What is the product?

Mathematics

1 answer:

Gennadij [26K]3 years ago

3 0

= -30a^5b + 12a^4b^3 + 6a^3b^2 + 10a^3b^2 - 4a^2b^4 - 2ab^3
= -30a^5b + 12a^4b^3 + 16a^3b^2 - 4a^2b^4 - 2ab^3

answer
B. -30a^5b + 12a^4b^3 + 16a^3b^2 - 4a^2b^4 - 2ab^3

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Slope of the line that passes through the pair of points (-6.1,1.25) and (0.35,8.9)

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2 years ago

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$