Answer:
Bryce is wrong in step 1 because he did not distribute 3 over 5/3
Explanation
Given the steps taken by bryce as shown, we are to find where he made an error
Given the expression;
Step 1:Expand the bracket using the distributive law;
8/3 = 3c + 3(5/3)
<em>Simplify</em>
8/3 = 3c + 15/3
Step 2: Subtract 15/3 from both sides
8/3 - 15/3 = 3c+15/3-15/3
(8-15)/3 = 3c
-7/3 = 3c
Step 3: Multiply both sides by 1/3
-7/3 * 1/3 = 3c * 1/3
-7/9 = c
Swap
c = -7/9
From the calculation, we can see that Bryce is wrong in step 1 because he did not distribute 3 over 5/3 thereby making his solution incorrect
Quadratic is in the form
ax^2+bx+c=0
so distribute and stuff and simplify
remember
a(b+c)=ab+ac
(x+2)^2+5(x+2)-6=0
remember order of opertaions
(x+2)(x+2)+5(x+2)-6=0
x^2+4x+4+5x+10-6=0
add like terms
x^2+9x+8=0
Answer:
1 solution
Step-by-step explanation:
s² - 35 = -35
Add 35 to both sides: <em>s² = 0</em>
<em>s can only be 0, and 0 has no positives nor negatives</em>, so one number.
Answer:
and 
Step-by-step explanation:
A simple way to solve this problem is to plug the corresponding x and y into the function. We need only one pair since all the functions are quasi-linear (y=kx) and the increase is proportional.
In when x=3, y=15/4≈2.14
In 
when x=3, y=1.8
In when x=3, y≈2.33
In 
when x=3, y≈1.90
We can observe that in two cases, and , y is greater than 2.
