We need to find, how many 1/3s are in 1 2/3 the hexagon represents 1 whole.
In order to find that value, we need to divide 1 2/3 by 1/3.
Let us first convert 1 2/3 into improper fraction first.
1 2/3 = (1*3+2)/3 = 5/3.
Therefore,
5/3 ÷ 1/3.
Changing division sign into multiplication and flipping the second fraction, we get
5/3 × 3/1
3's cross out from top and bottom, we get
= 5.
<h3>Therefore, there are 5 times of 1/3s in 1 2/3.</h3>
<h3><u>
Answer:</u></h3>
Option: D
Horizontal stretching.
<h3><u>
Step-by-step explanation:</u></h3>
We have to find the effect on the graph of the function f(x)=2x when it is replaced by f(0.5 x).
We know that when a parent function f(x) is replaced by f(kx) then either the graph is stretched horizontally or shrinked horizontally.
if k>1 then the graph is shrinked horizontally.
if k<1 then the graph is stretched horizontally.
Hence here k=0.5<1 so the graph of the function is stretched horizontally.
Answer:
y = (3 / (x-7)) - 5
Step-by-step explanation:
The original equation has the asymptotes at x = 0 and y = 0.
We want to translate to the right by 7 units so that the asymptote is at x = 7 and downwards 5 units so that the asymptote is y = -5.
Now, in the first case, to move it to the right, you must subtract 7 from the independent variable (i.e. x)
And in the second case, to move a function down 5 units, subtract 5 units from the entire function.
Applying the above, the new function would be:
y = (3 / (x-7)) - 5
