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4 years ago

Anyone know the answer to this?

Mathematics

1 answer:

Sedaia [141]4 years ago

5 0

6x^4-x^3+5x because that is the answer

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Helppppppp 3 Σ (2n – 3) n=0

Nataly [62]

Since there are only 4 terms in the sum, it's not too much work to expand it as

$\displaystyle \sum_{n=0}^3 (2n-3) = -3 + (-1) + 1 + 3 = \boxed{0}$

Alternatively, we can use the well-known formulas

$\displaystyle \sum_{n=1}^N 1 = \underbrace{1 + 1 + \cdots + 1}_{N \text{ 1s}} = N$

$\displaystyle \sum_{n=1}^N n = 1+2+\cdots+N = \frac{N(N+1)}2$

These sums start at n = 0, so in our given sum we will keep track of the 0-th term separately:

$\displaystyle \sum_{n=0}^3 (2n-3) = -3 + \sum_{n=1}^3 (2n-3) = -3 + 2 \sum_{n=1}^3 n - 3 \sum_{n=1}^3 1 = -3 + 3\times4 - 3\times3 = 0$

as expected.

8 0

2 years ago

Item 14 Solve 2x−3y=62x−3y=6 for yy.

Arada [10]

Please can you write a whole question

4 0

4 years ago

If f(×)=1/3×^2+5find f(-9)

Snezhnost [94]

Answer:

I think this is the answer

$f( - 9) = \frac{1}{3( - 9) {}^{2} } + 5 \\ = \frac{1216}{243} \\ = 5.0041$

5 0

3 years ago

24" 4 * 3 = Х I really need help the help with this

GREYUIT [131]

288. 4 times 3 is 12. Multiply that with 24 and thats your answer.

8 0

3 years ago

Peter the optician sees 16 patients one Tuesday.4 are neither short nor long sighted,the total number of short sighted patients

My name is Ann [436]

The number of short-sighted patients is 7

The number of long-sighted patients is 5

The number of patients requiring a varifocal lens is 4

Step-by-step explanation:

Stepwise solution-

Throughout the question, only 3 types of patients are mentioned. Hence, it would be safe to assume that he saw/sees these 3 types of patients on Tuesday.

As per the given details-

4 are neither short-sighted patients nor long-sighted patients. Then it means no of varifocal lens patients is 4.

Hence remaining total patients=16-4=12

As given, no of short-sighted patients is 2 more than the number of long-sighted patients.

Let us assume the number of long-sighted patients as “x”

Thus, the number of short-sighted patients would be “x+2”

x+(x+2)=12 (total remaining patients)

2x=10

x=5

Hence, No. of long-sighted outpatients = 5

No. of short-sighted outpatients =7

5 0

4 years ago