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3 years ago

I need help please.

Mathematics

1 answer:

pishuonlain [190]3 years ago

7 0

Step-by-step explanation:

2(x - 7) = 3(x + 9)

2x - 14 = 3x + 27

2x - 3x = 27 + 14

-x = 41

x = - 41

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What are the zeros of f(x) = x2 - 8x + 16?

dolphi86 [110]

x² - 8x + 16 = 0

(x - 4)^2 = 0

x = 4

Answer: One zero, x=4

8 0

3 years ago

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

Please help with these question, giving brainliest :)

garri49 [273]

Answer:

Step-by-step explanation:

3 0

2 years ago

Will mark brainliest !!!

Wittaler [7]

Answer:

just mark points it is very easy

5 0

3 years ago

CAN SOMEONE HELP ME PLEASE ASAP!?

kramer

Answer:

1418.03

Step-by-step explanation:

86.50 is 6.1% of what amount?

Multiply 86.50 x 100 = 8650

Divide by 6.1 = 1418.03

5 0

2 years ago

Read 2 more answers