Consider the function y=x^2+5x-2 . What would happen to the graph if (x + 5) was substituted in place of the x?A. The graph would shift 5 units to the right.
B. The graph would shift down 5 units.
C. The graph would shift up 5 units.
D. The graph would shift 5 units to the left.
Consider the function y=x^2+5x-2 . What would happen to the graph if (x + 5) was substituted in place of the x?
A. The graph would shift 5 units to the right.
B. The graph would shift down 5 units.
C. The graph would shift up 5 units.
D. The graph would shift 5 units to the left.
Answers
Answer:
Option: D is the correct answer.
D. The graph would shift 5 units to the left.
Step-by-step explanation:
We know that the transformation of a function f(x) to f(x+a) is a shift of the function either to the left or to the right by a units depending on the sign of a i.e. if a>0 then the shift is a units to the left
and if a<0 then the shift is a units to the right.
Here we have:
is converted to:
Here we have a=5>0
Hence, the shift is 5 units to the left.
Related Questions
Simplify (3n – 2m)2 = ? a. 6n2 – 12mn – 4m2 b. 9n2 – 12mn – 4m2 c. 9n2 12mn 4m2 d. 9n2 – 12mn 4m2
Easy math problem, please help.O is the center of the circle. Assume that lines that appear to tangent are tangent. What is the value of x?A. 56B. 62C. 248D. 304
Choose the correct simplification of the expression a to the 7th power times b to the 8th power all over a to the 4th power times b to the 4th power
John had $21 to spend on 2 notebooks. After buying them he had $13. How much did each notebook cost?
Answers
we need to find m - gradient and c - intercept of the graph
since we have been given 2 points we know the x and y coordinates of these 2 points
point A , x = 0 and y = -4
point B , x = 6 and y = 2
substituting these x and y values in y = mx + c form
-4 = m*0 + c
-4 = c
then intercept is -4
substituting x and y in point B
2 = m * 6 + c
since c = -4
2 = 6m - 4
6m = 6
m = 1
since we know m=1 and c = -4
the equation for line AB is
y = 1*x - 4
y = x - 4
Answer: the answer is y + 4 = x
B.1 cm³
C.1 m³
D.1 mm
select all that apply
Answers
I believe it’s A and B
Answers
Answer:
Step-by-step explanation:
To draw the plan of the class foundation using a scale of 5m to 4cm, you'll need to create a scaled-down representation of the room, including the door and windows. Here are the steps to draw the plan:
1. Determine the size of your drawing area. Since the scale is 5m to 4cm, you need to calculate the dimensions of your drawing area.
Length of the class: 20m
Breadth of the class: 10m
Using the scale, for every 5 meters in reality, you will represent it as 4 centimeters in your drawing. So, you'll need a drawing area that can accommodate these dimensions, and the scale conversion.
Length of drawing area = (20m / 5m) * 4cm = 16cm
Breadth of drawing area = (10m / 5m) * 4cm = 8cm
Therefore, your drawing area should be approximately 16cm by 8cm.
2. Draw the outline of the class: Using a ruler and a pencil, draw a rectangle with dimensions 16cm by 8cm to represent the classroom. This rectangle represents the foundation of the class.
3. Draw the door: The door is 8 meters long and 4 meters wide. Using your scale, you'll represent it as 4cm by 2cm in your drawing. Draw a rectangle within the classroom rectangle to represent the door. The top edge of the door should align with one of the longer sides of the classroom.
4. Draw the windows: The windows are 5 meters long and 3 meters wide. Using your scale, you'll represent each window as 4cm by 2.4cm in your drawing. Place the windows where they would be on the classroom walls, leaving space between them and the door.
5. Label the drawing: You can label the door and windows to indicate their dimensions if needed.
Answers
Answers
so 851=23W
Divide both sides by 23 to get the width by itself and you get
37=W or the width is 37 meters
Standard Deviation:________ a1
Answers
Mean = total sum of points / # of data
[Stantard deviation] ^ 2 = {Sum of [every data - mean]^2 } / [number of data - 1]
Procedure:
Mean = total sum of points / # of data
Total sum of points = point of pass + point of fail
points of pass = 90*4 = 360
point of fail = 10*0 = 0
Total sum of points = 360
Number of data = 100
Mean = 360/100 = 3.6
[Stantard deviation] ^ 2 = {Sum of [every data - mean]^2 } / [number of data - 1]
[Sum of every data - mean]^2 = 90*[4 -3.6]^2 + 10* [0 - 3.6]^2 = 14.4 + 129.6 = 144
[Stantard deviation] ^ 2 = [144]/[100-1] =144/99
Standar deviation = √(144/99) = 1.206