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2 years ago

At Norman's Newsstand, 5 magazines cost $20.00. How many magazines could you buy with $44.00?

Mathematics

2 answers:

kobusy [5.1K]2 years ago

8 0

Answer: 11 magazines

Step-by-step explanation:

1) 20/5 = $4 per magazine

2) 44/4 = 11 magazines

labwork [276]2 years ago

5 0

Answer:

10 magazines

Step-by-step explanation:

5 magazines cost $20.00

You have $44.00

If 5 magazines cost $20.00

Then 10 magazines would cost $40.00

Since the maximum amount of money you can spend is $44.00, you can only buy 10 magazines for $40.00 and be left with $4.00.

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A rectangle is 8 inches long and 4 inches wide. A similar rectangle is 16 inches long. What is the width of the second rectangle

Shkiper50 [21]

8 × 2 = 16 so..... 4 × 2 = 8

the width is 8 inches

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3 years ago

Ms. F needs you to walk her dogs Ella and Aries. She will pay you $10 for each time you take Ella and Aries out for a walk. The

snow_tiger [21]

Answer: $70/week

Step-by-step explanation:

E = 10w

w = 7 (1 time per day for 7 days = 7)

E = 10(7)

E = $70/week

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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

2 years ago

Simplify: 3(8) ° . (3) -2

jek_recluse [69]

3 x 8 = 24

3 x -2 = -6

reply if you this was wrong or you needed a different answer pls >:(

3 0

3 years ago

3. In a triangle, two angles have measures of 45°

Colt1911 [192]

Answer:

Triangles always have 180°, split in some manner between the three angles. With two of the three angles given, you can set up a relatively simple equation, where the third angle is assigned to the variable x:

45°+85°+x=180°

130°+x=180°

x=50°

5 0

3 years ago