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grigory [225]
3 years ago
13

Solve for x 2(X-5) +6=7(X-1)-2 (x -6)

Mathematics
1 answer:
e-lub [12.9K]3 years ago
7 0

Answer:

x = - 3

Step-by-step explanation:

2x - 10 + 6 = 7x - 7 - 2x + 12

2x - 4 = 5x + 5

- 5 - 4 = 5x - 2x

-9 = 3x

x = - 9/3

x = - 3

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Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in s
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5 0
3 years ago
Can someone help me with all this? Lol idk any of this and E-learning sucks I need to show my work as well
Margarita [4]

Answer:

1a) 230 feet

1b) 3.75s

1c) t=7.54s

2) t=2s

3) t=2.32s

Step-by-step explanation:

The equation that models Brett's last home run is  h(t)=-16t^2+120t+5

We need to complete the square to obtain the function in the vertex form:

h(t)=-16(t^2-7.5t)+5

h(t)=-16(t^2-7.5t+3.75^2)+5+-16(-3.75)^2

h(t)=-16(t-3.75)^2+230

1a) The vertex is (3.75,230).The maximum value is the y-coordinate of the vertex, which is 230 feet.

1b)  The time it takes to reach the maximum  value is the x-coordinate of the vertex. It reach the maximum heigth after t=3.75 seconds

1c) To find the time the ball hit the ground, we equate the h(t)=0 and solve for t.

-16(t-3.75)^2+230=0

-16(t-3.75)^2=-230

(t-3.75)^2=14.375

t-3.75=\pm \sqrt{14.375}

t=3.75\pm \sqrt{14.375}

t=3.75\pm 3.79

t=3.75-3.79\:\:or\:t=3.75+3.79

t=3.75-3.79\:\:or\:t=3.75+3.79\\t=-0.04\:\:or\:t=7.54

The time is positive, so the ball hit the ground after 7.54 seconds.

Question 2)

The function that models the amusement ride is h(t)=-16t^2+64t+60

We want to find the time it takes for the riders to reach the maximum height.

This time is given by: t=-\frac{b}{2a}

Comparing  h(t)=-16t^2+64t+60 to h(t)=at^2+bt+c we have a=-16, b=64, c=60.

We substitute to obtain:

t=-\frac{64}{2*-16} \\t=-\frac{64}{-32} \\t=2

Hence it took 2 seconds to rech the maximum height.

Question 3)

The equation that models the height f the catridge is h(t)=-16t^2+35t+5

To find the  time that the catridge will land, we equate the function to zero and solve for t.

-16t^2+35t+5=0

This is a quadratic equation with =-16, b=35, an c=5

The solution is given by:

t=\frac{-b\pm \sqrt{b^2-4ac} }{2a}

We substitute the values to get:

t=\frac{-35\pm \sqrt{35^2-4*-16*5} }{2*-16}

This gives:

t=\frac{35-\sqrt{1545} }{32} \:or\: t=\frac{35+\sqrt{1545} }{32}

This simplifies to:

t=-0.13\:or\: t=2.32

The time it takes to land must be positive thefore t=2.32 seconds

7 0
3 years ago
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