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3 years ago

A partial sum of an arithmetic sequence is given. Find the sum.

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

3 0

Answer:

72.3

Step-by-step explanation:

Given that:

The arithmetic sequence:

$\sum \limits ^{14}_{k=0} (3 + 0.26k)$

The first term of the sequence $a_0$ = 3 since k = 0

i.e (3 + 0.26(0)) = 3

The last term of the sequence $a_{14}$ = 6.64

i.e (3+ 0.26(14)

= (3 + 3.64)

= 6.64

Total no of terms = 15 i.e from 0 to 14

∴

The partial sum of the arithmetic sequence = $\dfrac{Total \ no \ of \ terms }{2} \times (a_o+a_{15})}$

$=\dfrac{15 }{2} \times (3+6.64)}$

= 72.3

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Show with work please.

kolbaska11 [484]

Answer:

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

Step-by-step explanation:

The identity you will use is:

$\csc \left(x\right)=\frac{1}{\sin \left(x\right)}$

So,

$\csc \left(\theta-\frac{\pi }{2}\right)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{\sin \left(-\frac{\pi }{2}+\theta\right)}$

Now, using the difference of sin

Note: state that $\text{sin}(\alpha\pm \beta)=\text{sin}(\alpha) \text{cos}(\beta) \pm \text{cos}(\alpha) \text{sin}(\beta)$

$\csc \left(\theta-\frac{\pi }{2}\right)=\frac{1}{-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)}$

Solving the difference of sin:

$-\cos \left(\theta\right)\sin \left(\frac{\pi }{2}\right)+\cos \left(\frac{\pi }{2}\right)\sin \left(\theta\right)$

$-\cos \left(\theta\right) \cdot 1+0\cdot \sin \left(\theta\right)$

$-\text{cos} \left(\theta\right)$

Then,

$\csc \left(\theta-\frac{\pi }{2}\right)=-\frac{1}{\cos \left(\theta\right)}$

Once

$\text{sec}(-\theta)=\text{sec}(\theta)$

And, $\text{sec}(\theta)=-0.73$

$-\frac{1}{\cos \left(\theta\right)}=-\text{sec}(\theta)$

$-\frac{1}{\cos \left(\theta\right)}=-(-0.73)$

$-\frac{1}{\cos \left(\theta\right)}=0.73$

Therefore,

$\csc \left(\theta-\frac{\pi }{2}\right)=0.73$

3 0

3 years ago

Use the distributive property then combine any like terms<br><br> 4(7p+4-2p)

ivolga24 [154]

Answer:20p+16

Step-by-step explanation:

Combine like terms

4(7p+4-2p)

4(5p+4)

Distribute

4(5p+4)

20p+16

7 0

3 years ago

Read 2 more answers

Suppose p" must approximate p with relative error at most 10-3 . Find the largest interval in which p* must lie for each value o

goblinko [34]

Answer:

$[p-|p|*10^{-3} \, , \, p+|p|* 10^-3]$

Step-by-step explanation

The relative error is the absolute error divided by the absolute value of p. for an approximation p*, the relative error is

r = |p*-p|/|p|

we want r to be at most 10⁻³, thus

|p*-p|/|p| ≤ 10⁻³

|p*-p| ≤ |p|* 10⁻³

therefore, p*-p should lie in the interval [ - |p| * 10⁻³ , |p| * 10⁻³ ], and as a consecuence, p* should be in the interval [p - |p| * 10⁻³ , p + |p| * 10⁻³ ]

8 0

3 years ago

Doreen is flipping two fair coins. What is the probability that both coins land on heads?

andrezito [222]

Answer:

There are four different possible outcomes: both coins are heads, the red coin is heads and the blue coin is tails, the red coin is tails and the blue coin is heads, or both coins are tails. Each outcome has equal probability. So the probability of both being heads is 1/4.

Step-by-step explanation:

6 0

3 years ago

Need help tremendously

Alina [70]

<span>basically since 30 minutes is half of an hour, to get to a full hour you can multiply by 2, so you do that for both, 2/3 included so you get 2/3 * 2/1 4/3 is 1 1/3</span>

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3 years ago