}

= \frac{2\pi}{\sqrt{3} }

6 0
3 years ago
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