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irina1246 [14]
3 years ago
14

Three times a number, less 4 is 2

Mathematics
1 answer:
tekilochka [14]3 years ago
3 0

Answer:

2

Step-by-step explanation:

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One of the roots of the equation 3x^2+7x−q=0 is −5. Find the other root and q.
statuscvo [17]

Answer: The answer is \textup{The other root is }\dfrac{8}{3}~\textup{and}q=40.Step-by-step explanation:  The given quadratic equation is[tex]3x^2+7x-q=0\\\\\Rightarrow x^2-\dfrac{7}{3}x-\dfrac{q}{3}=0.

Also given that -5 is one of the roots, we are to find the other root and the value of 'q'.

Let the other root of the equation be 'p'. So, we have

p-5=-\dfrac{7}{3}\\\\\\\Rightarrow p=5-\dfrac{7}{3}\\\\\\\Rightarrow p=\dfrac{8}{3},

and

p\times(-5)=-\dfrac{q}{3}\\\\\\\Rightarrow \dfrac{8}{3}\times 5=\dfrac{q}{3}\\\\\\\Rightarrow q=40.

Thus, the other root is \dfrac{8}{3} and the value of 'q' is 40.

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3 years ago
2/3 x 4/7 write the answer in the simplest form
trapecia [35]

Answer:

8/21

I don't think that it can be simplified further....

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2 years ago
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Free points have a good week!!! Stay safe.
jeka57 [31]

Answer:

you too

Step-by-step explanation:

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2 years ago
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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
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A) Mental Map .......
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