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DaniilM [7]
3 years ago
13

Jake throws a stone off a bridge into the river. The stones height in meters above the water is based on the time in seconds aft

er jake threw the stone. The quadratic equation y=-5^2+10x+15 represents the scenario
Mathematics
1 answer:
aev [14]3 years ago
5 0

Answer:

The stone will hit the water after 3 seconds

Step-by-step explanation:

Given

y = -5x^2 + 10x +15

Required

At what time will the stone enter the water

When the stone enters the water;

y = 0

So, we have:

-5x^2 + 10x +15 = 0

Divide through by -5

x^2 -2x -3 = 0

Expand

x^2 +x -3x -3 = 0

Factorize

x(x +1) -3(x +1) = 0

Factor out x + 1

(x -3) (x +1) = 0

Solve for x

x - 3 = 0\ or\ x + 1 = 0

x  = 3\ or\ x = -1

x (time) cannot be negative.

So:

x  = 3

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3 years ago
A box witxh a height (x+5) cm has a square base with side x com. A second box with height (x+2) com has a square base with side
Eva8 [605]

Answer:

x=5.37 cm

Step-by-step explanation:

we know that

The volume of the box is

V=Bh

where

B is the area of the base of the box

h is the height of the box

<em>Box 1</em>

we have that

The area of the base is

B=x^2\ cm^2

h=(x+5)\ cm

The volume of the Box 1

is equal to

V_1=x^2(x+5)\ cm^3

V_1=(x^3+5x^2)\ cm^3

<em>Box 2</em>

we have that

The area of the base is

B=(x+1)^2\ cm^2

h=(x+2)\ cm

The volume of the Box 2

is equal to

V_2=[(x+1)^2(x+2)]\ cm^3

V_2=[(x^2+2x+1)(x+2)]\ cm^3

V_2=(x^3+2x^2+x+2x^2+4x+2)\ cm^3

V_2=(x^3+4x^2+5x+2)\ cm^3

Equate the equation of Volume 1 to the equation of Volume 2

(x^3+5x^2)=(x^3+4x^2+5x+2)

(5x^2)=(4x^2+5x+2)

5x^2-4x^2-5x-2=0

x^2-5x-2=0

Solve the quadratic equation by graphing

using a graphing tool

The solution is x=5.37 cm

see the attached figure

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4 years ago
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State the vertical asymptote of the rational function. f(x) =((x-9)(x+7))/(x^2-4)
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D:x^2-4\not=0\\&#10;D:x^2\not=4\\&#10;D:x\not=-2 \wedge x\not =2\\\\&#10;\displaystyle&#10;\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\&#10;\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\&#10;\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty&#10;

\displaystyle&#10;\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\&#10;\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\&#10;\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\

So, the vertical asymptotes are x=\pm 2
5 0
4 years ago
Read 2 more answers
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