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3 years ago

If a truck weighs 3 1/2 tons, how many pounds does it weigh?

Mathematics

1 answer:

Gennadij [26K]3 years ago

4 0

1 ton is 2000 pounds your answer is 7000 pounds

Hope this helps

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Simplify: 5j - (2j - 6). Thanks!

xxMikexx [17]

Answer:

3j + 6

Step-by-step explanation:

5j - (2j - 6) Distribute the negative inside the parentheses

5j - 2j + 6 Combine like terms

3j + 6

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3 years ago

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HELP PLEASE I WILL GIVE THE BRAINLIEST ANSWER Parking meter that is 1.6 m tall cast a shadow 3.6 m long at the same time a tree

Julli [10]

Answer:

I am not able to answer your question because I am unsure of what to solve for. :/

Step-by-step explanation:

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3 years ago

A ball is thrown downward from a cliff. Its position at time t seconds is given by the formula s(t) = 16t2 + 32t, where s is in

Rom4ik [11]

Answer:

It takes 2.8 seconds for the ball to fall 215 ft.

Step-by-step explanation:

We are given a position function s(t) where s stands for the number of feet the ball has fallen, so we have to replace s with the given value of 215 ft and solve for the time t.

Setting up the equation.

The motion equation is given by

$s(t) =16t^2+32t$

We can replace there s = 215 ft to get

$215=16t^2+32t$

Solving for the time t.

From the previous equation we can move all terms in one side to get

$16t^2+32t-215=0$

At this point we can solve for t using quadratic formula.

$t = \cfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

where a, b and c are the coefficients of the quadratic equation

$at^2+bt+c=0$

So we get

$a=16\\b=32\\c=-215$

Replacing on the quadratic formula we get

$t = \cfrac{-32\pm \sqrt{32^2-4(16)(-215)}}{2(16)}$

Using a calculator we get

$t=-4.8 , t = 2.8$

Physically speaking the only result that makes sense is to move forward in time that give us t = 2.8 seconds.

We can conclude that it takes 2.8 seconds for the ball to fall 215 ft.

3 0

3 years ago

Consider a series circuit consisting of a resistor of R ohms, an inductor of L henries, and variable voltage source of V(t) volt

ser-zykov [4K]

Answer:

$I(t)=\frac{1}{3}(1-e^{-30t})$

Step-by-step explanation:

We are given that

$\frac{dI}{dt}+\frac{R}{L}I=\frac{V(t)}{L}$

R=150 ohm

L=5 H

V(t)=10 V

$P=\frac{R}{L}=\frac{150}{5}=30$

$I.F=e^{\int Pdt}=e^{\int 30 dt}=e^{30 t}$

$I(t)\times I.F=\int e^{30 t}\times 10 dt+C$

$I(t)\times e^{30 t}=\frac{10}{30}e^{30 t}+C$

$I(t)=\frac{1}{3}+Ce^{-30 t}$

I(0)=0

Substitute t=0

$0=\frac{1}{3}+C$

$C=-\frac{1}{3}$

Substitute the values

$I(t)=\frac{1}{3}-\frac{1}{3}e^{-30 t}$

$I(t)=\frac{1}{3}(1-e^{-30t})$

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3 years ago

Correctly name the following angle.

chubhunter [2.5K]

Answer: ∠EDC or ∠CDE

Step-by-step explanation:

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3 years ago