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Vitek1552 [10]
3 years ago
10

What is the areas of triangle

Mathematics
1 answer:
SashulF [63]3 years ago
6 0

135 is the answer hope this helps

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Sketch seven points. Then, connect the points to form a closed plane figure. What kind of polygon did you draw?
weqwewe [10]

It’s a heptagon, which has seven points.

7 0
3 years ago
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Let y be a random variable with p(y) given in the accompanying table. find e(y ), e(1/y ), e(y 2 − 1), and v(y ). y 1 2 3 4 p(y)
Studentka2010 [4]
<span>E[Y] = 0.4·1 + 0.3·2 + 0.2·3 + 0.1·4 = 2 E[1/Y] =0.4·1/1 + 0.3·1/2 + 0.2·1/3 + 0.1·1/4 = 0.4 + 0.15 + 0.0666 + 0.025?0.64 V[Y] =E[Y2]-E[Y]2= (0.4)·12+(0.3)·22+(0.2)·32+(0.1)·42-22= 0.4+1.2+1.8+1.6-4= 5-4 = 1</span>
8 0
3 years ago
The temperature is -5°C at sunrise. By noon, the temperature changes by 6°C what could be the temperature at noon? answer asap !
Citrus2011 [14]

Answer:

the temperature is -5C which is negative so its behind the zero on a number line  and 6C is positive which is infront of the zero so it would be 1C by noon

Step-by-step explanation:

3 0
3 years ago
Find the following integral
ololo11 [35]

There's nothing preventing us from computing one integral at a time:

\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2

\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2

\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx

Expand the integrand completely:

x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x

Then

\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}

4 0
2 years ago
Round 66,315,254 to the nearest ten-million
Flauer [41]

Answer:

Step-by-step explanation:

66,315,254.....rounded to the nearest 10 million =

70,000,000

4 0
3 years ago
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