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3 years ago

Need help please please

Mathematics

1 answer:

masya89 [10]3 years ago

4 0

Answer:

A No

B Yes

C No

D No

Step-by-step explanation:

y = x + 8 is the only linear function. Its graph is a straight line.

The others are not linear. Their graphs are curved

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Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo

Zepler [3.9K]

Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period 2 is 9

Forecast can be given by

$F_{t+1}=F_t+\alpha (D_t-F_t)$

where

$F_{t+1}=Future Forecast$

$F_t=Present\ Period\ Forecast$

$D_t=Present\ Period\ Demand$

$\alpha =smoothing\ constant$

$F_{t+1}=5+0.2(7-5)$

$F_{t+1}=5.4$

Forecast for Period 3

$F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})$

$F_{t+2}=5.4+0.2\cdot (9-5.4)$

$F_{t+2}=6.12$

8 0

3 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago

Malik deposited 1050 in a savings account and it earned 241.50 in simple interest after four years. Find the interest after four

kap26 [50]

Answer:

if it's true simple interest then each year's interest is the same so interest for one year is a quarter of 241.50 which give you the rate but if each year's interest in left in the account the result will be different , it is a really poorly worded question

7 0

3 years ago

Find the four arithmetic means between -21 and -36.

lara31 [8.8K]

The numbers given in the problem above are part of an arithmetic sequence with first and sixth terms equal to -21 and -36, respectively. Firstly, calculate for the common difference (d).

d = (-36 - -21) / (6 - 1) = -3

The arithmetic mean is calculated by adding -3 to the term prior to it.

a2 = -21 + -3 = -24 a3 = -24 + -3 = -27
a4 = -27 + -3 = -30 a5 = -30 + -3 = -33

Thus the four arithmetic means are -24, -27, -30, and -33.

4 0

3 years ago

Triangle PQR has vertices , , and . It is translated according to the rule . What is the y-value of ?

steposvetlana [31]

Answer:

-10 is the correct answer to the given question .

Step-by-step explanation:

Missing information:

Following question is incomplete there is no information about the vertices and the rules .Following are the complete question that is mention below

Triangle PQR has vertices $P(-2, 6), \ Q(-8, 4), and\ R(1, -2).$ It is translated according to the rule $(x, y)\ -> \ (x\ - \ 2, y\ - 16)$ . What is the y-value of P'?

Now coming to the solution as already mention in the question

The translated rule is $(x, y)\ -> (x - 2, \ y -16).$

Now calculated the vertices P value according to the rule of translated

$P(-2, 6)\\Now \ apply \ the\ translated\ rule \ in\ P\ vertices\\P(-2, 6)->P1(-2-2,\ 6-16)\\P1->(-4,-10)$

So -10 is the value of y in P vertices .

6 0

3 years ago

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