If EM=11, AM=16, and CF=27, what are the lengths of the following sides?

After two six sided dice are rolled, what is the probability that the dice faces add up to either a 4 or 6?

VladimirAG [237]

Answer:

hope this helps you

option B

Step-by-step explanation:

ecf-zutd-cpt.

8 0

2 years ago

Read 2 more answers

The function, y(r)=2.76r , is used to convert Russian rubles in Japanese yen, and the function, P(y)=0.16y , is used to convert

timurjin [86]

They probably want 880 pesos, since they asked you do intermediate rounding

but a more accurate answer would be
883.20 pesos

To combine the functions, you'd substitute in 2.76r for y in 0.16y. This would give you 0.16(2.76r), which simplifies to 0.4416r, but they want you round this to the nearest hundredth, so its really 0.44r. Substitute in the 2000 for r and you get 0.44 * 2000 which is 880 pesos

5 0

3 years ago

Find the mass of the triangular region with vertices (0, 0), (3, 0), and (0, 1), with density function ρ(x,y)=x2+y2.

ololo11 [35]

Since density is the ratio of mass to (in this case) area, we can find the mass of the triangular region $\mathcal T$ by computing the double integral of the density function over $\mathcal T$ :

$\mathrm{mass}=\displaystyle\iint_{\mathcal T}\rho(x,y)\,\mathrm dx\,\mathrm dy$

The boundary of $\mathcal T$ is determined by a set of lines in the $x,y$ plane. One way to describe the region $\mathcal T$ is by the set of points,

$\mathcal T=\left\{(x,y)\mid0\le x\le 3\,\land\,0\le y\le1-\dfrac x3\right\}$

So the mass is

$\mathrm{mass}=\displaystyle\int_{x=0}^{x=3}\int_{y=0}^{y=1-x/3}(x^2+y^2)\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2y+\frac{y^3}3\right)\bigg|_{y=0}^{y=1-x/3}\,\mathrm dx$

$=\displaystyle\int_{x=0}^{x=3}\left(x^2\left(1-\frac x3\right)+\frac{\left(1-\frac x3\right)^3}3\right)\,\mathrm dx$

$=\displaystyle\frac1{81}\int_0^3(27-27x+90x^2-28x^3)\,\mathrm dx=\frac52$

6 0

3 years ago

Show in each case that the series converges or diverges

Montano1993 [528]

I can't make out the summand in (d), and I addressed (c) in your other question.

(a) $\displaystyle\sum_{n\ge1}\frac{\cos n\pi}{n\sqrt n}$

We have for positive integers $n$ that $\cos n\pi=(-1)^n$ . We also are aware that the series

$\displaystyle\sum_{n\ge1}\frac1{n^{3/2}}$

converges, since it is a $p$ -series with $p=\dfrac32>1$ . Since the $p$ -series converges in absolute value, the alternating series must also converge by comparison.

- - -

(b) $\displaystyle\sum_{n\ge1}(-1)^n\frac{\ln n}n$

By the alternating series test, this series will converge if the absolute value of the summand is increasing for some large enough $N$ and approaches zero.

We have

$\left|(-1)^n\dfrac{\ln n}n\right|=\dfrac{\ln n}n>0$

for all $n\ge1$ , and we also have that

$\displaystyle\lim_{n\to\infty}\frac{\ln n}n=\lim_{m\to\infty}\frac m{e^m}=0$

(where we substituted $m=\ln n$ , so that $e^m=n$ ).

Therefore (b) also converges.

4 0

2 years ago

For the bake sale, Helen made cookies and brownies. The cookies sold for 95 cents, and the brownies sold for 60 cents. If she ba

Natalka [10]

Let us suppose that 'm' cookies and 'n' brownies were baked.

Since, Helen baked 31 items in total.

So, m+n = 31 (Equaion 1)

So, m = 31-n

Now, since cookies are sold for 95 cents, and the brownies sold for 60 cents.

Since, 1 $ = 100 cents

So, cookies are sold for $0.95, and the brownies sold for $0.60.

Helen sold the cookies and brownies at $23.85

So, 0.95m + 0.60n = 23.85 (Equaion 2)

Substituting the value of 'm' in equation 2, we get as

$0.95(31-n)+0.60n = 23.85$

29.45 - 0.95n +0.60n = 23.85

-0.35n = -5.6

So, n = 16

Since, m = 31-n

m = 31-16

m = 15

So, Helen sold 15 cookies and 16 brownies.

6 0

3 years ago