Answer:
The exponential function that passes through (2,36) is:
.
Step-by-step explanation:
We are asked to find which function passes through the point (2,36).
i.e. we will put the input value '2' in the following given functions and check which gives the output value as '36'.
1)
now we put x=2.
hence option 1 is correct.
2)
Now we put x=2.
Hence, option 2 is incorrect.
3)
Now we put x=2
Hence, option 3 is incorrect.
4)
Now we put x=2.
Hence, option 4 is incorrect.
Hence, option 1) is correct.
i.e. The exponential function that passes through (2,36) is:
Answer:
Option A. is the answer.
Step-by-step explanation:
In this question we can get the correct option by plugging in the coordinates of point (2, 36) in the functions given in all options.
Option A.
For (2, 36),
36 = 4×9
36 = 36
It's true so the function passes through the point (2, 36).
Option B.
For, (2, 36)
36 = 4×8
36 = 32
Which is not true.
Therefore, option B is not the answer.
Option C.
For(2, 36)
36 = 6×9
36 = 54
It's not true.
Therefore, option C is not the nswer.
Option D.
For (2, 36),
36 = 6×9
36 = 54
Which is not true.
Therefore, option D is not the answer.
Answer:
Mr.Smith paid $29.4.
Step-by-step explanation:
Mr.Smith paid $24.5 plus additional 20% of $24.5, and to figure out the total amount that he paid, we need to know what is 20% of $24.50.
20% of $24.50 is:
.
Mr.Smith paid additional $4.9; therefore, the total amount he paid was
$24.5 + $4.9 = $29.4.
Mr. smith paid $29.4.
Equation 2: 3m = 4 + 4n
Step 1:
−3(m) = −3(8 + 2n) [Equation 1 is multiplied by −3.]
3m = 4 + 4n [Equation 2]
Step 2:
−3m = −24 − 6n [Equation 1 in Step 1 is simplified.]
3m = 4 + 4n [Equation 2]
Step 3:
−3m + 3m = −24 − 6n + 4n [Equations in Step 2 are added.]
Step 4:
0 = −24 − 2n
Step 5:
n = −12
In which step did the student first make an error?
Step 4
Step 3
Step 2
Step 1
The student first makes an error in the Step 3 where he addsequations in Step 2 to use the elimination method.
To create an equation in one variable using the elimination method, you can either add or subtract the equations. To eliminate a variable, add the equations when the coefficients of one variable are in opposition, and subtract the equations when the coefficients of one variable are in equality.
How to solve this problem?
Notice that the student uses the elimination method to solve the equations. In Step 1, he makes the coefficients of m equal in both equations. In Step 2, he simplifies the previous step. In Step 3, he wants to add both equations to create an equation in one variable. But He forgot to add 4 of Equation 2. It's a mistake.
Therefore the student first makes an error in the Step 3 where he addsequations in Step 2 to use the elimination method.
Know more about the elimination method here -
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Answer:
triangle inequality: the sum of the length of two sides of a triangle must always be greater than the length of the third side.
we know 2 sides whose sum = 5+8 = 13cm
the 3rd side must be <13 included
the length of the third side must be less than 13 cm, i.e. between 1 and 13 (it cannot be equal to 0 because in this case it does not exist)
we are given points A and B
A=(3,150)
B=(4,200)
we can find points
x1=3 , y1=150
x2=4 , y2=200
(a)
the change in y-values from Point A to Point B is
now, we can plug values
...........Answer
(b)
the change in x-values from Point A to Point B is
now, we can plug values
...........Answer
(c)
the rate of change of the linear function in feet per second is
now, we can plug values
.................Answer
The straight line graph indicates that rate of change of the distance traveled with time by the red kangaroo is a constant
Reasons:
The y-values at point A = 150 feet
The y-values at point B = 200 feet
Change in y-values from point A to point B, Δy is given as follows;
Δy = y-values at point B - y-values at point A
Δy = 200 feet - 150 feet = 50 feet
Change in y-values from point A to point B, Δy = 50 feet
The x-values at point A = 3 seconds
The x-values at point B = 4 seconds
Change in x-values from point A to point B, Δx is given as follows;
Δx = x-values at point B - x-values at point A
Δx = 4 seconds - 3 seconds= 1 second
Change in x-values from point A to point B, Δx = 1 second.
The rate of change of the linear function is , therefore, we have;
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