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Ilia_Sergeevich [38]
2 years ago
15

PLEASE HELPPP i would really appreciate it :)

Mathematics
1 answer:
qaws [65]2 years ago
3 0

Answer:

Y = - 12.55X1 + 276.52

Step-by-step explanation:

Given the data:

Age of child :(X) : 4, 6, 10, 13, 14, 17

Time (seconds) (Y) : 230, 205, 132, 119, 108, 62

Using technology, the linear regression equation obtained using the data Given is :

Y = - 12.55X1 + 276.52

Where :

Slope = - 12.55

Intercept = 276.52

This also indicates that a negative slope is present and hence a negative relationship between age of Child and time in seconds.

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Is 0.304 meters less than 1 meter
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PlS HELP QUICKLY :One of the roots of the equation x2−bx+c=0 is equal to 5. Find c in terms of b.:
Grace [21]

Answer:First suppose that the roots of the equation

x2−bx+c=0(1)

are real and positive. From the quadratic formula, we see that the roots of (1) are of the form

b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2.

For the root or roots to be real, we require that b2−4c≥0, that is, b2≥4c. In order for them to be positive, we require that

b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0.

This immediately tells us that b>0, but we can go further. We can rearrange this to get

b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,

which (assuming that b>0) is true if and only if

b2>b2−4c,

since both sides of the inequality are positive so we may square. But then

4c>0.

That is, if the roots are real and positive then b>0 and b2≥4c>0.

Now suppose that b>0 and b2≥4c>0.

Then the roots of (1) are real since b2−4c≥0, and b>0 guarantees that the root b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√ is positive.

So it remains to show that b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√>0. We have that

4c>0,

so that

b2>b2−4c,

then square rooting shows that

b>b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√,

so the roots of (1) are real and positive, as required.

Approach 2

the curve y equals x squared minus b x plus c showing two positive roots for y equals zero

This is intended to be a proof without words! We have from the diagram that:

If c, b and b2−4c are all positive, there are two real positive roots for x2−bx+c=0 (if b2=4c, we have two real positive equal roots).

If there are two real positive roots for x2−bx+c=0, then c and b are positive and b2−4c is non-negative.

(Why is the distance between the roots b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√?)

Approach 3

When solving problems about the roots of polynomials, it is often useful to find expressions those roots must satisfy and see if this tells us anything new. If α and β denote the roots of the equation, then

x2−bx+c=(x−α)(x−β)=x2−(α+β)x+αβ

and so α+β=b and αβ=c.

We also know that the roots of a quadratic equation are real if and only if the discriminant is non-negative, that is, if and only if b2−4c≥0.

Using these facts, if α and β are both real and positive, then b=α+β>0, c=αβ>0 and b2≥4c, as above.

Conversely, if b>0 and b2≥4c>0, then we know the discriminant is positive and hence both roots are real. We also have that

αβ>0(2)

and

α+β>0.(3)

As α and β are both real, by (2), we know that α and β are either both positive or both negative. However, if α and β were both negative, then (3) could not possibly hold. Hence α and β are both positive, as required.

We now sketch on a graph the region where b>0, c>0 and b2≥4c:

The curve b squared = 4 c as a quadratic with the c-axis vertical and the b-axis horizontal. The region below it is shaded.

The region of the b-c plane for which b>0, c>0 and b2≥4c

Sketch the region of the b-c plane in which the roots of the equation are real and less than 1 in magnitude.

We know that in order for the roots to be real we need b2≥4c as in the first part. We now need to find the region where

−1<b±b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.(4)

We have b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2≤b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2 so we only need to consider the values for which both

−1<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2andb+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√2<1.

Firstly, we will consider the values for which

b+b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2.

Rearranging gives

b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<2−b.

So b<2, as the square root is non-negative, and we can square both sides to get

b2−4c<4−4b+b2,

which we may rearrange to find c>b−1.

We will now consider the values for which

−2<b−b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√.

Similarly, we can rearrange to get

b2−4c⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√<b+2.

So b>−2, and we can, as before, square to get

b2−4c<b2+4b+4,

and hence c>−b−1.

To sketch the graph, we start by considering the boundary curves b2=4c, c=b−1 and c=−b−1, and the points at which they intersect. We can see that the two lines only intersect when b=0 and c=−1, and the lines intersect the curve when

b2=4b−4andb2=−4b−4

which rearrange to

(b−2)2=0and(b+2)2=0.

This tells us that each line intersects with the curve in only one place and so these lines must be tangent.

Is there a way we could have deduced this directly from (4)?

What do the lines being tangent signify in terms of our equation x2−bx+c=0?

Is this a representation of a well-known property of these equations?

Sketching the graph then yields the following picture:

The graph with the previous curve and the lines 4 c = 4 b minus 4 and 4 c = minus 4 b - 4. Each line touches the curve once and the two lines intersect at (0, minus 1). The region between the three lines/curves is shaded.

The shaded region is where b2≥4c, c>b−1, c>−b−1 and −2<b<2

We might notice that when we derived the inequalities c>b−1 and c>−b−1, a lot of the work we did was very similar.

We might also notice that the graph above is symmetric about the c axis. Why is this?

Does the graph give us any ideas about how we might deduce one inequality from the other?

Step-by-step explanation:

7 0
3 years ago
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