All categories

3 years ago

Find two mixed numbers so that the sum is 15 3/10 and the difference is 8 5/10

1 answer:

8 0

11 9/10 and 3 2/5

$\left \{ {{ 11\frac{9}{10}+3\frac{2}{5}=15\frac{3}{10}} \atop { 11\frac{9}{10}-3\frac{2}{5}=8\frac{5}{10}}} \right.$

To solve for this, you can create a series of two equations
$\left \{ {{y+x=15 \frac{3}{10} } \atop {y-x= 8\frac{5}{10} }} \right.$

After that, you solve one of the equations for x and plug it into the other equation.
$\left \{ {{x=15 \frac{3}{10}-y } \atop {y-x= 8\frac{5}{10} }} \right.$

${2y-15 \frac{3}{10}= 8\frac{5}{10}$

Then, solve for variable y
${2y=23 \frac{8}{10}$
${y=11\frac{9}{10}$

Then, you plug in this value for y into one of the original equations

$11\frac{9}{10}+x=15 \frac{3}{10}$

$x=15 \frac{3}{10} -11\frac{9}{10}$

$x=3 \frac{2}{5}$

You might be interested in

vredina [299]

No, the sequence is algebraic.

7 0

3 years ago

Faye Tallity earns simple interest on her savings account. She deposits $1,200 into her account and will have $1,344 after 2 yea

Marat540 [252]

Faye will have $1704.00 total money in her account.
She earned $504.00 interest for 7 years.
(She earned $144 for 2 years, divide in half $72 a year= $72×7=$504

8 0

3 years ago

Please! Quick, how may days are in between 12/1/16 and 7/20/20???

lozanna [386]

1,327 days are in between 12/1/16 and 7/20/20

8 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

2 years ago

The change in the value of an account when given to the nearest dollar 1/16 inch

Feliz [49]

This would be 2 because of the rounding rules of it being 5 and up round up and below 5 round down. I hope I helped!

7 0

2 years ago