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2 years ago

Select two answers please!!!!!

Mathematics

1 answer:

Alex2 years ago

6 0

Expanding using the distributive property, we get (x-5)*(x-5)=x^2-5x-5x+25=x^2-10x+25. Multiplying that by -6, we get -6x^2+60x-150 . We then add 12 at the end to get -6x^2+60x-138. Using the quadratic formula (since we want the function to equal 0 to find the zeros), we get x=(-60+-sqrt(3600-3312))/(-6*2)=(-60+-sqrt(288))/-12. For sqrt(288), since 12^2=144 and 144*2=288, we can rewrite it as 12*sqrt(2). Putting that back in, we get
(-60+-(12*sqrt(2))/-12. Factoring it out, we get -5+-sqrt(2)=-5+sqrt(2) or -5-sqrt(2) which is C

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What is the value of the 4 in 432?

amm1812

The value of 4 is Hundreds.
Your welcome

7 0

2 years ago

Read 2 more answers

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

4 0

3 years ago

Consider a simple example of moral hazard. Suppose that Woodrow goes into a casino to make one bet a day. The casino is very bas

Neko [114]

Answer:

The expected value of the safe bet equal $0

Step-by-step explanation:

$S=\left\{s_1,s_2,...,s_n\right\}$

is a finite numeric sample space and

$P(X=s_k)=p_k$ for k=1, 2,..., n

is its probability distribution, then the expected value of the distribution is defined as

$E(X)=s_1P(X=s_1)+s_2P(X=s_2)+...+s_nP(X=s_n)X)$

What is the expected value of the safe bet?

In the safe bet we have only two possible outcomes: head or tail. Woodrow wins $100 with head and “wins” $-100 with tail So the sample space of incomes in one bet is

S = {100,-100}

Since the coin is supposed to be fair,

P(X=100)=0.5

P(X=-100)=0.5

and the expected value is

E(X) = 100*0.5 - 100*0.5 = 0

6 0

3 years ago

What’s the answerrr..........

Afina-wow [57]

Answer:

D.90

the answer is 90 because it only flips once

8 0

2 years ago

Please can someone help me

Sidana [21]

If the answer is D. the equation to represent the ages of students is:
n + n+1 + n+2 = 45
it is showing that n is the age of one student and the other student has age one year more than the first one and third students has age of 2 years more than first student and one year greater than the second student. if we solve this equation for the value of n, we get;
3n + 3 = 45
3n = 45 -3
3n = 42
n = 14
so the age of one student is 14, second student 15 and third student 16 years.
and if we add there ages, 14 + 15 + 16 = 45

4 0

3 years ago