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Diano4ka-milaya [45]
3 years ago
14

Please help this is timed! I’ll give brainliest.

Mathematics
1 answer:
lakkis [162]3 years ago
4 0

Answer:

a. Function 1

b. Function 3

c. Function 2, Function 3 and Function 4

Step-by-step explanation:

✔️Function 1:

y-intercept = -3 (the point where the line cuts across the y-axis)

Slope, using the two points (0, -3) and (1, 2):

\frac{y_2 - y_1}{x_2 -x_1} = \frac{2 -(-3)}{1 - 0} = \frac{5}{1} = 5

Slope = 5

✔️Function 2:

y-intercept = -1 (the value of y when x = 0)

Slope, using the two points (0, -1) and (1, -4):

\frac{y_2 - y_1}{x_2 -x_1} = \frac{-4 -(-1)}{1 - 0} = \frac{-3}{1} = -3

Slope = -3

✔️Function 3: y = 2x + 5

y-intercept (b) = 5

Slope (m) = 2

✔️Function 4:

y-intercept = 2

Slope = -1

Thus, the following conclusions can be made:

a. The function's graph that is steepest is the function whose absolute value of its slope is greater. Therefore Function 1 is the steepest with slope of 5

b. Function 3 has a y-intercept of 5, which is the farthest from 0.

c. Function 2, Function 3, and Function 4 all have y-intercept that is greater than -2.

-1, 5, and 2 are all greater than -2.

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It is 4 times greater.

Step-by-step explanation:

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2 years ago
Trey drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 12 hours. When Trey drove
USPshnik [31]

Question was Incomplete;Complete question is given below;

Trey drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 12 hours. When Trey drove home, there was no traffic and the trip only took 8 hours. If his average rate was 20miles per hour faster on the trip home, how far away does Trey live from the mountains?

Do not do any rounding.

Answer:

Trey lives 480 miles from the mountain.

Step-by-step explanation:

Given:

Time taken to drove the mountain =12 hours

Time taken to return back from mountain = 8 hours.

Let the speed at which he drove to mountain be denoted by 's'.

Speed on the trip to home = s+20

We need to find the distance Trey live from the mountains.

Solution:

Let the distance be denoted by 'd'.

Now we know that;

Distance is equal to speed times Time.

framing in equation form we get;

distance from home to mountain d=12s

Also distance from mountain to home d = (s+20)8=8s+160

Now distance is same for both the trips;

so we can say that;

12s=8s+160

Combining the like terms we get;

12s-8s=160\\\\4s=160

Dividing both side by 4 we get;

\frac{4s}{4}=\frac{160}{4}\\\\s=40\ mph

Speed while trip to mountain = 40 mph

Speed while trip to home = s+20=420+20=60\ mph

So Distance d=12s=12\times40 = 480\ miles

Hence Trey lives 480 miles from the mountain.

6 0
2 years ago
Una lancha que viaja a 10 m/s pasa por debajo de un puente 3 segundos después que ha pasado un bote que viaja a 7 m/s, ¿después
ExtremeBDS [4]

Answer:

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

Step-by-step explanation:

Sea el punto debajo del puente el punto de referencia y que ambas lanchas se desplazan a velocidad a continuación, las ecuaciones cinemáticas para cada embarcación son presentadas a continuación:

Bote a 7 metros por segundo

x_{A} = x_{o}+v_{A}\cdot t (Ec. 1)

Lancha a 10 metros por segundo

x_{B} = x_{o}+v_{B}\cdot (t-3\,s) (Ec. 2)

Donde:

x_{o} - Posición debajo del puente, medido en metros.

x_{A}, x_{B} - Posición final de cada embarcación, medido en metros.

v_{A}, v_{B} - Velocidad de cada embarcación, medida en metros por segundo.

t - Tiempo, medido en segundos.

Para determinar la posición en la que ambas embarcaciones se encuentran, se debe determinar el instante en que ocurre a partir de la siguiente condición: x_{A} = x_{B}

Igualando (Ec. 1) y (Ec. 2) se tiene que:

v_{A}\cdot t = v_{B}\cdot (t-3\,s)

Ahora despejamos el tiempo:

3\cdot v_{B} = (v_{B}-v_{A})\cdot t

t = \frac{3\cdot v_{B}}{v_{B}-v_{A}}

Si sabemos que v_{B} = 10\,\frac{m}{s} y v_{A} = 7\,\frac{m}{s}, entonces:

t = \frac{3\cdot \left(10\,\frac{m}{s} \right)}{10\,\frac{m}{s}-7\,\frac{m}{s}}

t = 10\,s

Ahora, la posición de encuentro es: (x_{o} = 0\,m, v_{A} = 7\,\frac{m}{s} y t = 10\,s)

x_{A} = 0\,m + \left(7\,\frac{m}{s} \right)\cdot (10\,s)

x_{A} = 70\,m

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

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