Answer:
See below.
Step-by-step explanation:
a.
The first figure has 1 square. The second figure has a column of 2 squares added to the left. The third figure has a column of 3 squares added to the left. Each new figure has a column of squares added to the left containing the same number of squares as the number of the figure.
b.
Figure 10 has 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55 squares.
c.
The formula for adding n positive integers starting at 1 is:
1 + 2 + 3 + ... + n = n(n + 1)/2
For figure 55, n = 55.
n(n + 1)/2 = 55(56)/2 = 1540
d.
Let's use the formula set equal to 190 and solve for n. If n is an integer, then we can.
n(n + 1)/2 = 190
n(n + 1) = 380
We know that 380 = 19 * 20, so n = 19.
Answer: yes
e.
Use the formula above,
S = n(n + 1)/2, where S is the sum.
f.
n(n + 1) = 1478
38 * 39 = 1482
37 * 38 = 1406
There will 80 characters per each line hold. Hope it help!
I’m not exactly sure what you mean by “write it in math,” however I can tell you what I think.
4y=x
(It says that x is 4 times the value of y which means that y times 4 is x.)
For the answer to the question above, If you've noticed, the statement mentioned "a number". Let's assign x to be the number. The first statement is twice a number (2x). Then, decreased by the quotient of that number and 2 (-x/2). Lastly, the statement at least 12 is written as >12.
So the answer is
2x - (x/2) >12
This is an exponential function.
If x = 0, 2^x = 2^0 = 1. The beginning value of 2^x is 1 and the beginning value of 51*2^x is 51.
Make a table and graph the points:
x y=51*2^x point (x,y)
-- --------------- ---------------
0 51 (0,51)
2 51*2^2 = 51(4) = 204 (2,204) and so on.
The graph shows up in both Quadrants I and II. Its y-intercept is (0,51). Its slope is always positive.
