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2 years ago

Consider p(x) = 3x^3+ax-5a

Mathematics

2 answers:

german2 years ago

7 0

Answer:

3aw 3aw 3aw 3aw 3aw 3aw 3aw

kotegsom [21]2 years ago

3 0

Have you solve this

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Is 17.31 REPEATING at 31 irrational or rational

Amiraneli [1.4K]

Answer: Repeating per irrational.

8 0

2 years ago

What is the best approximation for the measure of angle ABC?

SVETLANKA909090 [29]

Cos(x) = 10.5 / 20
cos(x) = 0.525

answer
C.
x = 58.3"

3 0

2 years ago

A test has a 50 questions. Two are essay and the remaining are multiple choice and fill in the blank at a 5:1 ratio respectively

o-na [289]

Answer:If the ratio is 5:1 doesn’t mean that it will be in order but there will be 8 fill in the blank and 40 multiple choice

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

You are situated 300 feet from the base of Tower Glitz Plaza watching an external elevator descend down the side of the building

nordsb [41]

Answer:

18.66 ft/s

Step-by-step explanation:

The distance between you and the elevator is given by:

$h=\sqrt{x^2+y^2}$

The rate of change for the distance between you and the elevator is given by:

$\frac{dh}{dt}=\frac{dh}{dy}*\frac{dy}{dt}$

$-16=\frac{dh}{dy}*\frac{dy}{dt}$

$\frac{dh}{dy}=\frac{d}{dy} (\sqrt{x^2+y^2})\\$

Applying the chain rule:

$u=x^2+y^2\\\frac{dh}{dy}=\frac{d\sqrt u}{du} *\frac{du}{dy}\\\frac{dh}{dy}=\frac{1}{2\sqrt u} *2y\\\frac{dh}{dy}=\frac{y}{\sqrt {(x^2+y^2)}}$

Therefore, at x=300 and y = 500, dy/dt is:

$-16=\frac{y}{\sqrt {(x^2+y^2)}}*\frac{dy}{dt}\\-16=\frac{500}{\sqrt {(300^2+500^2)}}*\frac{dy}{dt}\\\frac{dy}{dt}=-18.66\ ft/s$

The elevator is descending at 18.66 ft/s.

7 0

3 years ago

A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee

katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

$^nC_r=\dfrac{n!}{(n-r)!r!}$

Given : Total multiple-choice questions = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= $^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84$

No. of ways to answer 4 open-ended problems

= $^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15$

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= $84\times15=1260$

Hence, the correct answer is option A) 1260

5 0

3 years ago