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3 years ago

Find the area of a triangle with a base and height of 1/3 and 1/4

Mathematics

1 answer:

balandron [24]3 years ago

4 0

Answer: 1/12

Step-by-step explanation:

Multiply 1/3 by 1/4. Or just multiply 3 by 4.

3 x 4 = 12

Therefore, the answer is 1/12.

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the average temperature for febuary was -4.03 in chicago and -7.14 in michigan.which of the following represent the difference b

ozzi

To put it simply, the answer is 3.11.

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3 years ago

I need helppp on my test and i can't get it wrongggg

Artyom0805 [142]

Answer:

Perimeter = 16x + 16

Area = 40x + 15

Step-by-step explanation:

Perimeter = the sum of all the side lengths

Perimeter = 2(8x+3) + 2(5) = 16x + 6 + 10 = 16x + 16

Area = product of base times height

Area = (8x+3)*5 = 40x + 15

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asambeis [7]

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4 years ago

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fiasKO [112]

3/x = 4/3
4x = 9
x = 9/4

5/y = 4/3
4y = 15
y = 15/4 (B)

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