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3 years ago

PLEASE HELP !!!!!!!!!!!

Mathematics

1 answer:

8_murik_8 [283]3 years ago

5 0

Answer:

4 units right and 3 units up

Step-by-step explanation:

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I will fail if i dont pass this test plzz help!

Fiesta28 [93]

Answer:

Step-by-step explanation:

${2}^{2} + 2(2) + 4 = 12$

8 0

3 years ago

A water faucet leaks 3/5 of a liter of water in 3/4 of an hour. Which TWO statements are correct for this situation

Svetach [21]

Answer:

A & C

Step-by-step explanation:

Please just trust me

4 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

Use the definition of the derviative to compute the derivative of f(x)= 1- 7x^2 at the specific point x=2.

GaryK [48]

Answer:

$f'(x) = -28 \ at \ x = 2$

Step-by-step explanation:

$f'(2) = \lim_{h \to 0}\frac{f(2+h)) -f(2)}{h} \\\\f(2+h) = 1 - 7(2+h)^2 = 1 - 7(4 +h^2 +4h) = 1 - 28 - 7h^2 - 28h = - 7h^2 -28h - 27\\\\f(2) = 1 - 7(2^2) = 1 - 28 = -27\\\\f(2+h) - f(2) = -7h^2 - 28h - 27 - (-27) = -7h^2 -28h -27 + 27 = -7h^2 -28h\\\\f'(2) = \lim_{h \to 0}\frac{-7h^2 - 28h}{h} \\\\$

$= \lim_{h \to 0} \frac{h(-7h -28)}{h}\\\\= \lim_{h \to 0} (-7h -28)}\\\\= -7 (0) - 28\\\\= -28$

4 0

3 years ago

Two six-sided dice are tossed 288 times. how many times would you expect to get a sum of 5?

dusya [7]

First find the probability of getting 2 and 3 or 1 and 4 and toss the dice. multiply this by 288 ;)

4 0

3 years ago