Answer:
A. 7
Step-by-step explanation:
Since the triangle sides are not defined, therefore let the base side have the 12 unit and one of the slant side be 7 unit.
Therefore height of the triangle, h can gotten from the Pythagoras theorem as thus:
7² = (12/2)² + h²
49 = 36 + h²
h² = 49 - 36 = 13
h = √13
knowing the height of the triangle, we can apply same rule to the other unknown slant side, x as thus:
x² = (12/2)² + h²
x² = 36 + 13 = 49
x = √49
x = 7
Answer:
Yes, the bathroom has enough water and shampoo for all of them.
Step-by-step explanation:
70L+ 60S < 5600
Putting 8 into L and 7 into S, gives:
70(8) + 60(7) = 560 + 420 = 980
That is definitely less than 5600, so water is OK.
Now,
0.02L + 0.01S
Putting 8 into L and 7 into S, gives:
0.02(8) + 0.01(7) = 0.16 + 0.07 = 0.23
That's definitely less than 2.5 liters, so shampoo is OK as well.
Hence, bathroom has enough water and shampoo for them.
Procedure:
1) Integrate the function, from t =0 to t = 60 minutues to obtain the number of liters pumped out in the entire interval, and
2) Substract the result from the initial content of the tank (1000 liters).
Hands on:
Integral of (6 - 6e^-0.13t) dt ]from t =0 to t = 60 min =
= 6t + 6 e^-0.13t / 0.13 = 6t + 46.1538 e^-0.13t ] from t =0 to t = 60 min =
6*60 + 46.1538 e^(-0.13*60) - 0 - 46.1538 = 360 + 0.01891 - 46.1538 = 313.865 liters
2) 1000 liters - 313.865 liters = 613.135 liters
Answer: 613.135 liters
Step-by-step explanation:
8-3 = 5,
3-5 = -2 = the x coordinate of B
-2-4 =-6,
-2-6 =-8 = the y coordinate of B
(-2,-8) is the point B
A is 5 to the right of the midpoint, so B is 5 to the left of the midpoint
A is 6 up from the midpoint so B is 6 down from the midpoint
Answer:
• zero: -4, -4/3, 7
• positive: -4 < x < -4/3 . . . or 7 < x
• negative: x < -4 . . . or -4/3 < x < 7
Step-by-step explanation:
Zeros of the function are at x=-4, -4/3, +7. These are the values that make each of the individual factors be zero. For example, x-7=0 when x=7.
The function will be negative for x-values left of an odd number of zeros. It will be positive for x-values left of an even number of zeros (including left of no zeros, which is to say right of all zeros). This is because the sign of the factor giving rise to the zero changes for x-values on either side of that zero. (This is not true for zeros with even multiplicity, as the sign does not change at those.)
