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2 years ago

What is a21 of the arithmetic sequence for which a7=−19 and a10=−28? A. -35 B. 35 C. -58 D. -61

Mathematics

2 answers:

shepuryov [24]2 years ago

8 0

Answer:

a21 = -61

Step-by-step explanation:

$a_{n}=a_{1}+(n-1)d$

$-19=a_{1}+(7-1)d$

$-28=a_{1}+(10-1)d$ (subtract to eliminate a₁)

9 = -3d

d = -3

-19 = a₁ + (6)(-3)

-1 = a

a21 = -1 + (21 - 1)(-3)

= -61

zhuklara [117]2 years ago

5 0

Answer:

-61 (Answer D)

Step-by-step explanation:

The general formula for an arithmetic sequence with common difference d and first term a(1) is

a(n) = a(1) + d(n - 1)

Therefore, a(7) = -19 = a(1) + d(7 - 1), or a(7) = a(1) + d(6) = -19

and a(10) = a(1) + d(10 - 1) = -28, or a(1) + d(10 - 1) = -28

Solving the first equation a(1) + d(6) = -19 for a(1) yields a(1) = -19 - 6d. We substitute this result for a(1) in the second equation:

-19 - 6d + 9d = -28. Grouping like terms together, we get:

3d = -9, and so d = -3.

Going back to an earlier result: a(1) = -19 - 6d.

Here, a(1) = -19 - 6(-3), or a(1) = -1.

Then the formula specifically for this case is a(n) = -1 - 3(n - 1)

and so a(21) = -1 - 3(20) = -61 (Answer D)

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Now, solve this problem. 106% of $ _______ = $7.95

Alex787 [66]

Answer:

7.5

Step-by-step explanation:

106% of $x = $7.95

In other words,

$\frac{106}{100}$ of $x = 7.95

Multiplying both sides of the equation by $\frac{100}{106}$

$\frac{106}{100} * \frac{100}{106}$ of x = 7.95 * $[\frac{100}{106}]$

=> x = $[\frac{750}{100}]$

=> x = $[\frac{7.50}{1}]$

=> x = 7.5

Validation: 106% of 7.5 = 7.95

4 0

2 years ago

Darcie wants to crochet a minimum of 3 blankets blankets to donate to a homeless shelter. Darcie crochets at a rate of 1/15 of a

bearhunter [10]

Answer:

The inequality to determine the number of days, $s$ , Darcie can skip crocheting and still meet her goal is:

$s\leq 15$

Thus, Darcie can skip a maximum of 15 days.

Step-by-step explanation:

Question

Darcie wants to crochet a minimum of 3 blankets to donate to a homeless shelter. Darcie crochets at a rate of 1/15 of a blanket per day. She has 60 days until when she wants to donate the blankets, but she also wants to skip crocheting some days so she can volunteer in other ways. Write an inequality to determine the number of days, s, Darcie can skip crocheting and still meet her goal.

Given:

Darcie has a target to crochet a minimum of 3 blankets.

Rate at which Darcie crochets = $\frac{1}{15}$ of a blanket per day.

Darcie has 60 days to crochet the blankets

To write an inequality to determine the number of days, $s$ , Darcie can skip crocheting and still meet her goal.

Solution:

The number of days Darcie can skip crocheting is represented by $s$

So, number of days left for Darcie to crochet = $60-s$

In 1 day Darcie crochets $\frac{1}{15}$ of a blanket.

So, in $(60-s)$ days she will crochet = $\frac{1}{15}(60-s)$ blankets.

Her aim is to crochet at least 3 blankets.

Thus, the inequality can be given as:

$\frac{1}{15}(60-s)\geq 3$

Solving for $s$

Multiplying both sides by 15 to remove fractions.

$15\times\frac{1}{15}(60-s)\geq 3\times 15$

$60-s\geq 45$

Subtracting both sides by 60.

$60-60-s\geq 45-60$

$-s\geq -15$

Dividing both sides by -1.

$\frac{-s}{-1}\leq \frac{-15}{-1}$ [ On dividing by negative number the sign of the inequality is reversed]

∴ $s\leq 15$

Thus, Darcie can skip a maximum of 15 days.

