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Marta_Voda [28]
3 years ago
10

Solve each equation.

Mathematics
1 answer:
Marrrta [24]3 years ago
8 0

(1)\\\\|2x-4| =10\\\\\implies 2x-4 = 10~~ \text{or}~~ 2x -4 = -10\\\\\implies x = \dfrac{14}2 = 7 ~~\text{or}~~ x = -\dfrac{6}2=-3\\\\\\(2)\\\\-5|-4x+4|-7 = -67\\\\\implies -5|-4x+4| = -67 +7 = -60\\\\\implies |-4(x-1)| = \dfrac{-60}{-5}=12\\\\\implies 4|x-1| = 12\\\\\implies |x-1| =\dfrac{12}4 = 3\\\\\implies x-1 = -3~~\text{or}~~ x-1=3\\\\\implies x = -2~~ \text{or}~~ x = 4

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Find values of x, HJ, and JK.​
xz_007 [3.2K]

Answer:

x = 2, HJ = JK = 11

Step-by-step explanation:

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5 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



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3 years ago
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abruzzese [7]
9x-72=
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