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Anestetic [448]
3 years ago
14

Noah measured the length of three pieces of cloth. The measurements were 4.29 feet, 3.6 feet, and 2.34 feet. what was the total

length of the three pieces of cloth ?
Mathematics
1 answer:
iragen [17]3 years ago
4 0
10.23 feet is the answer because when you line up the decimals and add, its 10.23
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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
2 years ago
What is the answer <br> Ps I need the answers quick
Ghella [55]

Answer:

30%

Step-by-step explanation:

first it's on sale for half off

then u have a coupon for 20%

50%+20%=70%

100%-70%=30%

3 0
3 years ago
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