Yes, you are correct. The answer to your problem is indeed 4.
Answer:
see attached
Step-by-step explanation:
y-2=-2(x+4)
y -2 = x(-2) + 4(-2)
y -2 = -2x - 8
y = -2x - 8 + 2
y = -2x - 6
When x = 0, y = -6. Hence (0,-6) is a point on the line
when y = 0, x = -3. Hence (-3,0) is a point on the line.
Plot these 2 points and connect with a line (see attached)
Answer:
4th graph
6th graph
3rd graph
Step-by-step explanation:
Linear = y = x + 3
it is 4th graph
when you substitute x = 0 y is 3
and when y = 0 x is -3
Quadratic = y = 3x^2
It is 6th graph
for both positive and negative values of x, y is rapidly increasing
Exponential = y = 3^x
it is 3rd graph
for x = 0 y value is 1 because 3^0 = 1
for positive values of x, y is exponentially rising
for negative values of x it is nearly almost touches zero
Given :-
- a² - 2a - b² = 0
- 2b + 2ab = 0
To find :-
Solution :-
<u>Taking</u><u> </u><u>second</u><u> </u><u>equation</u><u>:</u><u>-</u>
- 2b + 2ab = 0
- 2b ( 1 + a ) = 0
- 2b = 0 or (1+a) = 0
- b = 0 , a = -1
<u>Substitute</u><u> </u><u>in </u><u>first </u><u>equation</u><u> </u><u>:</u><u>-</u><u> </u>
<u>When </u><u>b </u><u>=</u><u> </u><u>0</u><u> </u><u>,</u>
- a² - 2a - 0² = 0
- a² - a = 0
- a( a -1) =0
- a = 0 , 1
<u>When </u><u>a </u><u>=</u><u> </u><u>-</u><u>1</u><u> </u><u>,</u>
- (-1)² - 2*(-1) - b² = 0
- 1 + 2 - b² = 0
- b² = 3
- b = ±√3
<u>Answer </u><u>:</u><u>-</u><u> </u>
- a = 0,1 ; b = 0
- a = -1 , b = ±√3