All categories

2 years ago

If a triangle has a side that is 5 and a side that is 12, what is the third side

Mathematics

1 answer:

8090 [49]2 years ago

7 0

Is this multiple chose? if so I would like to see the answers. But just using my brain, I would say, 5, or 7.

You might be interested in

Assume the exponential growth model A(t)equals=Upper A 0 e Superscript ktA0ekt and a world population of 5.95.9 billion in 2006

Whitepunk [10]

Answer:

1.4% is the maximum acceptable annual rate of growth such that the population must stay below 24 billion during the next 100 years.

Step-by-step explanation:

We are given the following in the question:

The exponential growth model is given by:

$A(t) = A_0e^{kt}$

where k is the growth rate, t is time in years and $A_0$ is constant.

The world population is 5.9 billion in 2006.

Thus, t = 0 for 2006

$A_0 = 5.9\text{ billions}$

We have to find the maximum acceptable annual rate of growth such that the population must stay below 24 billion during the next 100 years.

Putting these values in the growth model, we have,

$24 = 5.9e^{100k}\\\\k = \dfrac{1}{100}\ln \bigg(\dfrac{24}{5.9}\bigg)\\\\k = 0.01403\\k = 0.01403\times 100\% = 1.4\%$

1.4% is the maximum acceptable annual rate of growth such that the population must stay below 24 billion during the next 100 years.

8 0

2 years ago

Which products result in a difference of squares? Check all that apply.

Amanda [17]

Answer:

2nd - (w - 5)(w + 5)

4th - (-4v - 9)(-4v + 9)

Step-by-step explanation:

1. The first option shows an expression multiplied by its opposite(x -1), so therefore, it does not show the difference of squares

2. The second option does show the difference of squares because it is in the form (a + b)(a - b)

3. The third option is just a square because the same expression is multiplied by itself.

4. The fourth option is the difference of squares because it is in the form (a + b)(a - b). a equals -4v and b equals 9 in this case.

5. The fifth option is not the difference of squares. No term in common in both expressions

6. The sixth option is just a square because the same expression is multiplied by itself.

In all, there are two options that are the difference of squares, the 2nd and 4th.

5 0

2 years ago

How many sqaure feet of carpet do you need will you need

kati45 [8]

12 ft I had the question the other day on online school :)

5 0

2 years ago

Shade 1/3 of this pie

padilas [110]

Split the pie into 3 sections and shade only one of them

3 0

2 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