Answer:
D) 14 seconds
Step-by-step explanation:
First we will plug 500 in for y:
500 = -4.9t² + 120t
We want to set this equal to 0 in order to solve it; to do this, subtract 500 from each side:
500-500 = -4.9t² + 120t - 500
0 = -4.9t²+120t-500
Our values for a, b and c are:
a = -4.9; b = 120; c = -500
We will use the quadratic formula to solve this. This will give us the two times that the object is at exactly 500 meters. The difference between these two times will tell us when the object is at or above 500 meters.
The quadratic formula is:
Using our values for a, b and c,
The two times the object is at exactly 500 meters above the ground are at 5 seconds and 19 seconds. This means the amount of time it is at or above 500 meters is
19-5 = 14 seconds.
Answer:
(12 2/3)/( 2 1/3)= 38/7 or 5 3/7 or 5.43 decimal
Step-by-step explanation:
Simplify the following:
(12 + 2/3)/(2 + 1/3)
Put 2 + 1/3 over the common denominator 3. 2 + 1/3 = (3×2)/3 + 1/3:
(12 + 2/3)/((3×2)/3 + 1/3)
3×2 = 6:
(12 + 2/3)/(6/3 + 1/3)
6/3 + 1/3 = (6 + 1)/3:
(12 + 2/3)/((6 + 1)/3)
6 + 1 = 7:
(12 + 2/3)/(7/3)
Put 12 + 2/3 over the common denominator 3. 12 + 2/3 = (3×12)/3 + 2/3:
((3×12)/3 + 2/3)/(7/3)
3×12 = 36:
(36/3 + 2/3)/(7/3)
36/3 + 2/3 = (36 + 2)/3:
((36 + 2)/3)/(7/3)
36 + 2 = 38:
(38/3)/(7/3)
Multiply the numerator by the reciprocal of the denominator, (38/3)/(7/3) = 38/3×3/7:
(38×3)/(3×7)
(38×3)/(3×7) = 3/3×38/7 = 38/7:
Answer: 38/7
Answer:
(4) y=(x-3)²-4
I dont have much evidence to give
Answer:
<h2>From least to greatest: C, B and A</h2>
Step-by-step explanation:
To order each Kicker correctly, we have to find the success rate of each one, which is obtained by dividing and then multiplying by 100 to obtained the percentage:
<h3>Success rate of A:</h3>
<h3>Success rate of B:</h3>
<h3>Success rate of C:</h3>
So, comparing all three, from least to greatest we have:
C with 79%, then B with 83%, and A with 86%.
Answer:
Tanto Pedro como Jorge deberán pintar 1/4 de la habitación cada uno.
Step-by-step explanation:
Dado que el cuarto entero es igual a 1, y que Arturo por ser el mayor de los hermanos tiene que pintar 1/2 de él, la parte que falta pintar es igual a 1/2 restante.
De ese 1/2, los hermanos menores, Pedro y Jorge, deben pintar dicha fracción en partes iguales. Por lo tanto, dado que (1/2) / 2 es igual a 1/4, la fracción que tanto Pedro como Jorge deberán pintar es de 1/4 de la habitación.
