To answer the question presented above, the conversion factor for km to feet is 1 kilometers is equal to 3280.84 feet. From the given,
(12 km) x (3280.84 ft / 1 km) = 39370.08 feet
Thus, Joe traveled an approximate distance of 39370 feet.

6 0

3 years ago

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Shelly paid $2.00 per kg for the item that weighs 25% of the total weight of the 4 items calculate the amount of money she paid

madam [21]

1.000 is the answers 2000 25 100

6 0

3 years ago

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

Leona [35]

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

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

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

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

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

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

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

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

3 0

3 years ago