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Ilya [14]
3 years ago
9

The radius of the wheel of a bicycle is 14 inches. What is the distance, in feet, that the bicycle covers after ten full rotatio

ns of the wheels? Use π = 3.14.
Mathematics
1 answer:
GREYUIT [131]3 years ago
8 0

the circumference of a circle is C=2*pi*r

so for the tire C=2*pi*14 or 28pi

so 10 rotation would be 10 times the circumference

28pi*10=280pi or 879.65ft

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50 squared equals 2 + 25 squared equals 2 * 5 equals 10 what is wrong
Klio2033 [76]

Convert to an equation:

50^2=2+25^2=2*5=10

Simplify:

2500=625=10=10

What is wrong is that 2500 does not equal 625 which does not equal 10.

7 0
3 years ago
Write a recursive formula for the sequence 1, 5, 9, 13
11Alexandr11 [23.1K]

Answer:

\bold{a(n) \ = \ a(n \ + \ 1) \ + \ 4}

Explanation:

Sequence: 1, 5, 9, 13

  • Add(ing) 4 every time

\bold{a(1) \ = \ 1}

\bold{\bold{a(n) \ = \ a(n \ + \ 1) \ + \ 4}}

5 0
3 years ago
A rock is thrown up from the ground at a velocity of 84 feet per second. The formula h = -16t2 + 84t gives the rock's height in
kenny6666 [7]

Answer:

h = 110.24 m

Step-by-step explanation:

It is given that,

A rock is thrown up from the ground at a velocity of 84 feet per second. The formula h = -16t^2 + 84t gives the rock's height in feet after t second.

We need to find the maximum height of the rock. For maximum height put \dfrac{dh}{dt}=0

So,

\dfrac{d( -16t^2 + 84t)}{dt}=0\\\\-32t+84=0\\\\t=\dfrac{84}{32}\\\\t=2.62\ s

Put t = 2.62 s in the equation of height. So,

h = -16(2.62)^2 + 84(2.62)\\\\h=110.24\ m

So, the maximum height of the rock is 110.24 m.

7 0
3 years ago
Need help with these 2 problems on my homework.
poizon [28]
4. 55 + 0.65*55 =  $90.75

5. 90.75 * 0.25*90.75 = $113.44

6. 113.44 - 0.33*113.44 = $76.00
4 0
4 years ago
Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of
Maru [420]

Parameterize S{/tex] by[tex]\vec s(u,v)=u\,\vec\imath+v\,\vec\jmath+(8-u^2-v^2)\,\vec k

with 0\le u\le1 and 0\le v\le1.

Take the normal vector to S to be

\vec s_u\times\vec s_v=2u\,\vec\imath+2v\,\vec\jmath+\vec k

Then the flux of \vec F across S is

\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^1\int_0^1\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1(uv\,\vec\imath+v(8-u^2-v^2)\,\vec\jmath+u(8-u^2-v^2)\,\vec k)\cdot(2u\,\vec\imath+2v\,\vec\jmath+\vec k)\,\mathrm du\,\mathrm dv

=\displaystyle\int_0^1\int_0^1\bigg(2u^2v+(u+2v^2)(8-u^2-v^2)\bigg)\,\mathrm du\,\mathrm dv=\boxed{\frac{1553}{180}}

6 0
3 years ago
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