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2 years ago

What is the general term equation, a_n , for the arithmetic sequence 5, 1,–3, –7…and what is the 21st term of this sequence?

1 answer:

4 0

The difference between each term is -4, so we know that part of the general term is -4n. Now we need to adjust -4n so at u1 it is 5. Therefore -4+x=5 and we rearrange to see x is 9.

Therefore our general sequence is:
$\boxed{u_n=-4n+9}$

Now we can substitute easily into the second equation with the common difference of 4:
$u_n=u_1+(n-1)d \\ u_{21}=5+(21-1)*-4 = 5-80 \\ \boxed{u_{21}=-75}$

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6a_6b+3a+4b pleas help me and cool

Ahat [919]

Answer:

9a - 2b

Step-by-step explanation:

6a - 6b +3a +4b . . . . . . . we assume your _ is supposed to be -

= a(6 +3) +b(-6 +4)

= 9a -2b

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3 years ago

There is an ample supply of identical blocks (as shown). Each block is constructed from four 1 × 1 × 1 unit-cubes glued whole-fa

bogdanovich [222]

Answer:

9+9=1-6+9=890 so the awnser would be 8947

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2 years ago

Read 2 more answers

Today, both the soccer team and the basket ball team had games (starting on Monday). The soccer team plays every 3 days and the

yarga [219]

I think eight because if you add those days together they will play again in eight days so that they play on the same day.

Sorry if I got it wrong I tried and that's what counts, right?

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2 years ago

What is the answer to this question?

Degger [83]

Answer:

$y = -\frac{19}{9}x + \frac{5}{3}$

Step-by-step explanation:

Given:

19x + 9y = 15

Required:

Rewrite in slope-intercept form, y = mx + b

SOLUTION:

19x + 9y = 15

Subtract 19x from both sides

9y = -19x + 15

Divide both sides by 9

y = -19x/9 + 15/9

$y = -\frac{19}{9}x + \frac{5}{3}$

5 0

2 years ago

Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distribute

balu736 [363]

Answer:

The area of the shaded region is 0.9082.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value, which is the area of the shaded region, is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 100, \sigma = 15, X = 120$

Area of the shaded region:

pvalue of Z

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{120 - 100}{15}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

The area of the shaded region is 0.9082.

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3 years ago