<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>
<em><u>Solution:</u></em>
Given that,
<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>
<em><u>At what times will the object be at a height of 56 feet</u></em>
<em><u>Substitute h = 56</u></em>
Solve the above equation by quadratic formula
Thus, at 2 second and 1.75 second, the object be at a height of 56 feet
Answer:
2x^2-6x^4-2
Step-by-step explanation:
<h3>
Answer: Approximately 6.58 years old</h3>
The more accurate value is 6.57881347896059, which you can round however you need. I picked two decimal places.
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Explanation:
Let's pick a starting value of the car. It doesn't matter what the starting value, but it might help make the problem easier. Let's say A = 1000. Half of that is 1000/2 = 500.
So we want to find out how long it takes for the car's value to go from $1000 to $500 if it depreciates 10% per year.
The value of r is r = 0.10 as its the decimal form of 10%
t is the unknown number of years we want to solve for
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y = A(1 - r)^t
500 = 1000(1 - 0.1)^t
500 = 1000(0.9)^t
1000(0.9)^t = 500
0.9^t = 500/1000
0.9^t = 0.5
log( 0.9^t ) = log( 0.5 )
t*log( 0.9 ) = log( 0.5 )
t = log( 0.5 )/log( 0.9 )
t = 6.57881347896059
Note the use of logs to help us isolate the exponent.
Answer:
y=0.5x-2
Step-by-step explanation:
if it is perpendicular to the line y=-2x-6, then you know that its slope is the negative reciprocal of that line, and it has a different y intercept which you need to solve for using the point given. You solve by plugging in the x and y values from the point and plugging in the slope into the standard equation, and solving for b, the y intercept
y=0.5x+b
1=0.5(6)+b
1=3+b
-2=b
