The numerical length of JK is 21 units.
Given that, point J is on line segment IK.
We need to determine the numerical length of JK.
In geometry, a line segment is a part of a line that is bounded by two distinctend points and contains every point on the line that is between its endpoints.
Given JK=x+6, IJ=9, and IK=2x.
Now, IK=JK+IJ
⇒2x=x+6+9
⇒2x=x+15
⇒x=15
Now, JK=x+6=21
Therefore, the numerical length of JK is 21 units.
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The distance of the line segment is solved and length of JK = 21 units.
Given data:
Since point J is on the line segment IK, the sum of lengths JK and IJ should be equal to the length of the whole line segment IK.
IJ + JK = IK
IJ = 9
IK = 2x
Now, substitute the values and solve for JK:
9 + JK = 2x
To find JK, we need to isolate it on one side of the equation. Subtract 9 from both sides:
JK = 2x - 9
Now, we are also given that JK = x + 6, so we can set these two expressions equal to each other:
2x - 9 = x + 6
Subtract x from both sides:
2x - x = 6 + 9
x = 15
Now, substitute the value of x back into the expression for JK:
JK = 2(15) - 9
JK = 30 - 9
JK = 21
Hence, the numerical length of JK is 21 units.
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Answer:
8
Step-by-step explanation:
Of those 9 even integers, 8 can be written as the sum of two prime numbers.
4 = 2+2; 6 = 3+3; 8 = 3+5; 10 = 5+5; 12 = 5+7; 14 = 7+7; 16 = 5+11; 18 = 7+11
b. x^2 - 2x + 2
c. x^2 + 2x – 12
d. 2x^2 + 4x – 24
Answer:
(2x²+4x-24) in.
Step-by-step explanation:
.
Answer: $16
Step-by-step explanation:
Given: Marie has renters insurance that she must pay twice a year.
The amount of each payment = $96
So, the total payment in the year =
Since, in one year = 12 months
Therefore, the amount of money she should set aside each month to cover her renters insurance=
Hence, She should set aside $16 each month to cover her renters insurance.
Answer: $24.50 is the price for each adult ticket.
Step-by-step explanation:
1. First we find out the price of the childrens' tickets.
$18.50 x 3 = $55.50
2. Subtract kids tickets from total cost to get adults tickets.
104.50 - 55.50 = 49.00
3. $49 is the price for both adult tickets so we need to divide this by 2.
49/2= 24.50
The equation for the aircraft's flight is a quadratic equation representing the height of the aircraft at any given time. By rearranging the equation to isolate time and applying the quadratic formula, we can find the time at which the aircraft reaches its maximum height, which in this case is 3.79 minutes.
The flight of an aircraft from Toronto to Montreal is modeled by the equation h = -2.5t2 + 200t where t represents time in minutes and h represents height in meters. This is fundamentally a quadratic equation which is utilized in physics to characterize motion under constant acceleration. In this case, it models the height of the aircraft at any given time.
To find the time at which the airplane's maximum height is achieved, we must solve the equation for t. By rearranging the equation, we can isolate t, yielding a quadratic equation as follows: 0 m = 0 m + (10.0 m/s) t + (2.00 m/s2) t2. This simplifies to 200 = 10t + t2.
Applying the quadratic formula, we find two solutions for t, 3.79 s and 0.54 s. The time it takes the aircraft to reach its maximum height would be the longer solution, which is 3.79 minutes in this case.
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The question provides a quadratic equation to model the flight of an aircraft. This equation can be used to calculate the height of the aircraft at a specific time or to determine when the aircraft reaches its maximum height.
The question is asking about the trajectory of an aircraft as modelled by a quadratic equation, and specifically, how time influences height. The equation given is h = -2.5t²+200t. Quadratic equations are frequently used to describe the motion of objects when the acceleration is constant. This equation tells us that the height of the aircraft is dependent on the time squared and the time.
To solve for a specific time (t), we can plug the desired time into the equation to find the height of the aircraft at that time. For instance, if we want to find out the height of the aircraft 10 minutes into the flight, we would substitute t=10 into the equation, giving us h=-2.5 × (10)²+200 × (10). Simplifying this equation would provide the height of the aircraft 10 minutes into the flight.
Additionally, this equation could also be used to find the maximum height of the aircraft. The maximum height is reached when the derivative of the equation equals zero. Taking the derivative of h = -2.5t²+200t and setting it equal to zero will provide the time when the maximum height is reached.
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