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WITCHER [35]
3 years ago
9

The length that a hanging spring stretches varies directly with the weight placed at the end of the spring. If a weight of 15lb

15⁢lb stretches a certain spring 19in. 19⁢in. , how far will the spring stretch if the weight is increased to 30lb 30⁢lb ? (Leave the variation constant in fraction form. Round off your final answer to the nearest in.)
Mathematics
1 answer:
Ann [662]3 years ago
7 0

Answer: the spring would be stretched by 38 inches.

Step-by-step explanation:

The length that a hanging spring stretches varies directly with the weight placed at the end of the spring. Let L represent the length of the spring and let w represent the weight placed at the end of the spring. Introducing a constant of variation, k, the expression would be

L = kW

If a weight of 15lb stretches a certain spring 19in, the expression would be

19 = 15k

k = 19/15

The equation would be

L = 19w/15

if the weight is increased to 30lb, then the length by which the spring would be stretched is

L = 19 × 30/15

L = 38 inches

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Step-by-step explanation:

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FromTheMoon [43]

Answer:

a. y-intercept:(0,  -6), x-intercepts: (3, 0) and (-2, 0). vertex: (0.5, -6.25)

b. y-intercept: (0, 6), x-intercepts(3, 0) and (-2, 0). vertex: (0.5, 6.25)

Step-by-step explanation:

a:

 So finding the y-intercept is really easy and is simply when x=0. If you plug in 0 as x it makes y=(0)^2-0-6 which simplifies to -6, which is the y-intercept. As for the x-intercepts you can calculate that by using the quadratic equation x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\. In this case a=1, b=-1, c=-6. So plugging those values in you get x=\frac{-(-1)\pm\sqrt{(-1)^2-4(1)(-6)}}{2(1)}, which simplifies to x=\frac{1\pm5}{2}. This gives you the x-intercepts 6/2 and -4/2 which are 3 and -2. The vertex can be calculated by manipulating the equation so it's in the form of y=(x-h)^2+k where (h, k) is the vertex of the parabola. This is done by moving c to the other side and then completing the square and the isolating y. So the first step will be

Move c to the other side

y+6=x^2-x

Complete the square by adding (b/2)^2

y+6+0.25 = x^2-x+0.25

Rewrite as square binomial

y+6.25 = (x-0.5)^2

Isolate y

y=(x-0.50)^2-6.25

(h, k) = 0.50, -6.25 which is the vertex

b: To identify the y-intercept you plug in 0 as x which will only leave c which in this case is 6 which is the y-intercept. (0, 6). To identify the x-intercepts you can simplify plug in the values a, b, c into the quadratic equation which was stated in the previous answer. In this case a, b, c = -1, 1, 6. Plugging these values in gives the equation y=\frac{-(1)\pm\sqrt{1^2-4(-1)(6)}}{2(-1)}. which simplifies to x=\frac{-1\pm5}{-2} which gives the values -2 and 3. To find the vertex it's the same process as before

Factor out -1

y=-(x^2-x-6)

Add 6 to both sides (on the left side add -6 since -1 was factored out).

y-6=-(x^2-x)

Complete the square by adding (b/2)^2 to both sides (add -(b/2)^2 to left side since -1 was factored out)

y-6-0.25 = -(x^2-x+0.25)

Rewrite as square binomial

y-6.25=-(x-0.5)^2

Add 6.25 to both sides

y=(x-0.50)^2+6.25

(h, k) = (0.50, 6.25)

When you graph the parabolas you'll notice there just flipped relative to the x-axis. This can be deduced by simply looking at the two equations, since the two equations have the same absolute value coefficients, the signs are just different, and more specifically they're all opposite. If you took the first equation and multiplied the entire right side by -1 you would get the same equation. And since that equation really represents the value of y (since it's equal to y) you're reflecting it across the x-axis.

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Rama09 [41]
The associative property of addition is pretty much just any numbers in place of these: a + (b + c) = (a + b) + c. Therefore, your answer would be d) (6 + 1) + 9 = 6 + (1 + 9)
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