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Anna35 [415]
2 years ago
8

Can someone please help ?

Mathematics
1 answer:
den301095 [7]2 years ago
4 0

5x+4y=-2

3x+4y=10


Make one with negative coefficient to find first what is x then solve for y .


-1(5x+4y)= -1 (-2)

3x+4y=10


-5x-4y=2

+ 3x+4y=10

_________

-x = 12


Divide both sides by -2

-2x/-2=12/-2


X= -6


Now substitute your x into one of the equations to find y .


-5x-4y=2

-5(-6)+4y=2


30+4y=2


Subtract 30 both sides


-4y= 2-30


Divide both sides by -4


Y= -28/-4


Y= 7



Your answer is in point form (-6,7) meaning x is -6 and y is 7.

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Solve for <em>q</em> :

3.2 + <em>q</em> = 0.60 (8 + <em>q</em>)

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The students in Mr. Wilson's Physics class are making golf ball catapults. The
Mnenie [13.5K]

Answer:

Part a) About 48.6 feet

Part b) About 8.3 feet

Part c) The domain is 0 \leq x \leq 48.6\ ft and the range is 0 \leq y \leq 8.3\ ft

Step-by-step explanation:

we have

y=-0.014x^{2} +0.68x

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum

where

x is the ball's  distance from the catapult in feet

y is the flight of the balls in feet

Part a)  How far did the ball fly?

Find the x-intercepts or the roots of the quadratic equation

Remember that

The x-intercept is the value of x when the value of y is equal to zero

The formula to solve a quadratic equation of the form

ax^{2} +bx+c=0

is equal to

x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}

in this problem we have

-0.014x^{2} +0.68x=0

so

a=-0.014\\b=0.68\\c=0

substitute in the formula

x=\frac{-0.68(+/-)\sqrt{0.68^{2}-4(-0.014)(0)}} {2(-0.014)}

x=\frac{-0.68(+/-)0.68} {(-0.028)}

x=\frac{-0.68(+)0.68} {(-0.028)}=0

x=\frac{-0.68(-)0.68} {(-0.028)}=48.6\ ft

therefore

The ball flew about 48.6 feet

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Find the maximum (vertex)

y=-0.014x^{2} +0.68x

Find out the derivative and equate to zero

0=-0.028x +0.68

Solve for x

0.028x=0.68

x=24.3

<em>Alternative method</em>

To determine the x-coordinate of the vertex, find out the midpoint  between the x-intercepts

x=(0+48.6)/2=24.3\ ft

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y=-0.014(24.3)^{2} +0.68(24.3)

y=8.3\ ft

the vertex is the point (24.3,8.3)

therefore

The ball flew above the ground about 8.3 feet

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we know that

A  reasonable domain is the distance between the two x-intercepts

so

0 \leq x \leq 48.6\ ft

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so

we have the interval -----> [0,8.3]

0 \leq y \leq 8.3\ ft

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