Answer:
0 = x2 - 9x + 20
Step-by-step explanation:
y + 5x = x² + 10 Simplify this equation
- 5x - 5x
y = x² - 5x + 10
Set both equations equal to each other since they both equal y
x² - 5x + 10 = 4x - 10
-4x - 4x
x² - 9x + 10 = -10
+ 10 + 10
x² - 9x + 20 = 0
If this answer is correct, please make me Brainliest!
The name of place value of 9 in 199 is ones.
The second 9 has a value of tens.
The number 1 has a place value of hundreds.
Answer:
- correct answer is C
- Haley incorrectly applied the distributive property
Step-by-step explanation:
If you simplify the given equation, you find it matches choice C.
__
Haley's error seems to be failing to distribute the 1/2 properly when she eliminated parentheses. Apparent, she incorrectly decided that ...
1/2(6 -x) ⇒ 3 -x . . . . instead of 3 -1/2x
Then when -x was added to +3x, she got 2x. Had she done it properly, she would have added -1/2x to +3x to get 5/2x.
_____
<em>Additional comment</em>
It is a common error to "distribute" the factor outside parentheses to the first term only, as Haley apparently did. Another common error is to fail to distribute minus signs properly. The distributive property requires you apply the outside factor to <em>all</em> of the terms inside parentheses.
First let’s establish a few things, The first car can travel 300 miles with 20 gallons of gas, if 14 gallons of gas costs 49.49 then for ever gallon of gas Sam is paying 3 dollars and 53.5 cents. That means for 20 gallons, Sam will need around 71$ for the entire trip if he uses the first car.
What are you trying to do here?
Solve the graph, or make it appear as something else?
First, we're going to take one sec (x) out so that we get:
sec (x) (2sec (x) -1 -1) = 0
sec (x) (2sec (x) -2) = 0
Then we're going to separate the two to find the zeros of each because anything time 0 is zero.
sec(x) = 0
2sec (x) - 2 = 0
Now, let's simplify the second one as the first one is already.
Add 2 to both sides:
2sec (x) = 2
Divide by 3 on both sides:
sec (x) = 1
I forgot my unit circle, so you'd have to do that by yourself. Hopefully, I helped a bit though!
