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TiliK225 [7]
3 years ago
10

(5k − 1 + 2k3) − (6k3 − 3k2 − 8) + (8k3 − 2k2 − 7k)

Mathematics
2 answers:
Sedaia [141]3 years ago
4 0
It is 45/79k I beileve so
Vera_Pavlovna [14]3 years ago
3 0

Answer:

4k^3 + k^2 - 2k + 7

Step-by-step explanation:

Hope this helps, and good luck

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When an electric current passes through two resistors with resistance r1 and r2, connected in parallel, the combined resistance,
kondaur [170]

Answer:

a)

The combined resistance of a circuit consisting of two resistors in parallel is given by:

\frac{1}{R}=\frac{1}{r_1}+\frac{1}{r_2}

where

R is the combined resistance

r_1, r_2 are the two resistors

We can re-write the expression as follows:

\frac{1}{R}=\frac{r_1+r_2}{r_1r_2}

Or

R=\frac{r_1 r_2}{r_1+r_2}

In order to see if the function is increasing in r1, we calculate the derivative with respect to r1: if the derivative if > 0, then the function is increasing.

The derivative of R with respect to r1 is:

\frac{dR}{dr_1}=\frac{r_2(r_1+r_2)-1(r_1r_2)}{(r_1+r_2)^2}=\frac{r_2^2}{(r_1+r_2)^2}

We notice that the derivative is a fraction of two squared terms: therefore, both factors are positive, so the derivative is always positive, and this means that R is an increasing function of r1.

b)

To solve this part, we use again the expression for R written in part a:

R=\frac{r_1 r_2}{r_1+r_2}

We start by noticing that there is a limit on the allowed values for r1: in fact, r1 must be strictly positive,

r_1>0

So the interval of allowed values for r1 is

0

From part a), we also said that the function is increasing versus r1 over the whole domain. This means that if we consider a certain interval

a ≤ r1 ≤ b

The maximum of the function (R) will occur at the maximum value of r1 in this interval: so, at

r_1=b

6 0
3 years ago
John took out a loan for 18,550. To pay it back he will make 60 monthly payments of 389. How much will he pay in interest?
Debora [2.8K]

Answer:

He paid an of 4790

Step-by-step explanation:

If John took a loan of 18550

Then in 60 months, we will pay back 389 multiply by 60

389 × 60 = 23340

Therefore he paid an interest of 23340 minus 18550, that is

23340 - 18550 = 4790

3 0
3 years ago
At a raffle, 1000 tickets are sold for $5 each. There are 20 prizes of $25, 5 prizes of $100, and 1 grand prize of $2000. Suppos
postnew [5]
Ducks can fly so good.....ITS 3/4
5 0
3 years ago
BRAINLIEST FOR SOLVING THESE
slavikrds [6]

Answer:

1st a 2nd is c give me brianliest please pls pls

6 0
3 years ago
A space S is defined as S = {1, 3, 5, 7, 9, 11} , and three subsets as A = {1, 3, 5} , B = {7, 9, 11} , C = {1, 3, 9, 11} . Assu
Lina20 [59]

Answer:

a) 0.5

b) 0.5

c) 0.67

d) 1

e) 0.83

f) 0.67

Step-by-step explanation:

a)

P(A)=n(A)/n(S)

n(A)=number of outcomes in event A=3

n(S)= Total number of outcome of an experiment=6

P(A)=3/6=1/2=0.5

b)

P(B)=n(B)/n(S)

n(B)=number of outcomes in event B=3

P(B)=3/6=1/2=0.5

c)

P(C)=n(C)/n(S)

n(C)=number of outcomes in event C=4

P(C)=4/6=2/3=0.67

d)

A∪B= {1,3,5} ∪ {7,9,11}={1,3,5,7,9,11}

P(A∪B)=n(A∪B)/n(S)=6/6=1

e)

A∪C={1,3,5}∪{1,3,9,11}={1,3,5,9,11}

P(A∪C)=n(A∪C)/n(S)=5/6=0.83

f)

A-C={1,3,5}-{1,3,9,11}={5}

(A-C)∪B={5}∪{7,9,11}={5,7,9,11}

P((A-C)∪B)=n((A-C)∪B)/n(S)=4/6=2/3=0.67

8 0
3 years ago
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