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Mathematics

1 answer:

sammy [17]4 years ago

8 0

if the line is parallel then that means it has the same slope, jjust a different y-intercept. so you'd have to substitute both the x and y value into the equation and keep the y-intercept a variable then solve for it, you'll get 4. so the equation is y=x/3 +4

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<h3>Option B</h3><h3>At 2 second and 1.75 second, the object be at a height of 56 feet</h3>

<em><u>Solution:</u></em>

Given that,

<em><u>The height h(t) in feet of an object t seconds after it is propelled straight up from the ground with an initial velocity of 60 feet per second is modeled by the equation:</u></em>

$h(t) = -16t^2 + 60t$

<em><u>At what times will the object be at a height of 56 feet</u></em>

<em><u>Substitute h = 56</u></em>

$56 = -16t^2 + 60t\\\\16t^2 - 60t + 56 = 0\\\\Divide\ the\ equation\ by\ 4\\\\4t^2 - 15t + 14=0$

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$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=4,\:b=-15,\:c=14\\\\x =\frac{-\left(-15\right)\pm \sqrt{\left(-15\right)^2-4\cdot \:4\cdot \:14}}{2\cdot \:4}\\\\Simplify\\\\x = \frac{15 \pm \sqrt{1}}{8}\\\\x = \frac{15 \pm 1}{8}\\\\We\ have\ two\ solutions\\\\x = \frac{15+1}{8} \text{ and } x = \frac{15-1}{8}\\\\x = 2 \text{ and } 1.75$

Thus, at 2 second and 1.75 second, the object be at a height of 56 feet

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