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Veronika [31]
3 years ago
8

12. Complete the two-column proof.

Mathematics
1 answer:
Reil [10]3 years ago
8 0

Answer:

supplementary

vertical angles

sub

Step-by-step explanation:

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Answers are below!

Step-by-step explanation:

Well it is easy.

a) -3-8/1-2=-11/-1=11

b)-2-0/-6+4=-2/-2=1

c)-7+2/4-0=-5/4

All you need to do is doing the needed action.

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Suppose u and v are functions of x that are differentiable at x=0 and that u(0)=4, u′(0)=7, v(0)=2 and v′(0)=1. Find the values
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Answer with Step-by-step explanation:

We are given that u and v are functions of x and both are differentiable at x=0

u(0)=4,u'(0)=7,v(0)=2,v'(0)=1

a.We have to find the values of \frac{d(uv)}{dx}

\frac{d(u\cdot v)}{dx}=u'v+uv'

Using this formula

Then , we get

[\frac{d(uv)}{dx}]_{x=0}=u'(0)v(0)+u(0)v'(0)=7(2)+4(1)=14+4=18

[\frac{d(uv)}{dx}]_{x=0}=18

b.\frac{d(u/v)}{dx}=\frac{u'v-uv'}{v^2}

[\frac{d(u/v)}{dx}]_{x=0}=\frac{u'(0)v(0)-u(0)v'(0)}{v^2(0)}=\frac{7(2)-4(1)}{2^2}=\frac{14-4}{4}=\frac{10}{4}=\frac{5}{2}

[\frac{d(u/v)}{dx}]_{x=0}=\frac{5}{2}

c.

[\frac{d(v/u)}{dx}]_{x=0}=\frac{v'(0)u(0)-v(0)u'(0)}{u^2(0)}=\frac{1(4)-7(2)}{4^2}

[\frac{d(v/u)}{dx}]_{x=0}=\frac{-10}{16}=\frac{-5}{8}

d.\frac{d(-6v-9u)}{dx}=-6v'-9u'

[\frac{d(-6v-9u)}{dx}]_{x=0}=-6v'(0)-9u'(0)=-6(1)-9(7)=-6-63=-69

[\frac{d(-6v-9u)}{dx}]_{x=0}=-69

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