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3 years ago

Could I get an explanation and some help?

Mathematics

1 answer:

True [87]3 years ago

7 0

Think of it as a normal linear equation first. Let's find the slope.
m = rise/run = (3-1)/(0-1) = -2
We know the slope is negative now, so we can immediately get rid of the first two answers. Now, we know that the solutions must be under the line itself, so we can try figuring it out by testing some points. Let's use (0,0).
Is 0 </> 0+3? Since it's <, then we know the last answer is correct (y < -2x + 3).

You might be interested in

If you plotted the points in the table and connected

castortr0y [4]

Answer:

U shaped.

Step-by-step explanation:

When x = 0 , f(x) = 6

when x =1 yf(x) = 0

when x = 2 f(x) = -2

x = 3 f(x) = 0

x = 4 f(x) = 6.

So the graphs falls from the left and rises to the right in the form of a U.

You can draw a rough graph to confirm this.

3 0

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

Riley Kate had of a doughnut to

NeTakaya

Both child would receive 1/2

4 0

3 years ago

Read 2 more answers

OMG plz help me. IDK how to do thisI need help asap

larisa [96]

1)1.10w+2.35(172-s)=294.20
When you solve
1.10x+2.35(172-x)=294.20 solve for x
X water=88
S=172-88=84
2)2t+4(40-f)=100
When you solve
2x+4(40-x)=100 solve for x
X=30 for t
40-30=10 for f
Hope it helps

5 0

4 years ago

Solve each equation by completing the square. n^2 - 6n + 6 = -2

Law Incorporation [45]

<h3><u>Explanation</u></h3>

Separate the constant out of expression.

$( {n}^{2} - 6n) + 6 = - 2$

Find the constant that makes the expression able to be squared. You have to subtract the separated constant by new constant as well.

$( {n}^{2} - 6n + 9) + 6 - 9 = - 2 \\ {(n - 3)}^{2} - 3 = - 2 \\ {(n - 3)}^{2} = 1$

Square Root both sides to get rid the squared expression. Make sure to write plus or minus.

$\sqrt{ {(n - 3)}^{2} } = + \sqrt{1} \\ \sqrt{ {(n - 3)}^{2} } = - \sqrt{1} \\ n - 3 = 1 \\ n - 3 = - 1$

$n - 3 = 1 \: \: \: or \: \: \: n - 3 = - 1 \\ n = 1 + 3 \: \: \: or \: \: \: n = - 1 + 3 \\ n = 4 \: \: \: or \: \: \: n = 2$

<h3><u>Answer</u></h3>

$\large \boxed {n = 4,2}$

3 0

3 years ago