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sergeinik [125]

2 years ago

6

A square field has an area of 10 square miles. What is the approximate length, to the nearest thousandth, of one side of this fi

eld?
a.2.5 miles
b.3 miles
c.3.162 miles
d.3.612 miles

2 answers:

VladimirAG [237]2 years ago

6 0

Sea is the correct answer

blsea [12.9K]2 years ago

4 0

$A=10 [mi^{2}]\\
A=s*s=s^{2}\\
s= \sqrt{A} \\
s= \sqrt{10} =3.162$

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3 years ago

Factor f(x)=x4-x3-7x2+13x-6 completely. Then Sketch the graph

Dennis_Churaev [7]

ANSWER TO QUESTION 1

$f(x)=(x-2)(x+3)(x-1)^2$ .

EXPLANATION

The function given to us is,

$f(x)=x^4-x^3-7x^2+13x-6$

According to rational roots theorem,

$\pm1,\pm2,\pm3,\pm6$ are possible rational zeros of

$f(x)=x^4-x^3-7x^2+13x-6$ .

We find out that,

$f(-3)=(-3)^4-(-3)^3-7(-3)^2+13(-3)-6$

$f(-3)=81+27-63-39-6$

$f(-3)=6-6$

$f(-3)=0$

Also

$f(2)=(2)^4-(2)^3-7(2)^2+13(2)-6$

$f(2)=16-8-28+26-6$

$f(2)=6-6$

$f(2)=0$

This implies that

$x-2 and x+3$ are factors of

$f(x)=x^4-x^3-7x^2+13x-6$ and hence $(x-2)(x+3)=x^2+x-6$ is also a factor.

We perform the long division as shown in the diagram.

Hence,

$f(x)=(x-2)(x+3)(x-1)^2$ .

ANSWER TO QUESTION 2

Sketching the graph

We can see from the factorization that the roots

$x=2$ and $x=-3$ have a multiplicity of 1, which is odd. This means that the graph crosses the x-axis at this intercepts.

Also the root $x=1$ has a multiplicity of 2, which is even. This means the graph does not cross the x-axis at this intercept.

Now we determine the position of the graph on the following intervals,

$x\le -3$

$f(-4)=(-4)^4-(-4)^3-7(-4)^2+13(-4)-6$

$f(-4)=150\:>0$

$-3\le x \le 1$

$f(0)=-6\:$

$1\le x\le 2$

$f(1.5)=-0.56\:$

$x \ge 2$

$f(3)=24\:>0$

We can now use these information to sketch the function as shown in diagram

4 0

3 years ago