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3 years ago

Pls answer this about mathhNote :: no links no explain no

Mathematics

2 answers:

statuscvo [17]3 years ago

7 0

Step-by-step explanation:

AC =BC

2x-6=x+8

2x-x=8+6

x=14

5) AC = 2x-6=2(14)-6=28-6= 22

6) BC = x+8= 14+8= 22

7) AB = AC+BC = 22+22= 44

AC =BC

4x+10=7x-5

4x-7x= -5-10

-3x= -15

x= -15/-3 = 5

8) AC = 4x+10=4(5) +10=20+10= 30

9) BC = 7x-5= 7(5) -5=35-5= 30

10) AB = AC+BC = 30+30= 60

Talja [164]3 years ago

4 0

<u>B) Solution</u>

AC = BC

→ 2x-6 = x+8

→ 2x-x = 8+6

→ x = 14

5:- AC = 2x-6 = 2(14)-6 = 28-6 = 22

6:- BC = x+8 = 14+8 = 22

7:- AB = AC+BC = 22+22 = 44

<u>C) Solution</u>

AC = BC

→ 4x+10 = 7x-5

→ 4x-7x = -5-10

→ -3x = -15

→ x = -15/-3

→ x = 5

8:- AC = 4x+10 = 4(5) +10 = 20+10 = 30

9:- BC = 7x-5 = 7(5)-5 = 35-5 = 30

10:- AB = AC+BC = 30+30 = 60

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zavuch27 [327]

Answer:

The equation contains exact roots at x = -4 and x = -1.

See attached image for the graph.

Step-by-step explanation:

We start by noticing that the expression on the left of the equal sign is a quadratic with leading term $x^2$ , which means that its graph shows branches going up. Therefore:

1) if its vertex is ON the x axis, there would be one solution (root) to the equation.

2) if its vertex is below the x-axis, it is forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will not have real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently:

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression: $x_v=\frac{-b}{2a}$

Since in our case $a=1$ and $b=5$ , we get that the x-position of the vertex is: $x_v=\frac{-b}{2a} \\x_v=\frac{-5}{2(1)}\\x_v=-\frac{5}{2}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = -5/2:

$y_v=f(-\frac{5}{2})\\y_v=(-\frac{5}{2} )^2+5(-\frac{5}{2} )+4\\y_v=\frac{25}{4} -\frac{25}{2} +4\\\\y_v=\frac{25}{4} -\frac{50}{4}+\frac{16}{4} \\y_v=-\frac{9}{4}$

This is a negative value, which points us to the case in which there must be two real solutions to the equation (two x-axis crossings of the parabola's branches).

We can now continue plotting different parabola's points, by selecting x-values to the right and to the left of the $x_v=-\frac{5}{2}$ . Like for example x = -2 and x = -1 (moving towards the right) , and x = -3 and x = -4 (moving towards the left.