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OleMash [197]
2 years ago
10

I dont understand what this is really telling me. can someone explain?

Mathematics
2 answers:
lukranit [14]2 years ago
5 0

Answer:

B

Step-by-step explanation:

It's where the line goes over the x line

Olenka [21]2 years ago
5 0

Answer:

B,

Step-by-step explanation:

The zeros are the x values  where the graph crosses the horizontal x-axis.

These are at x = 0 and x = 5.

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Can 7:13 be simplified
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Evaluate the following expression when x = 3 and y = 4:
JulijaS [17]
I don't understand.  What are the values given?  Are they values of an x,y plot?
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5 0
4 years ago
An object, which is at the origin at time t=0, has initial velocity V0= (-14.0i - 7.0j)m/s and constant acceleration a=(6.0i + 3
sesenic [268]
I am pretty sure the first thing we should do is to <span>integrate acceleration into velocity
And we will have: </span>a(t) = \ \textless \ 6,3\ \textgreater \v(t) = \ \textless \ (6t+C),(3t+C)\ \textgreater \ ; where    -- v(0) = \ \textless \ -14,-7\ \textgreater \
Then calculate v :&#10;v(t) = \ \textless \ (6t-14),(3t-7)\ \textgreater \
As you can see, velocity is zero and none of the objects is moving. It happens when <span> t=7/3 which means we can calculate this in the way we did (integrating) :
</span>p(t) = \ \textless \  (3t^2-14t+C) , ((3/2)t^2-7t+C) \ \textgreater \ &#10;
The product is <span>p(0) = 0 that makes us to reduce vector function to</span>p(t) = \ \textless \  (3t^2-14t) , ((3/2)t^2-7t) \ \textgreater \   where t = 7/3p(7/3) = \ \textless \  -49/3 , -49/6 \ \textgreater \<span>

</span>
7 0
4 years ago
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