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2 years ago

What digit is in the ten thousands place 1.03098 show how you got this and I will mark brainiest

Mathematics

1 answer:

Luden [163]2 years ago

3 0

The 0 is in the ten thousands

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Y = 4x - 3 y= 2x + 13

Serggg [28]

Answer:

x = 9, y = 29 or (8,29)

Step-by-step explanation:

Set the equations equal to each other because they both equal y:

4x - 3 = 2x + 13

Combine like terms:

2x = 16

Divide by 2:

x = 8

Plug 8 into the x for both equations and see if the y is the same:

y = 4(8) - 3 = 32 - 3 = 29

y = 2(8) + 13 = 16 + 13 = 29

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2 years ago

5 5/7 x 5 5/7 for a quiz

Sauron [17]

Answer:

61 36/49

Step-by-step explanation:

I took a test with that question. my answer was right

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2 years ago

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The diagram shows a cuboid. 4 cm 5 cm 9 cm What is the surface area of the cuboid?

Salsk061 [2.6K]

Answer:

69cm ^2

Step-by-step explanation:

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2 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

1 year ago

Y’all I need help fast

vitfil [10]

Answer:

All Real Numbers, this is because it has infinitely MANY solutions.

Step-by-step explanation:

7 0

2 years ago