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3 years ago

12

A line-mile truck is painting a line on the side of a roadway. The truck paints at a rate of 8 miles an hour. How many miles can

be painted in 1/4 hour?

1 answer:

Rzqust [24]3 years ago

7 0

Answer:

2 miles can be painted in 1/4 hour.

Step-by-step explanation:

The truck paints 8 miles an hour and you want to find the amount of miles that can be painted in 1/4 of an hour.

8/4 = 2

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In the figure below, m2 1=(x-12)° and mZ2 =5xº.<br> Find the angle measures.

erma4kov [3.2K]

Answer:

15 165

Step-by-step explanation:

m1 + m2 = 180

x-18 +5x = 180

6x = 198

x = 33

m1 = 15

m2 = 165

6 0

3 years ago

Who's is the solution set of the compound inequality 3.5x-10>-3 and 8x-9<39

velikii [3]

3.5x-10>-3 add 10 to both sides

3.5x>7 divide both sides by 3.5

x>2

... now for the other inequality:

8x-9<39 add 9 to both sides

8x<48 divide both sides by 3

x<6

So we have x>2 and x<6, so the compound inequality is:

2<x<6 and this means that the solution set is:

x=(2, 6)

4 0

3 years ago

Read 2 more answers

Solve for k: 3k - 8 = 22.<br><br><br> possible answers:<br> -10<br> 14/3<br> 15<br> 10

exis [7]

Answer:

???

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Rewrite in simplest terms: 5(0.9m+0.5)-3(0.4m+0.7)

hram777 [196]

Answer:

3.3m+1.4

Step-by-step explanation:

Multiply 5 and 0.9m

Multiply 5 and 0.5

5 times 0.9m is 4.5m

5 times 0.5 is 2.5

Multiply -3 and 0.4m

Multiply -3 and 0.7

-3 times 0.4m is -1.2m

-3 times -.7 is -2.1

4.5m+ 2.5 -1.2m-2.1

Which is 3.3m+1.4

Hope this helps!

3 0

2 years ago

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