All categories

4 years ago

Nilai x yang memenuhi persamaan linier dari 2x-7/2=3-6-5x/6 adalah

Mathematics

1 answer:

emmainna [20.7K]4 years ago

5 0

2x-7/2 = 3-6-5x/6
⇒ 2x+ 5x/6= -3+ 7/2
⇒ 2 5/6x= 1/2
⇒ x= (1/2)/ (2 5/6)
⇒ x= 3/17

The final answer is x= 3/17~

You might be interested in

Use the following problem to answer the question below:

kap26 [50]

According to the probles represented above, there is addition of two angles which makes it totally clear. I am pretty sure that the answer with the actual reason of this task is the second option <span>B. Angle Addition Postulate, as you can see a couple of new angles in there :)
Regards!</span>

6 0

3 years ago

Mr. Williams can grade 50 essays in 7 hours. His assistant, Mr. Wilfred, can do the same job in 14 hours. His other assistant, M

Elza [17]

Answer:

13.44 or 13 11/25 hours

Step-by-step explanation:

First use this format: number of essays/hours

- they are asking if all three work together, how many hours would it take, so this means addition:

50/7 + 50/14 + 25/6 = 200/x

Second, solve:

- least common multiple: 42

300/42 + 150/42 + 175/6 = 200/x

625/42 = 200/x

- cross-multiply

625x = 8,400

x = 336/25 = 13 11/25 = 13.44 (any one of these answers would work)

5 0

3 years ago

Given a quadratic function, y=x2+bx+c what happens to the graph when "a" is positive? A.the parabola opens down, the vertex is t

Llana [10]

Answer:

Step-by-step explanation:

The parabola opens up. Which means the vertex is the minimum value.

6 0

4 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

The factory produces 4 gallons of orange juice for every 24 gallons of apple juice but is use this ratio to answer the question

kkurt [141]

Answer:

For every 1 gallon of Orange Juice the factory makes 6 gallons of Apple Juice so id say the ratio is 1:6

Step-by-step explanation:

8 0

3 years ago