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3 years ago

Any easier way to solve this than plugging each x value in?

Mathematics

1 answer:

Evgesh-ka [11]3 years ago

6 0

Solving The Problem From Eaxh Side Is Easier

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Part A. Jose painted a square shaped mural. The length of each side of the canvas is 1 1/3 yards. What is the area of the mural

Pavel [41]

A. Its a square. This means each side times the side would give us the area (s^2)
4/3^2 or 1.3333 x 1.33333 = 16 / 9 area.

B. Really just pick any numbers you want.
16 yards long and 1/9 wide.
As long as they multiply to 16/9

4 0

3 years ago

Read 2 more answers

Describe a pattern in the sequence of numbers. Predict the next number.

marissa [1.9K]

Answer:

Pattern = Start at 256 and divide by 4 each time. Next number = 1.
Pattern = Start at 2 multiply by 3 each time. Next number = 162

3 0

3 years ago

18 + x = 10 x=____ help

hodyreva [135]

Answer:

X=-8

Step-by-step explanation:

18+x=10

-18 -18

x=-8

7 0

3 years ago

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Question 2*

lilavasa [31]

Answer:

The average rate of change between these times is 68 miles per hour

Step-by-step explanation:

Here, we are to determine the average rate of change between hour 2 and hour 7

The distance traveled at hour 2 is 140 miles

The distance traveled at hour 7 is 480 miles

So we can say we have two points and we want to know the rate of change between these points

Mathematically, we can represent the rate of change as Δ

Thus, between the two different times, we have;

Δ = (D7-D2)/(T7-T2)

where (T7,D7) = (7,480) and (T2,D2) = (2,140) represents the time and distance at hour 2 and hour 7 respectively

Now inputing the values into the equation, we have;

Δ = (480-140)/(7-2) = 340/5 = 68 miles/hour

3 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