Answer:
∠J = 60°
Step-by-step explanation:
The Law of Cosines tells you ...
j² = k² +l² -2kl·cos(J)
Solving for J gives ...
J = arccos((k² +l² -j²)/(2kl))
J = arccos((14² +80² -74²)/(2·14·80)) = arccos(1120/2240) = arccos(1/2)
J = 60°
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<em>Additional comment</em>
It is pretty rare to find a set of integer side lengths that result in one of the angles of the triangle being a rational number of degrees.
Answer:
18, 22
Step-by-step explanation:
Half of 40 is 20 right? Then one number has to be 2 less, while the other number is 2 more. So there is a difference of 4.
<h2>Hello!!!</h2>
Answer:
Vertex: (2,-2)
Axis of symmetry: x=2
Step-by-step explanation:
The vertex is the maximum or minimum point of a quadratic, which we can see on the graph is (2,-2).
The axis of symmetry is the line about which the graph of the parabola is symmetric, which is the line x=2.
Answer:
Jacqueline runs at 6.25 minutes per mile
Step-by-step explanation:
Here we have that the total training time for Jacqueline = 54 minutes
The time for which Jacqueline swims = 30 minutes
Therefore, the time in which Jacqueline runs = 54 min. - 30 min. = 20 minutes
Therefore the equation for Jacqueline's speed = Distance/Time we have
Jacqueline's pace = 3.2 miles/20 minutes = 0.16 miles/minute
Therefore Jacqueline's pace = 0.16 miles/minute
The equation for Jacqueline's pace in miles/minute is presented as follows
Jacqueline's pace in miles/minute = Time of running/Distance
Jacqueline's pace in miles/minute = 1/0.16 miles/minute = 6.25 minutes/mile.
Answer:
Step-by-step explanation:
Part 1:
Let
Q₁ = Amount of the drug in the body after the first dose.
Q₂ = 250 mg
As we know that after 12 hours about 4% of the drug is still present in the body.
For Q₂,
we get:
Q₂ = 4% of Q₁ + 250
= (0.04 × 250) + 250
= 10 + 250
= 260 mg
Therefore, after the second dose, 260 mg of the drug is present in the body.
Now, for Q₃ :
We get;
Q₃ = 4% of Q2 + 250
= 0.04 × 260 + 250
= 10.4 + 250
= 260.4
For Q₄,
We get;
Q₄ = 4% of Q₃ + 250
= 0.04 × 260.4 + 250
= 10.416 + 250
= 260.416
Part 2:
To find out how large that amount is, we have to find Q₄₀.
Using the similar pattern
for Q₄₀,
We get;
Q₄₀ = 250 + 250 × (0.04)¹ + 250 × (0.04)² + 250 × (0.04)³⁹
Taking 250 as common;
Q₄₀ = 250 (1 + 0.04 + 0.042 + ⋯ + 0.0439)
= 2501 − 0.04401 − 0.04
Q₄₀ = 260.4167
Hence, The greatest amount of antibiotics in Susan’s body is 260.4167 mg.
Part 3:
From the previous 2 components of the matter, we all know that the best quantity of the antibiotic in Susan's body is regarding 260.4167 mg and it'll occur right once she has taken the last dose. However, we have a tendency to see that already once the fourth dose she had 260.416 mg of the drug in her system, that is simply insignificantly smaller. thus we will say that beginning on the second day of treatment, double every day there'll be regarding 260.416 mg of the antibiotic in her body. Over the course of the subsequent twelve hours {the quantity|the quantity|the number} of the drug can decrease to 4% of the most amount, that is 10.4166 mg. Then the cycle can repeat.
