Answer:
90 nickels and 15 dimes
Step-by-step explanation:
If one is six times the other, then you would have 7 equal piles. Divide 105 by 7 to get 15 which is the number in each pile(number of dimes). Then 105 minus 15 gives you the total of the rest of the piles(number of nickels).
Check:
90 plus 15 equals 105, and
15 times 6 equals 90.
Answer:
y = 3/2x + 7
Step-by-step explanation:
The slops is rise over run, so in this case the slope is 3/2 because it goes up 3 units while going 2 units to the right.
The points (0,7) is on the y axis. This means that 7 is the y intercept
With this information, we can follow the format of the slope intercept equation: y = mx + b
m stands for the slope and b stands for the y intercept. Plugging the information in, our equation is:
y = 3/2x + 7
Any ratio which can be simplified to 1/7 is equivalent to it. So, for example, if you multiply this fraction by 2, you get 2/14, which is equivalent. Or 3/21, or 4/28, or 5/35...
Answer:
x = 35°
Step-by-step explanation:
assuming that the straight line at the bottom of the figure is actually a straight line, that means all the angles along that straight line must add up to 180 °
i.e.
40 + (2x+30) + 40 = 180
40 + 2x+30 + 40 = 180
110 + 2x = 180 (subtract 110 from both sides)
2x = 180 - 110
2x = 70
x = 35
You have to estimate the slope of the tangent line to the graph at <em>t</em> = 10 s. To do that, you can use points on the graph very close to <em>t</em> = 10 s, essentially applying the mean value theorem.
The MVT says that for some time <em>t</em> between two fixed instances <em>a</em> and <em>b</em>, one can guarantee that the slope of the secant line through (<em>a</em>, <em>v(a)</em> ) and (<em>b</em>, <em>v(b)</em> ) is equal to the slope of the tangent line through <em>t</em>. In this case, this would be saying that the <em>instantaneous</em> acceleration at <em>t</em> = 10 s is approximately equal to the <em>average</em> acceleration over some interval surrounding <em>t</em> = 10 s. The smaller the interval, the better the approximation.
For instance, the plot suggests that the velocity at <em>t</em> = 9 s is nearly 45 m/s, while the velocity at <em>t</em> = 11 s is nearly 47 m/s. Then the average acceleration over this interval is
(47 m/s - 45 m/s) / (11 s - 9 s) = (2 m/s) / (2 s) = 1 m/s²
