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3 years ago

SURFACE AREA!! can someone please help me get the answer on these two I’m stuck

Mathematics

1 answer:

elixir [45]3 years ago

4 0

Answer:

96 i think :)

Step-by-step explanation:

15x4=60

6x6=36

60+36=96

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Find the unit rate with the second given unit in the denominator. $57,256 for 586 color printers

Firlakuza [10]

The answer is 9.77 hope this helps

6 0

3 years ago

How do you do this question?

IRINA_888 [86]

Answer:

∑ (-1)ⁿ⁺³ 1 / (n^½)

∑ (-1)³ⁿ 1 / (8 + n)

Step-by-step explanation:

If ∑ an is convergent and ∑│an│is divergent, then the series is conditionally convergent.

Option A: (-1)²ⁿ is always +1. So an =│an│and both series converge (absolutely convergent).

Option B: bn = 1 / (n^⁹/₈) is a p series with p > 1, so both an and │an│converge (absolutely convergent).

Option C: an = 1 / n³ isn't an alternating series. So an =│an│and both series converge (p series with p > 1). This is absolutely convergent.

Option D: bn = 1 / (n^½) is a p series with p = ½, so this is a diverging series. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.

Option E: (-1)³ⁿ = (-1)²ⁿ (-1)ⁿ = (-1)ⁿ, so this is an alternating series. bn = 1 / (8 + n), which diverges. Since lim(n→∞) bn = 0, and bn is decreasing, then an converges. So this is conditionally convergent.

5 0

3 years ago

In a G.P the difference between the 1st and 5th term is 150, and the difference between the

liubo4ka [24]

Answer:

Either $\displaystyle \frac{-1522}{\sqrt{41}}$ (approximately $-238$ ) or $\displaystyle \frac{1522}{\sqrt{41}}$ (approximately $238$ .)

Step-by-step explanation:

Let $a$ denote the first term of this geometric series, and let $r$ denote the common ratio of this geometric series.

The first five terms of this series would be:

$a$ ,
$a\cdot r$ ,
$a \cdot r^2$ ,
$a \cdot r^3$ ,
$a \cdot r^4$ .

First equation:

$a\, r^4 - a = 150$ .

Second equation:

$a\, r^3 - a\, r = 48$ .

Rewrite and simplify the first equation.

$\begin{aligned}& a\, r^4 - a \\ &= a\, \left(r^4 - 1\right)\\ &= a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) \end{aligned}$ .

Therefore, the first equation becomes:

$a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right) = 150$ ..

Similarly, rewrite and simplify the second equation:

$\begin{aligned}&a\, r^3 - a\, r\\ &= a\, \left( r^3 - r\right) \\ &= a\, r\, \left(r^2 - 1\right) \end{aligned}$ .

Therefore, the second equation becomes:

$a\, r\, \left(r^2 - 1\right) = 48$ .

Take the quotient between these two equations:

$\begin{aligned}\frac{a\, \left(r^2 - 1\right) \, \left(r^2 + 1\right)}{a\cdot r\, \left(r^2 - 1\right)} = \frac{150}{48}\end{aligned}$ .

Simplify and solve for $r$ :

$\displaystyle \frac{r^2+ 1}{r} = \frac{25}{8}$ .

$8\, r^2 - 25\, r + 8 = 0$ .

Either $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ or $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ .

Assume that $\displaystyle r = \frac{25 - 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = -\frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= -\frac{1522\sqrt{41}}{41} \approx -238\end{aligned}$ .

Similarly, assume that $\displaystyle r = \frac{25 + 3\, \sqrt{41}}{16}$ . Substitute back to either of the two original equations to show that $\displaystyle a = \frac{497\, \sqrt{41}}{41} - 75$ .

Calculate the sum of the first five terms:

$\begin{aligned} &a + a\cdot r + a\cdot r^2 + a\cdot r^3 + a \cdot r^4\\ &= \frac{1522\sqrt{41}}{41} \approx 238\end{aligned}$ .

4 0

2 years ago

What is the value of z

vampirchik [111]

Answer:61

Step-by-step explanation:

Z+Z-11=Z+50

2Z-11=Z+50

Collect like terms

2Z-Z=11+50

Z=61

4 0

3 years ago

Points (1,3), (-4,7), and (-29,K) are collinear. Find K.<br><br> How do i solve this?

oee [108]

Co linear points have same slope
<span>Points (1,3), (-4,7), and (-29,K)
m1,2=7-3/-4-1=-4/5

m2,3=K-7/-29+4
m2,3=K-7/-25

-4/5=K-7/-25
4/5=K-7/25
4=K-7/5
20=K-7
K=27
</span>

3 0

3 years ago