The original volume of the shape was 250. You come up with this by multiplying all sides together. So, 5x5x10=250.
We do the same for the new shape, which now has a 7 in place of the 5. So 7x7x10=490.
Lastly, we subtract the volume of the new shape from the volume of the old shape to see the difference. So, 490-250=240.
So the answer to this is 240.
Since the definition of a parallelogram literally says that it's a quadrilateral with 2 parallel sides, we have AB || CD so the 2 other sides (AC and BD) need to be either similar or parallel (either works, due to that if they're similar we can prove that since AB || CD that they're parallel). B is our answer
<span>1.Describe how the graph of y = x2 can be transformed to the graph of the given equation.
y = (x+17)2
Shift the graph of y = x2 left 17 units.
2.Describe how the graph of y= x2 can be transformed to the graph of the given equation.
y = (x-4)2-8
Shift the graph of y = x2 right 4 units and then down 8 units.
.Describe how to transform the graph of f into the graph of g.
f(x) = x2 and g(x) = -(-x)2
Reflect the graph of f across the y-axis and then reflect across the x-axis.
Question 4 (Multiple Choice Worth 2 points)
Describe how the graph of y= x2 can be transformed to the graph of the given equation.
y = x2 + 8
Shift the graph of y = x2 up 8 units.
Question 5 (Essay Worth 2 points)
Describe the transformation of the graph of f into the graph of g as either a horizontal or vertical stretch.
f as a function of x is equal to the square root of x and g as a function of x is equal to 8 times the square root of x
f(x) = √x, g(x) = 8√x
vertical stretch factor 8
Plz mark as brainlest</span>
Answer:
The equation, y=3 x,
Step-by-step explanation:
y=Distance traveled
x =Total time
Also, in terms of straight line
Slope =3= uniform Velocity
Point (3,9) and (5,15) represents Distance traveled in 3 (unit of time) =9 unit, and 15 unit=Distance traveled in 5 (Unit of time).
→Alonso is moving with uniform speed=3 (unit of time), as velocity remains constant in the entire process.
