Answer:
1/10 th of a mile every minute
Answer:
−2 is a member of the domain of f.
f(0) > 0
Step-by-step explanation:
Statement 1 is TRUE. Domain value includes set of values of x of the function. -2 is plotted on the x-axis on the graph. Therefore, it is a member of domain of f.
Statement 2 is incorrect. Range includes all possible y-values of the function. On the graph, no y-value plotted is -2. Therefore, -2 is NOT a member of the range of f.
Statement 3 is TRUE.
f(0) is approximately 2.2 on the graph. i.e. at x = 0, y ≈ 2.2.
Therefore, f(0) > 0.
Statement 4 is INCORRECT.
f(2) = 1, that is at x = 2, y = 1 as seen in the graph. Therefore, f(2) is not greater than 2.
Answer:
i need a better veiw
Step-by-step explanation:
Answer:
m<N = 76°
Step-by-step explanation:
Given:
∆JKL and ∆MNL are isosceles ∆ (isosceles ∆ has 2 equal sides).
m<J = 64° (given)
Required:
m<N
SOLUTION:
m<K = m<J (base angles of an isosceles ∆ are equal)
m<K = 64° (Substitution)
m<K + m<J + m<JLK = 180° (sum of ∆)
64° + 64° + m<JLK = 180° (substitution)
128° + m<JLK = 180°
subtract 128 from each side
m<JLK = 180° - 128°
m<JLK = 52°
In isosceles ∆MNL, m<MLN and <M are base angles of the ∆. Therefore, they are of equal measure.
Thus:
m<MLN = m<JKL (vertical angles are congruent)
m<MLN = 52°
m<M = m<MLN (base angles of isosceles ∆MNL)
m<M = 52° (substitution)
m<N + m<M° + m<MLN = 180° (Sum of ∆)
m<N + 52° + 52° = 180° (Substitution)
m<N + 104° = 180°
subtract 104 from each side
m<N = 180° - 104°
m<N = 76°
Answer:
14 ohms
Step-by-step explanation:
The resistance varies directly with length, so increasing the length by a factor will increase the resistance by the same factor.
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The length of the wire increased by a factor of 1.4 from 100 ft to 140 ft. That means the resistance will increase by a factor of 1.4 from 10 ohms to 14 ohms. (The diameter did not change.)
The longer wire will have a resistance of 14 ohms.