3 0

1 year ago

Read 2 more answers

Find the greatest number which divides by 1280 and 1371 leaves a remainder in each case?

vladimir1956 [14]

1279???
Because the gcf does not leave any remainders behind. The gcf could be 1280, but that would not leave a remainder behind if you divided 1280 by 1280. So, I think it is 1279, but i am probably wrong

4 0

1 year ago

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The cost for a different taxi company is expressed with the equation y = 1.65x + 2.35, where x represents the miles driven and y

Rudiy27

To solve this, all you have to do is plug in the 14 where the x value is:
y=1.65(14) +2.35 which equals 25.45

9 0

2 years ago

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Solve the recurrence relation: hn = 5hn−1 − 6hn−2 − 4hn−3 + 8hn−4 with initial values h0 = 0, h1 = 1, h2 = 1, and h3 = 2 using (

musickatia [10]

(a) Suppose $h_n=r^n$ is a solution for this recurrence, with $r\neq0$ . Then

$r^n=5r^{n-1}-6r^{n-2}-4r^{n-3}+8r^{n-4}$
$\implies1=\dfrac5r-\dfrac6{r^2}-\dfrac4{r^3}+\dfrac8{r^4}$
$\implies r^4-5r^3+6r^2+4r-8=0$
$\implies (r-2)^3(r+1)=0\implies r=2,r=-1$

So we expect a general solution of the form

$h_n=c_1(-1)^n+(c_2+c_3n+c_4n^2)2^n$

With $h_0=0,h_1=1,h_2=1,h_3=2$ , we get four equations in four unknowns:

$\begin{cases}c_1+c_2=0\\-c_1+2c_2+2c_3+2c_4=1\\c_1+4c_2+8c_3+16c_4=1\\-c_1+8c_2+24c_3+72c_4=2\end{cases}\implies c_1=-\dfrac8{27},c_2=\dfrac8{27},c_3=\dfrac7{72},c_4=-\dfrac1{24}$

So the particular solution to the recurrence is

$h_n=-\dfrac8{27}(-1)^n+\left(\dfrac8{27}+\dfrac{7n}{72}-\dfrac{n^2}{24}\right)2^n$

(b) Let $G(x)=\displaystyle\sum_{n\ge0}h_nx^n$ be the generating function for $h_n$ . Multiply both sides of the recurrence by $x^n$ and sum over all $n\ge4$ .

$\displaystyle\sum_{n\ge4}h_nx^n=5\sum_{n\ge4}h_{n-1}x^n-6\sum_{n\ge4}h_{n-2}x^n-4\sum_{n\ge4}h_{n-3}x^n+8\sum_{n\ge4}h_{n-4}x^n$
$\displaystyle\sum_{n\ge4}h_nx^n=5x\sum_{n\ge3}h_nx^n-6x^2\sum_{n\ge2}h_nx^n-4x^3\sum_{n\ge1}h_nx^n+8x^4\sum_{n\ge0}h_nx^n$
$G(x)-h_0-h_1x-h_2x^2-h_3x^3=5x(G(x)-h_0-h_1x-h_2x^2)-6x^2(G(x)-h_0-h_1x)-4x^3(G(x)-h_0)+8x^4G(x)$
$G(x)-x-x^2-2x^3=5x(G(x)-x-x^2)-6x^2(G(x)-x)-4x^3G(x)+8x^4G(x)$
$(1-5x+6x^2+4x^3-8x^4)G(x)=x-4x^2+3x^3$
$G(x)=\dfrac{x-4x^2+3x^3}{1-5x+6x^2+4x^3-8x^4}$
$G(x)=\dfrac{17}{108}\dfrac1{1-2x}+\dfrac29\dfrac1{(1-2x)^2}-\dfrac1{12}\dfrac1{(1-2x)^3}-\dfrac8{27}\dfrac1{1+x}$

From here you would write each term as a power series (easy enough, since they're all geometric or derived from a geometric series), combine the series into one, and the solution to the recurrence will be the coefficient of $x^n$ , ideally matching the solution found in part (a).

3 0

2 years ago