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Mathematics

1 answer:

frosja888 [35]3 years ago

3 0

Hi king,

simplify and expand this polynomial:

$P= 14xy^{2}(3x^{3}y+5xy^{4}-6y^{3})$

$P=14xy^{2} x^{3} y-6*14xy^{2} y^{3} +14*5xy^{2} xy^{4} \\P=14*3x^{1+2}y^{2+1} -6*14xy^{2+3} + 14*5x^{2}y^{2+4}\\P=42x^{4}y^{3}+84xy^{5}+70x^{2}y^{6}$

The leading coefficient is 42
The degree is 4

Have a good day.

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**MARKING BRAINLIEST PLEASE HELP**

raketka [301]

Answer:

$\text{1. } v=16t+\frac{h-c}{t},\\\text{2. }90\:\mathrm{ft/s}, \\\text{3. Cannot be determined}$

Step-by-step explanation:

1. The initial equation given to us is $h=-16t^2+vt+c$ . Rearranging the equation to isolate $v$ , we have:

$vt=16t^2+h-c,\\\boxed{v=16t+\frac{h-c}{t}}$

2. Using the equation we rearranged in part 1, we can substitute given values:

$v=16(3)+\frac{131-5}{3},\\v=48+42=\boxed{90\:\mathrm{ft/s}}$

3. We see from our equation in part 1 ( $v=16t+\frac{h-c}{t}$ ) that when $t=0$ , the denominator of our fraction will be equal to 0. Since we cannot divide by 0, the velocity remains undefined and cannot be determined.