It is asking you to find the sum of k^2 - 1 from k=1 to k=4. Since that is only 4 numbers, calculating the sum by hand wouldn’t be that bad.

(1^2 - 1) + (2^2 - 1) + (3^2 - 1) + (4^2 - 1) = 26

The easier way to find the sum is to use a few simple formulas.

When we have a term that is just a constant c, the formula is c*n.

When we have a variable k, the formula is k*n*(n+1)/2.

When we have a squared variable, the formula is k*n*(n+1)*(2n+1)/6.

In this case, we have a squared variable k^2 and a constant of -1.

So plug in n=4 to the formulas:

4*5*9/6 - 1*4 = 26

The answer is 26

5 0

2 years ago

Find cotθ, cosθ, and secθ, where θ is the angle shown in the figure.

Darina [25.2K]

Answer:

$\cos( \theta) = \frac{5}{8} \\ {8}^{2} = {5}^{2} + {opp}^{2} \\ {opp}^{2} = 64 - 25 = 39 \\ opp = \sqrt{39} \\ \sin(\theta) = \frac{\sqrt{39}}{8} \\ \tan(\theta) = \frac{ \sqrt{39} }{5} \\ \cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{ \sqrt{39} }{5}} \\ \cot(\theta) = \frac{5}{ \sqrt{39} } \\ \csc( \theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{\sqrt{39}}{8}} \\ \csc( \theta) = \frac{8}{ \sqrt{39} } \\ \sec( \theta) = \frac{1}{\cos( \theta) } = \frac{1}{ \frac{5}{8} } \\ \sec( \theta) = \frac{8}{5}$

5 0

2 years ago

Can somone help me with number 23 and 24

tangare [24]

23.

store price = 14 + 0.5(14)

store price = 21

customer price = 21 + 0.08(21)

customer price = 21 + 1.68

customer price = $22.68

24.

formula: A = P(1 + rt)

A = 140(1 + (0.03)(2))

A = 140(1.06)

A = 148.40

interest = 148.40 = 140

interest = $8.40

Hope this helps!! :)

7 0

3 years ago

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