All categories

3 years ago

How do i do ghe work for -3/4=x/24

Mathematics

2 answers:

Dafna11 [192]3 years ago

8 0

Answer:

x = -18

Step-by-step explanation:

-3/4 x 24 = x

-72/4 = x

x = -18

prohojiy [21]3 years ago

6 0

Step-by-step explanation:

- 3/ 4 = x / 24

4x = - 3 * 24

x = - 72 / 4

x = - 18

Hope it will help :)

You might be interested in

Gene moved to a new school on the 24th day of the 2nd month of a common year. When his new teacher introduced him to the class,

Airida [17]

Answer:

Genes birthday is February 29th, or March 1st on a Leap Year

February has 29 days in a common year, so 24+7 is the 29th

February has 28 days in a Leap Year, so one less day would lead you to March 1st

7 0

3 years ago

caroline bought 6 pieces of gold ribbon. each piece was 3/5 of a yard long. How many yards of ribbon did she buy in all?

Zarrin [17]

Answer:

= 3 3/5 yards

Step-by-step explanation:

If 1 piece = 3/5 yard

What about 6 pieces = ?

= (6 x 3/5) ÷ 1

= 18/5 ÷ 1

= 18/5

= 3 3/5 yards

5 0

3 years ago

Help??( dont forget to explain)

Setler [38]

Answer:

You can decide where to put the first digit because you know 153 doesn't fit into 6 or 61. But you know it can fit into 613. You put it there.

5 0

3 years ago

Read 2 more answers

Today there is a 60 % of rain. What is the chance that they win their game today?

Jlenok [28]

75 give or take a few percent

7 0

3 years ago

Read 2 more answers

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