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Juliette [100K]
3 years ago
5

During 4 days, the price of the stock of PEV Corporation went up 1/4 of a point, down 1/3 of a point, down 3/4 of a point, and u

p 7/10 of a point. What was the net charge?
Mathematics
1 answer:
HACTEHA [7]3 years ago
8 0

The price of the stock of PEV Corporation varies as shown. we have to determine the net charge. Net change is just how much it changed overall, so we will look for all the ups minus all the downs.  

Here, "went up" goes with a + sign, and "went down" goes with a - sign.

So, Net charge = \frac{1}{4} - \frac{1}{3}-\frac{3}{4} + \frac{7}{10}

= \frac{1}{4}-\frac{3}{4}- \frac{1}{3} + \frac{7}{10}

= -\frac{2}{4}- \frac{1}{3} + \frac{7}{10}

= -\frac{1}{2}- \frac{1}{3} + \frac{7}{10}

LCM of 2,3 and 10 is 30

= \frac{-15-10+21}{30}

= \frac{-4}{30}

= \frac{-2}{15}

So, the net charge is \frac{-2}{15}

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Step-by-step explanation:

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2 years ago
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Suppose u and v are functions of x that are differentiable at x=0 and that u(0)= 7,u'(0)=-5,v(0)= -1, v'(0)= -4.
andriy [413]

This question is incomplete, the complete question is;

Suppose u and v are functions of x that are differentiable at x=0

and that { u(0) = 7, u'(0) = -5 }  { v(0)= -1, v'(0) = -4 }

Find the values of the following derivatives at x = 0.

a) \frac{d}{dx}( uv )

b)  \frac{d}{dx}( \frac{u}{v} )

c)  \frac{d}{dx}( \frac{v}{u} )

Answer:

a) \frac{d}{dx}( uv ) = -23  

b) \frac{d}{dx}( \frac{u}{v} )  = 33

c) \frac{d}{dx}( \frac{v}{u} ) = -32/49 or - 0.6531

Step-by-step explanation:

Given that;

{ u(0) = 7, u'(0) = -5 }  { v(0)= -1, v'(0) = -4 }

a)

\frac{d}{dx}( uv )

we differentiate

\frac{d}{dx}( uv )  = uv' + vu'

at x = (0), we substitute our values

\frac{d}{dx}( uv ) = ( 7 × -4 ) + ( -1 × -5)

\frac{d}{dx}( uv )  = -28 + 5

\frac{d}{dx}( uv ) = -23  

b)

\frac{d}{dx}( \frac{u}{v} )

we differentiate

\frac{d}{dx}( \frac{u}{v} ) = ( vu' - uv' ) / v²

at x=0, we substitute our values

\frac{d}{dx}( \frac{u}{v} ) = ( (-1 × -5) - (7 × -4 ) ) / (-1)²

\frac{d}{dx}( \frac{u}{v} ) = (( 5 - ( -28 )) / 1

\frac{d}{dx}( \frac{u}{v} )  = 33 / 1

\frac{d}{dx}( \frac{u}{v} )  = 33

c) \frac{d}{dx}( \frac{v}{u} )

we differentiate

\frac{d}{dx}( \frac{v}{u} )  = ( uv' - vu' ) / u²

at x=0, we substitute our values

\frac{d}{dx}( \frac{v}{u} )  = ( (7 × -4) - (-1 × -4) ) / (7)²

\frac{d}{dx}( \frac{v}{u} ) = ( -28 - ( 4 ) ) / 49

\frac{d}{dx}( \frac{v}{u} )  = ( -28 - 4 ) /49

\frac{d}{dx}( \frac{v}{u} )  = -32 / 49

\frac{d}{dx}( \frac{v}{u} ) = -32/49 or - 0.6531

6 0
3 years ago
A toy that originally sold for 12 dollars went of sale for 8 dollars what was the percent discount? Remember to show your work!
harkovskaia [24]
12———-100%
8 ———— x
X=(8*100)/12=800/12=66.66%
Then discount percent is 100-66.66=33.33%
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laila [671]
The answer is 45 okay bro good
5 0
2 years ago
Which relation is also a function?
hoa [83]
It’s d bec each x has one input
6 0
3 years ago
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