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4 years ago

10

Determine whether the ordered pair (1,4)is a solution of the linear inequality y-4x<0.

1 answer:

djyliett [7]4 years ago

6 0

It is not since the sign would have to be less than or equal to. I hope this helps. :)

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Setler [38]

Answer:

1. Y=-2x+3

2. Y=3x+2/3

I'm sorry I can't get them all but I know these are right

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3 years ago

Which expression could be used to find the perimeter of the triangle below ?

Free_Kalibri [48]

Please put the triangle below.. not possible to solve if i can’t see the triangle

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3 years ago

Pls answer fast my teacher is waiting!

a_sh-v [17]

Answer:

For company A, y = 24x + 42

For company B, y = 28x + 25

Step-by-step explanation:

x is the number of containers

y is the total cost

For company A, y = 24x + 42

For company B, y = 28x + 25

For the cost of both companies to be the same, then

24x + 42 = 28x + 25

28x - 24x = 42 - 25

4x = 17

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Mr Lycan would have to order about 4 containers so the cost would be the same from each company

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3 years ago

Tess had $1,063.40 in a savings account with simple interest. She had opened the account with $1,040 just 3 months earlier. What

JulsSmile [24]

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4 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