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2 years ago

Identify whether each of these functions contains a minimum or a maximum

Mathematics

1 answer:

iogann1982 [59]2 years ago

7 0

Answer:

I graphed the functions on the graph below.

Step-by-step explanation:

1) Minimum

2) Maximum

3) Minimum

4) Maximum

If these answers are correct, please make me Brainliest!

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7 • (6 • t) and (7 • 6) •t

xenn [34]

The answer to this question is 42t.

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2 years ago

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HELP ASAP!! WILL MARK BRAINLIEST

TiliK225 [7]

Answer:

Step-by-step explanation:

It's just B

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2 years ago

Consider the parabola r(t)equalsleft angle at squared plus 1 comma t right angle, for minusinfinityless thantless thaninfinity

kodGreya [7K]

Given:- $r(t)=< at^2+1,t>$ ; $-\infty < t< \infty$ , where a is any positive real number.

Consider the helix parabolic equation :

$r(t)=< at^2+1,t>$

now, take the derivatives we get;

$r{}'(t)=$

As, we know that two vectors are orthogonal if their dot product is zero.

Here, $r(t) and r{}'(t)$ are orthogonal i.e, $r\cdot r{}'=0$

Therefore, we have ,

$< at^2+1,t>\cdot < 2at,1>=0$

$< at^2+1,t>\cdot < 2at,1>=$

$=2a^2t^3+2at+t$

$2a^2t^3+2at+t=0$

take t common in above equation we get,

$t\cdot \left (2a^2t^2+2a+1\right )=0$

⇒ $t=0$ or $2a^2t^2+2a+1=0$

To find the solution for t;

take $2a^2t^2+2a+1=0$

The number $D = b^2 -4ac$ determined from the coefficients of the equation $ax^2 + bx + c = 0.$

The determinant $D=0-4(2a^2)(2a+1)$ $=-8a^2\cdot(2a+1)$

Since, for any positive value of a determinant is negative.

Therefore, there is no solution.

The only solution, we have t=0.

Hence, we have only one points on the parabola $r(t)=< at^2+1,t>$ i.e <1,0>

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2 years ago

A group of 485 people take a canoe trip. They fill up all avalible canoes that each hold 3 people with 273 people. The rest of t

Jobisdone [24]

Answer:
3c=273
c=91
2c= 212
C= 106
c= 91+ 106= 197
so they will need 197 canoes total

Why:
So first you have to determine a set variable to represent the number of canoes, I chose C. Then you make an equation to represent the number of canoes 273 people will use if they group into 3's, from this I got 3c=273. Solve for C and get 91.
The remainder of the group which is 485-273= 212 will use canoes in groups of 2's. To represent this, 2c=212. Solve for C and get 106. Combine 106 and 91 to get the total number of canoes.

4 0

2 years ago

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Suppose that one person in 10,000 people has a rare genetic disease. There is an excellent test for the disease; 98.8% of the pe

nirvana33 [79]

Answer:

A)The probability that someone who tests positive has the disease is 0.9995

B)The probability that someone who tests negative does not have the disease is 0.99999

Step-by-step explanation:

Let D be the event that a person has a disease

Let $D^c$ be the event that a person don't have a disease

Let A be the event that a person is tested positive for that disease.

P(D|A) = Probability that someone has a disease given that he tests positive.

We are given that There is an excellent test for the disease; 98.8% of the people with the disease test positive

So, P(A|D)=probability that a person is tested positive given he has a disease = 0.988

We are also given that one person in 10,000 people has a rare genetic disease.

So, $P(D)=\frac{1}{10000}$

Only 0.4% of the people who don't have it test positive.

$P(A|D^c)$ = probability that a person is tested positive given he don't have a disease = 0.004

$P(D^c)=1-\frac{1}{10000}$

Formula: $P(D|A)=\frac{P(A|D)P(D)}{P(A|D)P(D^c)+P(A|D^c)P(D^c)}$

$P(D|A)=\frac{0.988 \times \frac{1}{10000}}{0.988 \times (1-\frac{1}{10000}))+0.004 \times (1-\frac{1}{10000})}$

P(D|A)= $\frac{2470}{2471}$ =0.9995

P(D|A)= $0.9995$

A)The probability that someone who tests positive has the disease is 0.9995

(B)

$P(D^c|A^c)$ =probability that someone does not have disease given that he tests negative

$P(A^c|D^c)$ =probability that a person tests negative given that he does not have disease =1-0.004

=0.996

$P(A^c|D)$ =probability that a person tests negative given that he has a disease =1-0.988=0.012

Formula: $P(D^c|A^c)=\frac{P(A^c|D^c)P(D^c)}{P(A^c|D^c)P(D^c)+P(A^c|D)P(D)}$

$P(D^c|A^c)=\frac{0.996 \times (1-\frac{1}{10000})}{0.996 \times (1-\frac{1}{10000})+0.012 \times \frac{1}{1000}}$

$P(D^c|A^c)=0.99999$

B)The probability that someone who tests negative does not have the disease is 0.99999

8 0

2 years ago