Answer:
Step-by-step explanation:
y = 3x + 4
Plugin x = 3 in the equation,
y = 3*3 + 4
= 9 + 4
y = 13
Plugin x = 4 in equation,
y = 3*4 + 4
= 12 + 4
y = 16
Plugin x = 5 in the equation,
y = 3*5 + 4
= 15 + 4
y = 19
2) y =x - 7
Plugin x = 10 in the equation
y = 10 - 7
y = 3
Plugin x = 15 in the equation
y = 15 - 7
y = 8
Plugin x = 20 in the equation
y = 20 - 7
y = 13
Answer: <em>"7 is a solution to the original equation. The value –1 is an extraneous solution."</em>
Step-by-step explanation:
The equation can be solved by squaring both sides:
We can see that -1 and 7 are solutions, but make sure they are not extraneous by substituting them in the original equation:
The square root of 49 equals 7, but the square root of -1 is an imaginary number.
The correct choice is <em>"7 is a solution to the original equation. The value –1 is an extraneous solution."</em>
First, lets create a equation for our situation. Let
be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span>
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:
. So lets replace the values in our formula to find the average of our function:
We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
Answer:
75
Step-by-step explanation:
Number of necklace= k × amount of time spent
6= k × 2/5
K = 6÷2/5
K= 6×5/2
K= 15
Therefore the number of necklace made in 5 hours can be calculated as follows
15 = number of necklace ÷ 5
Number of necklace = 15 ×5
Number of necklace = 75
Answer:
Step-by-step explanation:
<u>Given Second-Order Homogenous Differential Equation</u>
<u>Use Auxiliary Equation</u>
<u />
<u>General Solution</u>
<u />
Note that the DE has two distinct complex solutions where and are arbitrary constants.