So first find a common denominator which is 100 in this question. Then change 5/10 into a denominator of 100 which is 50/100 you don't have to change 9/100 because it has a denominator of 100. So then add the fractions and the answer is 59/100
> 0
First, note that x is undefined at 5. / x ≠ 5
Second, replace the inequality sign with an equal sign so that we can solve it like a normal equation. / Your problem should look like:
= 0
Third, multiply both sides by x - 5. / Your problem should look like: 3x - 5 = 0
Forth, add 5 to both sides. / Your problem should look like: 3x = 5
Fifth, divide both sides by 3. / Your problem should look like: x =
Sixth, from the values of x above, we have these 3 intervals to test:
x <
< x < 5
x > 5
Seventh, pick a test point for each interval.
1. For the interval x <
:
Let's pick x - 0. Then,
> 0
After simplifying, we get 1 > 0 which is true.
Keep this interval.
2. For the interval
< x < 5:
Let's pick x = 2. Then,
> 0
After simplifying, we get -0.3333 > 0, which is false.
Drop this interval.
3. For the interval x > 5:
Let's pick x = 6. Then,
> 0
After simplifying, we get 13 > 0, which is ture.
Keep this interval.
Eighth, therefore, x <
and x > 5
Answer: x <
and x > 5
Answer:
- 10 - (-20 + 5)
Step-by-step explanation:
- 10 - (-20 + 5)
- 10 - (-15) = -10 + 15
-10 + 15 = 5
have a nice day and mark me brainliest! :)
Answer:
g(x)=3
Step-by-step explanation:
Let's find the answer.
W(f,g)=3e^x which can be written as:
W(f,g)=(3)*(e^x), notice that:
(e^x)=f(x) so:
W(f,g)=3*f(x), establishing:
W(f,g)=g(x)*f(x) then:
g(x)=3
In conclusion, g(x)=3.
For the first expression
15 + 2d
A possible word problem would be this:
A person saves $2 per day of money. Before he started saving, he had $15 dollars set aside. Look for the expression that expresses the total amount of money saved in terms of the number of days passed
The second expression is
200 - 2m
A possible word problem would be this:
The distance from school to the park is 200m. A kid riding a bike is traveling at a speed of 2m/s from the school to the park. Write an expression for the distance remaining between the park and the kid.<span />
