First find the decimal equivalent of square root 3: SQRT(3) = 1.732 ( roughly)
If the base and height were each 3, then the hypotenuse would need to be:
3^2 + 3^2 = C^2
9 + 9 = C^2
18 = C^2
C = SQRT(18) = 4.24
This is larger than sqrt(3), so this cannot be a right triangle.
If one leg was 3 and the other leg was sqrt(3) then the hypotenuse would be:
3^2 + 1.73^2 = C^2
9 + 3 = C^2
12 = C^2
C = SQRT(12) = 3.46
This is larger than 3, this cannot be a right triangle.
The answer is b) no.
Answer:
2
Step-by-step explanation:
Thats your answerrrrrr
5x+2y=20
Substitute 0.3 for x
5(0.3)+2y=20
Multiply 5 by 0.3
1.5+2y=20
Subtract 1.5 from both sides
2y=18.5
Divide 2 on both sides
Final Answer: y= 9.25
The question might have some mistake since there are 2 multiplier of t. I found a similar question as follows:
The population P(t) of a culture of bacteria is given by P(t) = –1710t^2+ 92,000t + 10,000, where t is the time in hours since the culture was started. Determine the time at which the population is at a maximum. Round to the nearest hour.
Answer:
27 hours
Step-by-step explanation:
Equation of population P(t) = –1710t^2+ 92,000t + 10,000
Find the derivative of the function to find the critical value
dP/dt = -2(1710)t + 92000
= -3420t + 92000
Find the critical value by equating dP/dt = 0
-3420t + 92000 = 0
92000 = 3420t
t = 92000/3420 = 26.90
Check if it really have max value through 2nd derivative
d(dP)/dt^2 = -3420
2nd derivative is negative, hence it has maximum value
So, the time when it is maximum is 26.9 or 27 hours
For a vertical projectile, the equation of motion is expressed h(t)=g+bt+c, where g is gravity, b is the initial upward velocity, and c is the starting height. The maxima of this equation is given by -b/2g, which in this case would be -56/-32, or 1.75 seconds after launch
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