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Rashid [163]
3 years ago
14

1. What are the possible numbers of solutions for equations in the form ax = ? How about

Mathematics
1 answer:
hram777 [196]3 years ago
5 0

Answer:

Step-by-step explanation:

as long as a and b are just numbers they can have infinitely many solutions.  Basically, take any number, you can multiply it by some other number and in return get another number you choose.  so if I want to multiply 3 by something to get literally any other number, I could.  The simplest way is to multiply by the real number n/3 where n s that number we want to get to.  Same if you add a b, you will just have to take that into account.  So we want ax + b to equal a number n?  make x be the number (n-b)/a  You just have to pick a number for a, b and n.

equations with just an x are known as linear, because theya re just a straight line.  maybe at an angle, but a straight line.  a quadratic is a "degree" above, which means it has an x^2.  Nothing higher, but it has to have an x^2.  It can have a plain old x as well.  so it is of the form ax^2 or ax^2 + bx or ax^2 + bx + c or even ax^2 + c.  needs that x^2 as the highest exponent a variable has.  also a cannot be 0 otherwise there won't be an x^2 and it'll just be linear again.

In relation to linear it is no longer straight, instead it has a single curve to it.  after x^2 they can have more than one curve, so for instance a "degree three" equation could have up to two curves.  degree three means the highest exponent on a variable is 3.  but it could also have just one as well, and this pattern continues.

So to summarize, the relation between linear and quadratic is the degree of the variable being a difference of 1 which in turn means a difference of one "curve"

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Answer:

(a) 2 feet.

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Step-by-step explanation:

We have been given that the velocity function v(t)=\frac{1}{\sqrt{t}} in feet per second, is given for a particle moving along a straight line.

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Since velocity is derivative of position function , so to find the displacement (position shift) from the velocity function, we need to integrate the velocity function.

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Therefore, the total displacement on the interval  1\leq t\leq 4 would be 2 feet.

(b). For distance we need to integrate the absolute value of the velocity function.

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Since square root is not defined for negative numbers, so our integral would be \int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt.

We already figured out that the value of \int\limits^4_1 {\frac{1}{\sqrt{t}}} \, dt is 2 feet, therefore, the total distance over the interval 1\leq t\leq 4 would be 2 feet.

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