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olga nikolaevna [1]

3 years ago

5

The two lines represent the amount of water, over time in two tanks that are the same size. Which container is filling more slow

ly

1 answer:

Brut [27]3 years ago

6 0

B is slowly and A is filling more quickly because the amount of water increases during a smaller period of time.

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Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

A box of crayons costs $1.75, including tax. Mr. Valentino wants to purchase boxes of crayons for his class and has a $25 budget

densk [106]

Answer:

1.75x ≤ 25

Step-by-step explanation:

8 0

3 years ago

The 4th and 8th term of a G.P. are 24 and 8/27 respectively. find the 1st term and common ratio

beks73 [17]

Answer:

see explanation

Step-by-step explanation:

The n th term of a geometric progression is

• $a_{n}$ = a₁ $r^{n-1}$

where a₁ is the first term and r the common ratio

given a₄ = 24, then

a₁ $r^{3}$ = 24 → (1)

Given a₈ = $\frac{8}{27}$ , then

a₁ $r^{7}$ = $\frac{8}{27}$ → (2)

Divide (2) by (1)

$r^{4}$ = $\frac{\frac{8}{27} }{24}$ = $\frac{1}{81}$

Hence r = $\sqrt[4]{\frac{1}{81} }$ = $\frac{1}{3}$

Substitute this value into (1)

a₁ × ( $\frac{1}{3}$ )³ = 24

a₁ × $\frac{1}{27}$ = 24, hence

a₁ = 24 × 27 = 648

4 0

3 years ago

Help ...............

sammy [17]

A straight line is 180° so....

9x+24 +6x - 24 = 180

15x=180. ( because you have to combine like terms.

X= 12

5 0

3 years ago

4(2x+8)=10x+2-2x+30 what is x?

Anastaziya [24]

Let's solve your equation step-by-step.

4(2x+8)=10x+2−2x+30

Step 1: Simplify both sides of the equation.

4(2x+8)=10x+2−2x+30

(4)(2x)+(4)(8)=10x+2+−2x+30(Distribute)

8x+32=10x+2+−2x+30

8x+32=(10x+−2x)+(2+30)(Combine Like Terms)

8x+32=8x+32

8x+32=8x+32

Step 2: Subtract 8x from both sides.

8x+32−8x=8x+32−8x

32=32

Step 3: Subtract 32 from both sides.

32−32=32−32

0=0

Answer:

All real numbers are solutions.

5 0

2 years ago