Answer:
approx. 6 friends brainiest rate 5 stars and thank mee : )
Step-by-step explanation:
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:
Answer:
70.8
Step-by-step explanation:
YOU'RE WELCOME
Answer:
D. 30 sq in.
Step-by-step explanation:
10+9+6+5=30 sq in.
Given:
The zeros of the polynomial are 
.
Degree = 4
Leading coefficient = 1
To find:
The polynomial.
Solution:
If c is a zero of a polynomial, then (x-c) must be a factor of the polynomial.
Here, -2,4,-5, 5, are zeros of the required polynomial, so (x+2), (x-4), (x+5), (x-5) are factors of required polynomial.
![[\because a^2-b^2=(a-b)(a+b)]](https://tex.z-dn.net/?f=%5B%5Cbecause%20a%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29%5D)
Using distributive property, we get
On combining like terms, we get
Here, the leading coefficient is 1. So, it is the required polynomial.
Therefore, the correct option is E.
