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3 years ago

Write a polynomial that fits the description:

Mathematics

1 answer:

rewona [7]3 years ago

4 0

Answer:

$-4x^3 + 2x^2-3$

Step-by-step explanation:

Degree is the highest exponent on the variable in a polynomial.

A leading coefficient is the first terms coefficient when the polynomial is written in standard form.

To write this polynomial, write $-4x^3$ as your first term. Its leading coefficient is -4 and it has a degree of 3. Then add whatever terms after it being sure to not add any more with exponent 3 or higher.

Example: $-4x^3 + 2x^2-3$

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Write 4 4/5 as an improper fraction

Rasek [7]

Step-by-step explanation:

4⅘=22/5.

hope this helps you.

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3 years ago

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Is on in which every member of the population has an equal chance of being selected for the sample?

Fittoniya [83]

This type of sampling would be Random Sampling.

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3 years ago

Find the sum of the first 10 terms for the sequence {21, 63, 189, . . . }.

antiseptic1488 [7]

The sum of n terms of a geometric sequence with first term "a" and common ratio "r" is given by
$S_{n}=a\dfrac{r^{n}-1}{r-1}$
You have a=21, r=3, so the sum of 10 terms is
$S_{10}=21\cdot \dfrac{3^{10}-1}{3-1}=620004$

The appropriate choice is the 2nd one:
620,004

3 0

4 years ago

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The 6-lb particle is subjected to the action of its weight and forces F1 = 52i + 6j - 2tk6 lb, F2 = 5t 2 i - 4tj - 1k6 lb, and F

olga2289 [7]

Answer:

r=294.9m

Step-by-step explanation:

The forces on the particle are

$W=mg\hat{j}\\F_{1}=52\hat{i}+6\hat{j}-2t\hat{k}\\F_{2}=5t^{2}\hat{i}-4t\hat{j}-1\hat{k}\\F_{3}=(5-2t)\hat{i}$

Now , we sum all these forces to get the net force

$F_{T}=W+F_{1}+F_{2}+F_{3}\\F_{T}=(52+5t^{2}+5-2t)\hat{i}+((6+6-4t)\hat{j}+(-2t-1)\hat{k}\\F_{T}=(57-2t+5t^{2})\hat{i}+(12-4t)\hat{j}+(-2t-1)\hat{k}\\$

we can use the fact F=m*a and integrate the acceleration

$a(t)=\frac{1}{m}F(t)\\\\v(t)=\int a(t)dt=\frac{1}{m}\int{F_{T}}dt\\\\v(t)=\frac{1}{m}[(57t-t^{2}+\frac{5}{3}t^{3})\hat{i}+(12t-2t^{2})\hat{j}+(-t^{2}-t)\hat{k}]\\\\r(t)=\int v(t)dt=\frac{1}{m}[(\frac{57}{2}t^{2}-\frac{1}{3}t^{3}}+\frac{5}{4}t^{4})\hat{i}+(6t^{2}-\frac{2}{3}t^{3})\hat{j}+(-\frac{1}{3}t^{3}-\frac{1}{2}t^{2})]$

and we evaluate in r(2) an we take the norm to obtain the distance

$r(2)=\frac{1}{m}[\frac{394}{3}\hat{i}+\frac{56}{3}\hat{j}-\frac{14}{3}\hat{k}]\\|r(2)|=\frac{1}{m}\sqrt{[(\frac{394}{3})^{2}+(\frac{56}{3})^{2}+(\frac{14}{3})^{2}]}\\|r(2)|=\frac{132.73}{0.45}=294.9m$

I hope this is useful for you

regards

8 0

4 years ago

A ball is launched into the sky at 54. 4 ft./s from a 268.8 m tall building. The equation for the ball’s height, h, at time t se

Crank

Answer:

t = 21

Step-by-step explanation:

The ball strikes the groujd when h = 0.

$-3.2t^2 + 54.4t+268.8=0 \\ \\ 32t^2 - 544t-2688=0 \\ \\ t^2 - 17t-84=0 \\ \\ (t-21)(t+4)=0 \\ \\ t=-4, 21$

Since time must be positive, t = 21.

6 0

2 years ago