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2 years ago

Help please... I'm close to be done soon... of my little sister homework...

Mathematics

2 answers:

harkovskaia [24]2 years ago

5 0

C' = (6,-1) or (-1,6)

D'= (6,6)

E'= (-4,1) or (1,-4)

Hope this helps somewhat

Dahasolnce [82]2 years ago

3 0

Step-by-step explanation:

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Answer:

It would take approximately 6.50 second for the cannonball to strike the ground.

Step-by-step explanation:

Consider the provided function.

$h(t)=-4.9t^2+30.5t+8.8$

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Substitute h(t) = 0 in above function.

$-4.9t^2+30.5t+8.8=0$

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$-49t^2+305t+88=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are: $x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute a = -49, b = 305 and c=88

$t=\frac{-305+\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}=-\frac{-305+\sqrt{110273}}{98}\\t = \frac{-305-\sqrt{305^2-4\left(-49\right)88}}{2\left(-49\right)}= \frac{305+\sqrt{110273}}{98}$

Ignore the negative value of t as time can't be a negative number.

Thus,

$t=\frac{305+\sqrt{110273}}{98}\approx6.50$

Hence, it would take approximately 6.50 second for the cannonball to strike the ground.

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