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4 years ago

Which is greater 8/10 or 80/100

Mathematics

2 answers:

balandron [24]4 years ago

6 0

They are equal because of you simplify 80/100 by dividing by 10 u get 8/10

Helga [31]4 years ago

5 0

8/10 and 80/100 are both equivalent to each other.

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Change equation into slope intercept form 4x+5y=20

asambeis [7]

the slope intercept form would be y = 4 /5 x-4

5 0

3 years ago

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URGENT, PLEASE HELP! (3/5) -50 POINTS- !please no wrong answers for the points.! A) <img src="https://tex.z-dn.net/?f=y%20%3D%20

shtirl [24]

Answer:

B y = -1/2x + 7/2

Step-by-step explanation:

We know that it has a negative slope since the points go from the top left to the bottom right

We can eliminate A and D

The y intercept is where it crosses the y axis

It should cross somewhere between 2 and 4

C has a y intercept of 9 which is too big

Lets verify with a point

x = -4

y = -4(-4)+9 = 16+9 = 25 (-4,25) not even close to being near the points on the graph

checking B

y = -1/2 (-4) +7/2

= 2 + 7/2 = 11/2 = 5.5 it seems reasonable

5 0

3 years ago

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y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

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

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

3 years ago

The sum of three consecutive integers is at least 86. what are the 3 integers?

pshichka [43]

They could be 28 + 29 + 30

3 0

4 years ago

Find an input that yields an output of 214. rule: y= -15x+79

Dafna1 [17]

Answer:

The input have to be -9 to get output 214 with the rule y=-15x+79

Step-by-step explanation:

x is the input, y is the output so:

214 = -15x + 79

214 - 79 = -15x

15x = -135

x = -9

6 0

4 years ago

Read 2 more answers