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3 years ago

I need to know this right now :(

Mathematics

2 answers:

Aleksandr [31]3 years ago

6 0

Answer:

ok so 1/2 is your answer 1/4 divide = 1//2 !

Step-by-step explanation:

kobusy [5.1K]3 years ago

3 0

Answer:

40000

400

Step-by-step explanation:

40000 because 4000 * 10 = 40000

400 because 1 / 10 of 4000 = 400

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What is the value of m squared minus 2 m n + n squared for m = negative 2 and n = 4?

topjm [15]

The answer is 36. Check work

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4 years ago

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What is the equation of a quadratic function p with rational coefficients that has a zero of 3-i

Kryger [21]

Answer:

p = x² − 6x + 10

Step-by-step explanation:

Complex roots come in conjugate pairs. So if 3−i is a root, then 3+i is also a root.

p = (x − (3−i)) (x − (3+i))

p = x² − (3+i)x − (3−i)x + (3−i)(3+i)

p = x² − 3x − ix − 3x + ix + (9 − i²)

p = x² − 6x + 10

You can check your answer using the quadratic formula.

x = [ -b ± √(b² − 4ac) ] / 2a

x = [ 6 ± √(36 − 40) ] / 2

x = (6 ± 2i) / 2

x = 3 ± i

6 0

3 years ago

Write an expression that is equivalent to (23.6-7.1a-4.2b)-(5.8b-9a)

KengaRu [80]

Let's simplify step-by-step.23.6−7.1a−4.2b−(5.8b−9a)Distribute the Negative Sign:=23.6−7.1a−4.2b+−1(5.8b−9a)=23.6+−7.1a+−4.2b+−1(5.8b)+−1(−9a)=23.6+−7.1a+−4.2b+−5.8b+9aCombine Like Terms:=23.6+−7.1a+−4.2b+−5.8b+9a=(−7.1a+9a)+(−4.2b+−5.8b)+(23.6)=1.9a+−10b+23.6

8 0

3 years ago

I need help with part b. I feel like there’s a catch, I want to do the first derivative test, however, I feel like there is a be

Sladkaya [172]

Answer:

The fifth degree Taylor polynomial of g(x) is increasing around x=-1

Step-by-step explanation:

Yes, you can do the derivative of the fifth degree Taylor polynomial, but notice that its derivative evaluated at x =-1 will give zero for all its terms except for the one of first order, so the calculation becomes simple:

$P_5(x)=g(-1)+g'(-1)\,(x+1)+g"(-1)\, \frac{(x+1)^2}{2!} +g^{(3)}(-1)\, \frac{(x+1)^3}{3!} + g^{(4)}(-1)\, \frac{(x+1)^4}{4!} +g^{(5)}(-1)\, \frac{(x+1)^5}{5!}$

and when you do its derivative:

1) the constant term renders zero,

2) the following term (term of order 1, the linear term) renders: $g'(-1)\,(1)$ since the derivative of (x+1) is one,

3) all other terms will keep at least one factor (x+1) in their derivative, and this evaluated at x = -1 will render zero

Therefore, the only term that would give you something different from zero once evaluated at x = -1 is the derivative of that linear term. and that only non-zero term is: $g'(-1)= 7$ as per the information given. Therefore, the function has derivative larger than zero, then it is increasing in the vicinity of x = -1