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topjm [15]
3 years ago
14

Arthur drops a ball from a height of 81 feet above the ground. Its height, h, is given by the equation h = –16t2 + 81, where t i

s the time in seconds. For which interval of time is the height of the ball less than 17 feet?
Mathematics
2 answers:
Lorico [155]3 years ago
6 0

Answer:

Step-by-step explanation:

We are given the position function and need to find the value of t when h<17.

Create an inequality that represents this situation:

-16t^2+81 The "less than" sign makes this very specifically a conjunction problem as opposed to a disjunction. That's important to the solution. But we'll get there.

The simplest way to solve this is to subtract 81 from both sides:

-16t^2 then divide both sides by -16:

t^2>4 Notice now that the sign is facing the other way since we had to divide by a negative number. Now it's a disjunction. The solution set to this inequality is that t>2 or t<-2. First and foremost, time will never be negative, so we can disregard the -2. Even if that was t<2, the more time that goes by, the greater the time interval is, not the lesser. It's the "<" that doesn't make sense, not only the -2. The solution to this inequality is

t > 2 sec. That means that after 2 seconds, the height of the ball is less than 17 feet.

MariettaO [177]3 years ago
3 0

Answer:

A on edg

Step-by-step explanation:

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Usimov [2.4K]

Answer:

2x+3

Step-by-step explanation:

-2x2 - 7x - 6

--------------------

-x-2

-2x2 - 7x - 6  =   -1 • (2x2 + 7x + 6)

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 2x2 + 7x + 6

x • (2x+3) and      (x+2)  •  (2x+3)

answer- 2x+3

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5 0
3 years ago
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sweet-ann [11.9K]

Answer:

<u>Third Option</u>: y = \frac{5}{4}x

Step-by-step explanation:

Given the points on the graph, (4, 5) and (-4, -5):

In order to determine the equation of the given graph in slope-intercept form, y = mx + b:

Use the given points to solve for the slope:

Let (x₁, y₁) = (-4, -5)

(x₂, y₂) = (4, 5)

m = (y₂ - y₁)/(x₂ - x₁)

m = \frac{5 - (5)}{4 - (-4)}  = \frac{5 + 5}{4 + 4}  = \frac{10}{8} = \frac{5}{4}

Therefore, the slope of the line is: m = \frac{5}{4}.

Next, use one of the given points on the graph, (4, 5) to solve for the y-intercept, b:

y = mx + b

5 = \frac{5}{4} (4) + b

5 = 5 + b

5 - 5 = 5 - 5 + b

0 = b

Therefore, the linear equation in slope-intercept form is: y = \frac{5}{4}x.  The correct answer is Option 3.

8 0
3 years ago
7. Are the ordered pairs below an arithmetic sequence or geometric
DochEvi [55]

Answer:

8. Arithmetic Progression

9. f(9) = 300

Step-by-step explanation:

Given

\{(1,55)\ (2, 45)\ (3, 35)\ (4, 25)\}

Solving (8): Arithmetic or Geometric

We start by checking if it is arithmetic by checking for common difference (d).

d = y_2 - y_1 = y_3 - y_2 = y_4 - y_3

This gives:

d = 45 - 55 = 35 - 45 = 25 - 35

d = -10 = -10 = -10

d=-10

<em>Because the common difference is equal, then it is an arithmetic progression</em>

<em></em>

Solving (8):

f(n) = f(1) + f(n-1)

To find f(9), we substitute 9 for n

f(9) = f(1) + f(9-1)

f(9) = f(1) + f(8)

We need to solve for f(8); substitute 8 for n

f(8) = f(1) + f(8 - 1)

f(8) = f(1) + f(7)

We need to solve for f(7); substitute 7 for n

f(7) = f(1) + f(7 - 1)

f(7) = f(1) + f(6)

We need to solve for f(6); substitute 6 for n

f(6) = f(1) + f(6 - 1)

f(6) = f(1) + f(5)

We need to solve for f(5); substitute 6 for n

f(5) = f(1) + f(5 - 1)

f(5) = f(1) + f(4)

From the function, f(4) = 25 and f(1) = 55.

So:

f(5)=55 + 25

f(5)=80

f(6) = f(1) + f(5)

f(6) = 55 + 80

f(6) = 135

f(7) = f(1) + f(6)

f(7) = 55 + 135

f(7) = 190

f(8) = f(1) + f(7)

f(8) = 55 + 190

f(8) = 245

f(9) = f(1) + f(8)

f(9) = 55 + 245

f(9) = 300

8 0
3 years ago
Find an upper limit for the zeroes 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0
erik [133]

<u>Answer-</u>

2 is the upper limit for the zeros.

<u>Solution-</u>

The given function f(x) is,

2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

For calculating the zeros,

\Rightarrow f(x)=0

\Rightarrow 2x^4 -7x^3 + 4x^2 + 7x - 6 = 0

\Rightarrow 2x^4-4x^3-3x^3+6x^2-2x^2+ 4x+3x-6=0

\Rightarrow 2x^3(x-2)-3x^2(x-2)-2x(x-2)+3(x-2)=0

\Rightarrow (x-2)(2x^3-3x^2-2x+3)=0

\Rightarrow (x-2)(x^2(2x-3)-1(2x-3))=0

\Rightarrow (x-2)(x^2-1)(2x-3)=0

\Rightarrow (x-2)(x+1)(x-1)(2x-3)=0

\Rightarrow x-2=0,\ x+1=0,\ x-1=0,\ 2x-3=0

\Rightarrow x=2,\ x=-1,\ x=1,\ x=\frac{3}{2}

From all the 4 roots, it can be obtained that 2 is the greatest zero.

7 0
3 years ago
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