Y> or equal 82P/100
P-maximum points of the tests
(-1.2,-2.0) and (1.9,2.2) are the best approximations of the solutions to this system.
Option B
<u>Step-by-step explanation:</u>
Here, we have a graph of two functions from which we need to find the approximate value of common solutions. Let's find this:
First look at where we have intersection points, In first quadrant & in third quadrant.
<u>At first quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point . After doing this we can clearly see that the perpendicular lines cut x-axis at x=1.9 and y-axis at y=2.2. So, one point is (1.9,2.2)
<u>At Third quadrant:</u>
Draw perpendicular lines from x-axis & y-axis from this point. After doing this we can clearly see that the perpendicular lines cut x-axis at x=-1.2 and y-axis at y= -2.0. So, other point is (-1.2,-2.0).
Answer:
It decreased by 40%
Step-by-step explanation:
Percent change = 42 - 70
|70|
x 100 = -28
70
x 100 = -40
Mark me brainliest if I helped:D
Answer:
Value of constant term c is (-4)
Step-by-step explanation:
The given table represents a function which is in the form of a quadratic equation,
y = ax² + bx + c
We choose three points (3, -10), (4, -16) and (5, -24) from the table and satisfy the equation to get the values of a, b, and c.
For point (3, -10)
-10 = a(3)² + 3b + c
9a + 3b + c = -10 -------(1)
For point (4, -16)
-16 = a(4)² + 4b + c
16a + 4b + c = -16 ------(2)
For point (5, -24)
-24 = a(5)² + 5b + c
25a + 5b + c = -24 -----(3)
Equation (1) - equation (2)
(9a + 3b + c) - (16a + 4b + c) = -10 + 16
-7a - b = 6
7a + b = -6 ------(4)
Equation (2) - equation (3)
(16a + 4b + c) - (25a + 5b + c) = -16 + 24
-9a - b = 8
9a + b = -8 -------(5)
Equation (4) - Equation (5)
(7a + b) - (9a + b) = -6 + 8
-2a = 2
a = -1
From equation (4),
-7(1) + b = -6
b = -6 + 7
b = 1
From equation (1)
9(-1) + 3(1) + c = -10
-9 + 3 + c = -10
c = -10 + 6
c = -4
Therefore, the value of constant term c is (-4).
Answer:
This is always ''interesting'' If you see an absolute value, you always need to deal with when it is zero:
(x-4)=0 ===> x=4,
so that now you have to plot 2 functions!
For x<= 4: what's inside the absolute value (x-4) is negative, right?, then let's make it +, by multiplying by -1:
|x-4| = -(x-4)=4-x
Then:
for x<=4, y = -x+4-7 = -x-3
for x=>4, (x-4) is positive, so no changes:
y= x-4-7 = x-11,
Now plot both lines. Pick up some x that are 4 or less, for y = -x-3, and some points that are 4 or greater, for y=x-11
In fact, only two points are necessary to draw a line, right? So if you want to go full speed, choose:
x=4 and x= 3 for y=-x-3
And just x=5 for y=x-11
The reason is that the absolute value is continuous, so x=4 works for both:
x=4===> y=-4-3 = -7
x==4 ====> y = 4-11=-7!
abs() usually have a cusp int he point where it is =0
Step-by-step explanation:
