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3 years ago

S = 2lw + 2wh + 2lh; solve for l

Mathematics

1 answer:

Pavel [41]3 years ago

4 0

Answer:

l = (S - 2wh)/(2w + 2h)

Step-by-step explanation:

S = 2lw + 2wh + 2hl

S - 2wh = 2lw + 2hl

S - 2wh = l (2w + 2h)

(S - 2wh)/(2w + 2h) = l

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How much money will you have if you started with $1200 and put it in an account that earned 7.3% every year for 10 years?

andrezito [222]

So an exponential equation is gonna be used for this problem. The formula for such is $y=ab^x$ , in which the a variable is the initial value and the b variable is the growth/decay rate.

For this situation, 1200 is gonna be your a variable since you've started out with it. And as for your b variable, its gonna be 107.3%, or 1.073, since it's increasing.

Your equation should look like this: $y=1200(1.073)^x$ and from here just plug in 10 into the x variable and solve.

Multiply (1.073)^10 (don't round answers until the end), and take that answer and multiply it with 1200, your answer should be $2427.61.

$y=1200(1.073)^{10} \\ y=2427.61$

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3 years ago

I NEED HELP ASAP

DIA [1.3K]

Answer:

$y = \frac{3}{2} x + 13$

Step-by-step explanation:

1) Use point-slope formula $y-y_1 = m (x-x_1)$ to find the slope-intercept form (or y = mx + b form) of the equation. m represents the slope, and $x_1$ and $y_1$ represent the x and y values of a point the line intersects. So, substitute $\frac{3}{2}$ for m, -4 for $x_1$ , and 7 for $y_1$ . Then, simplify and isolate y on the left side like so:

$y-(7) = \frac{3}{2} (x - (-4)) \\y -7 = \frac{3}{2} (x + 4) \\y - 7 = \frac{3}{2} x + 6\\y = \frac{3}{2} x + 13$

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2 years ago

Que número es 40% de 50?

uranmaximum [27]

Answer:

Step-by-step explanation:

5 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

PLASE ANSWER THIS QUESTION I GIVE YOU 18 POINTS

sukhopar [10]

Answer:

(a) 144

(b) 117

(d) 588

(e) 5472

Step-by-step explanation:

To find the LCM of two numbers, first find the prime factorization of each number. Then the LCM is the product of common and not common factors with the larger exponent.

(a)

$72 = 2^3 \times 3^2$