A car weighs 3.6 tons. Find its weight in pounds.

I have a question if u hay that is 123 IBS and get 12ibs of anything you can thing of and get it to maybe change how it feels ho

ioda

1476 is the answer hope this helps

4 0

3 years ago

A particle is moving on a straight line in such a way that its velocity v is given by v ( t ) = 2 t + 1 for 0 ≤ t ≤ 5 where t is

Lady_Fox [76]

Answer:

The total distance traveled by the particle is S = 30.

Step-by-step explanation:

Given that velocity,

v(t) = 2t + 1

To find the total distance travel, we integrate the velocity function, v(t), to obtain the distance function s(t), and evaluate the resulting distance at the interval given. That is at t = 0 to t = 5.

Integrating v(t) with respect to t, we have

s(t) = t² + t + C.

At t = 5

s(5) = 5² + 5 + C

= 25 + 5 + C

= 30 + C

At t = 0

S(0) = 0 + 0 + C

= C

The required distance is now

S(5) - S(0)

= 30 + C - C

= 30.

8 0

3 years ago

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Nate borrowed $62,000 at 9.4% interest per year if he owed a total of $910.90 in when he repaid the loan how many days do they k

OverLord2011 [107]

The formula for simple annual interest is:

I = Prt

where,

I = Interest accumulated = $910.90
P = Principal Amount = $62000
r = Interest rate = 9.4% = 0.094
t = time in years

Using the values in above equation, we get:

910.90 = 62000 x 0.094 x t

⇒ t = 910.90/(62000 x 0.094) = 0.156

This is the time in years. Since there are 365 days in a year, the time in days will be:

t = 0.156 x 365 = 57 (rounded to nearest day)

This means, Nate kept the borrowed money for 57 days

6 0

3 years ago

A wire b units long is cut into two pieces. One piece is bent into an equilateral triangle and the other is bent into a circle.

mezya [45]

1. Divide wire b in parts x and b-x.

2. Bend the b-x piece to form a triangle with side (b-x)/3

There are many ways to find the area of the equilateral triangle. One is by the formula A= $\frac{1}{2}sin60^{o}side*side= \frac{1}{2} \frac{ \sqrt{3} }{2} (\frac{b-x}{3}) ^{2}= \frac{ \sqrt{3} }{36}(b-x)^{2}$
A= $\frac{ \sqrt{3} }{36}(b-x)^{2}=\frac{ \sqrt{3} }{36}( b^{2}-2bx+ x^{2} )=\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}$

Another way is apply the formula A=1/2*base*altitude,
where the altitude can be found by applying the pythagorean theorem on the triangle with hypothenuse (b-x)/3 and side (b-x)/6

3. Let x be the circumference of the circle.

$2 \pi r=x$

so $r= \frac{x}{2 \pi }$

Area of circle = $\pi r^{2}= \pi ( \frac{x}{2 \pi } )^{2} = \frac{ \pi }{ 4 \pi ^{2} }* x^{2} = \frac{1}{4 \pi } x^{2}$

4. Let f(x)= $\frac{ \sqrt{3} }{36}b^{2}-\frac{ \sqrt{3} }{18}bx+ \frac{ \sqrt{3} }{36}x^{2}+\frac{1}{4 \pi } x^{2}$

be the function of the sum of the areas of the triangle and circle.

5. f(x) is a minimum means f'(x)=0

f'(x)= $\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x$ =0

$\frac{ -\sqrt{3} }{18}b+ \frac{ \sqrt{3} }{18}x+\frac{1}{2 \pi } x=0$

$(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) x=\frac{ \sqrt{3} }{18}b$

$x= \frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

6. So one part is $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$ and the other part is b- $\frac{\frac{ \sqrt{3} }{18}b}{(\frac{ \sqrt{3} }{18}+\frac{1}{2 \pi }) }$

4 0

3 years ago

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Solve the system using elimination.

Shalnov [3]

Answer:

After simplifying we get (x,y) as (1,3).

Step-by-step explanation:

Given:

$x+7y=22$ ,