Very simple.
Let's say you have an equation.
f(x) = x^2
You are asked to find the value for y when x equals 1.
The new equation is: f(1) = (1)^2
f(1) = 1
When x = 1, y = 1.
The same concept is applied here.
In the graph, where does x equal 0?
It equals zero at the origin.
Is there any y-value associated with 0?
Yes, there is.
Y equals five when x equals 0.
So
h(0) = 5
Answer:
The answer to the equation from question 7 is 14.
Step-by-step explanation:
In question 7, we are given an equation.
2³ + (8 - 5)² - 3
First, subtract 5 from 8 in the parentheses.
2³ + 3² - 3
Next, solve the exponents for 2³ and 3².
8 + 9 - 3
Add 8 to 9.
17 - 3
Subtract 3 from 17.
14
So, the answer to this equation from question 7 is 14.
Answer:-12(2)
Step-by-step explanation: the solution to the question-24 and 12 times 2 is 24 just add a negative to one number
2x-1=0
2x=1
x=1/2
x+5=0
x=-5
Final answer: B
Answer:
11.6
Step-by-step explanation:
<u>Solving with steps as below:</u>
- 31*r+1.1*(35−31)*r = 410.64
- 31r + 1.1*4r = 410.64
- 31r + 4.4r = 410.64
- 35.4r = 410.64
- r = 410.64/35.4
- r = 11.6
<u>Answer is:</u> r =11.6
