Answer:
a= domain = 3x range = 21-3 = 18
b= $200 As 0(B) = Point of origin balance = midway to $400
= 0 < 200 > 400 = midway 200.
c=function notation = 3x -3x + 6 = 21 we change the 3x to cancel out any balance.
d= 4th segment shows day 12 to day 15 across the x axis.
ok we see and count after reading this it has 21 days in total approx and 12th day it shows the function went to 12 days.
We can write F(x) = 12x+6 to equal the change in value then we need to distribute 12 x = 6
Which is the same as 9 days money = 6 days money
As 3 days was made reference to day 12- day 15 where she had zero in account. We can minus - 3 also to represent her last savings before adding the balance of 6.
ok no interest the maximum was $400 the zero y intercept (0.0)= $200
This makes everything so easy now.
We count 21 days = 21 this is what we will end with
Let 4 be x
4x -3 + 6
We count 5 change $ values 0, 1 , 2, 3, 4 on the graph and also 4 in repeat which makes the 400.
or could mean she started with 400 and the rest is interest. Now we know this we cna start writing a domain and i will keep this open and edit and try work this out for you.
Step-by-step explanation:
The last step for question one is converting 12 into 12
This means 4x = 3 so we change it to
3x -3 +6 = 21
Then you now can recognize and accept x = 4
Answer:
(a)
Domain: 0 to 21
Range: 0 to 400
(b)
B(0) = 200.
(c)
B(12) = 0
(d)
Segment 4. The value of B(d) in Segment 4 is 0.
Step-by-step explanation:
(a) The domain is the number of days. The range is the amount of money which is the number of dollars. The account was open for 3 weeks, and the graph shows the entire 3 weeks. 3 weeks = 21 days; the highest amount of money was $400.
Domain: 0 to 21
Range: 0 to 400
(b) B(0) is very close to half of the maximum value of B(d). Since the maximum value of B(d) = 400, B(0) = 200. B(0) means the value of the function at time 0. The account was opened with an initial deposit at time = 0. B(0) is the value of the account at time = 0 which is the initial deposit made to open the account.
(c)
B(12) = 0
(d)
There are 6 segments:
Segment 1: B(d) goes from $200 to $400.
Segment 2: B(d) remains at $400.
Segment 3: B(d) goes from $400 to $0.
Segment 4: B(D) remains at $0.
Segment 5: B(d) goes from $0 to $100.
Segment 6: B(d) goes from $100 to $300.
In Segment 4, the value of B(d) is 0, so Segment 4 represents days 12 through 15.
If the areas of two similar polygons are in the ration 64:81, then the ratio of their corresponding sides are
"A polygon, in geometry, any closed curve consisting of a set of line segments (sides) connected such that no two segments cross."
The simple polygons are triangles (3 sides), quadrilaterals (4 sides), and pentagons (5 sides).
Now if we take the given polygon as square (4 sides).
Then,
Let the sides of squares are a and b.
Hence, the ratio of the corresponding sides of polygon is
Learn more about the polygons here
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The required inequality for the least amount of money he needs to deposit to avoid a fee is 727.29 - 248.50 + x ≥ 500 and he needs to deposit at least $21.21 in her account to avoid a fee.
A statement of an order relationship-greater than, greater than or equal to, less than, less than or equal to- between two numbers or algebraic equations.
Now it is given that,
Amount in the account = $727.29
Amount in the check = $248.50
Amount need to maintain = $500
Now let x be the Tony needs to deposit.
So, total balance in the account = 727.29 - 248.50 + x
Since, he must maintain a $500 balance to avoid a fee. That means the balance can be greater than or equal to $500.
Thus the required inequality is:
727.29 - 248.50 + x ≥ 500
Adding alike terms,
468.79 + x ≥ 500
Subtracting 468,79 both the side we get,
x ≥ 500 - 468.79
Solving we get,
x ≥ 21.21
Therefore, he needs to deposit at least $21.21 in her account to avoid a fee.
Thus,the required inequality for the least amount of money he needs to deposit to avoid a fee is 727.29 - 248.50 + x ≥ 500 and he needs to deposit at least $21.21 in her account to avoid a fee.
To learn more about inequality:
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B)write an equation of the line tangent to the graph of f at the point where x=1
C) find the x coordinate of each point at which the line tangent to the graph of f is parallel to the line y=-2x+4
A convenient way to find the zeros of is by factoring.
a) The equation,
can be rewritten as,
We can think of this equation as a quadratic equation in , with .
Observe that .
We find two factors of 2 that adds up to . These are, .
Now let us split the middle term. to obtain;
We can factor to get,
We factor further to obtain;
Hence the zeroes of are;
b) To find the line tangent, we must first, find the slope using differentiation. That is,
At ,
Also, we need to determine the value at . That is;
Now we can use the slope and the point to write ythe equation of the line tangent.
c)
If the line tangent is parallel to the line , then
Since parallel lines have the same slope.
Hence the x-coordinates are,