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andreev551 [17]
3 years ago
14

Which of the following is the reciprocal of 5 3/7 38/7 7/38 22/3 3/22

Mathematics
2 answers:
elixir [45]3 years ago
6 0
The answer is 7/38.
jonny [76]3 years ago
4 0

Answer: 7/38

Step-by-step explanation:

You can turn 5 3/7 into an improper fraction by multiplying 5 by 7 and then adding it to the numerator to get 38/7. Then, using our knowledge that the reciprocal is the opposite of the fraction, we can flip the fraction 38/7, to get 7/38!

You might be interested in
What is 5 5/6 divided by 2/3
kherson [118]

Answer:

35/4 or 8,3/4

Step-by-step explanation:

Flip the second fraction and multiply.

8 0
2 years ago
Read 2 more answers
a manufacturer has determined that if his company sells x items per day then their profit is given by P=-x^2+ 700 x-10000 euros.
Radda [10]

Answer:

15

Step-by-step explanation:

To find the minimum amount of items, x, that need to be sold for the manufacturer to make a profit, we can use the quadratic formula

x = \frac{-b+\sqrt{b^{2}-4ac } }{2a}; x=  \frac{-b-\sqrt{b^{2}-4ac } }{2a}

x= \frac{-700+\sqrt{700^{2}-4(a)(-10000) } }{2(1)}; x= \frac{-700-\sqrt{700^{2}-4(a)(-10000) } }{2(1)}

x= \frac{-700+\sqrt{530000} }{2}; x= \frac{-700-\sqrt{530000} }{2}

x=\frac{-700+100\sqrt{53} }{2}; x=\frac{-700-100\sqrt{53} }{2};

x=-350 + 50\sqrt53 = 14.005; x = -350 - 50\sqrt53 = -714.005

In context of the problem, we can only rely on the positive value, as the negative value would lead to a loss of profit.

(-714.005)^2+700(-714.005)-10000=-0.360

Furthermore, we must round to the nearest whole number, as you cannot make part of an item.

Lastly, if you were to plug in 14 into the equation for Profit, you would still have a negative number (i.e. a negative profit), thus requiring the manufacturer to make no less than 15 items to make a profit:

(14)^2+700(14)-10000=-4\\\\(15)^2+700(15)-10000=725

3 0
1 year ago
Evaluate the integral x^3 (x 4 −8)^44 dx by making the substitution u = x^4 - 8
Eduardwww [97]

Answer:

(x^4-8)^45 /180 +c

Step-by-step explanation:

If u=x^4-8, then du=(4x^3-0)dx or du=4x^3 dx by power and constant rule.

If du=4x^3 dx, then du/4=x^3 dx. I just divided both sides by 4.

Now we are ready to make substitutions into our integral.

Int(x^3 (x^4-8)^44 dx)

Int(((x^4-8)^44 x^3 dx)

Int(u^44 du/4)

1/4 Int(u^44 dul

1/4 × (u^45 / 45 )+c

Put back in terms of x:

1/4 × (x^4-8)^45/45 +c

We could multiply those fractions

(x^4-8)^45 /180 +c

5 0
2 years ago
Help!! How do I solve this
tamaranim1 [39]
The answer for this question is Option B : 12.28
6 0
2 years ago
Read 2 more answers
Evaluate the surface integral:S
rjkz [21]
Assuming S does not include the plane z=0, we can parameterize the region in spherical coordinates using

\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle

where 0\le u\le2\pi and 0\le v\le\dfrac\pi/2. We then have

x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v
(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v

Then the surface integral is equivalent to

\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du

We have

\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle
\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle
\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle
\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v

So the surface integral is equivalent to

\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du
=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv
=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw

where w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv.

=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}
=\dfrac{243}2\pi
4 0
3 years ago
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