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Orlov [11]
3 years ago
5

Determine the maximum or minimum value of the following quadratic function. LaTeX: f\left(x\right)=-3\left(x+5\right)^2-1f ( x )

= − 3 ( x + 5 ) 2 − 1
Mathematics
1 answer:
Genrish500 [490]3 years ago
3 0

Answer:

Step-by-step explanation:

y=-3(x^2+10x+25)-1\\ \\ y=-3x^2-30x-76\\ \\ dy=-6x-30\\ \\ d^2y=-6\\ \\ \text{Since the second derivative is always negative, when the first derivative is equal to zero the function will be at a global maximum.}\\ \\ dy=0=-6x-30\\ \\ x=-5\\ \\ f(-5)=-1\\ \\ \text{So the global maximum occurs at the point (-5,-1)

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He temperature was -13 degrees and then rose 9 degrees.<br> -13 + 9<br> 13 + 9<br> 9 - (-13)
Aleonysh [2.5K]

Answer:

-13 + 9

Step-by-step explanation:

if you have -13° and then it rises 9°, it will be added since it’s rising and rise can be a way of addition. so -13 + 9 is your answer

3 0
3 years ago
Read 2 more answers
The annual tuition at a specific college was $20,500 in 2000, and $45,4120
nika2105 [10]

Answer: the tuition in 2020 is $502300

Step-by-step explanation:

The annual tuition at a specific college was $20,500 in 2000, and $45,4120 in 2018. Let us assume that the rate of increase is linear. Therefore, the fees in increasing in an arithmetic progression.

The formula for determining the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

a represents the first term of the sequence.

d represents the common difference.

n represents the number of terms in the sequence.

From the information given,

a = $20500

The fee in 2018 is the 19th term of the sequence. Therefore,

T19 = $45,4120

n = 19

Therefore,

454120 = 20500 + (19 - 1) d

454120 - 20500 = 19d

18d = 433620

d = 24090

Therefore, an

equation that can be used to find the tuition y for x years after 2000 is

y = 20500 + 24090(x - 1)

Therefore, at 2020,

n = 21

y = 20500 + 24090(21 - 1)

y = 20500 + 481800

y = $502300

6 0
3 years ago
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3 years ago
1-sin60÷cos60=1-Tan30÷1+Tan30
ipn [44]

Step-by-step explanation:

LHS=(1-sin60)/cos60

=(1-√3÷2)/1÷2

=2(1-√3÷2)

=2-√3

RHS=(1-tan30)/(1+tan30)

={1-(1÷√3)}/{1+(1÷√3)}

={(√3-1)/√3}/{(√3+1)/√3}

=(√3-1)/(√3+1)

={(√3-1)(√3-1)}/{(√3+1)(√3-1)}

=(3-√3-√3+1)/(3-1)

=(4-2√3)/2

=2-√3

Therefore LHS=RHS

8 0
3 years ago
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