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vichka [17]
3 years ago
12

How can you decompose the composite figure to determine its area?

Mathematics
1 answer:
In-s [12.5K]3 years ago
6 0

Answer:Área y perímetro de figuras compuestas

por Luz | Dic 5, 2014 | Geometría, Matematicas | 39 Comentarios

Obtener el área o perímetro es de las habilidades más fáciles y básicas en geometría. Sin embargo, al momento de combinar varias figuras planas se forma figuras compuestas, que pueden incluir: cuadrados, círculos, triángulos, rectángulos, trapecio, etc.

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Using 5,7,2 to equal 6
choli [55]

Answer:

here my answer

Step-by-step explanation:

5+1=6

7-1=6

3x2=6

hope it helps :)

8 0
3 years ago
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help me plz you'll get brainliest if u show work and the only way for u to help me is if u show your work so plz do
Usimov [2.4K]

Answer:

D.

Step-by-step explanation:

This will be D because the beaker already has 0.2 mL of water. If Jovita needs to add more water to the beaker, and<em> then </em>it is 0.8. In D, the diagram shows you 0.2 + x = 0.8, which makes it the right diagram.

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5 0
2 years ago
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∆ABC transforms to produce ∆A'B'C'. Which transformation did NOT take place?
never [62]

Answer: The answer is (D) Reflection across the line y = -x.


Step-by-step explanation:  In figure given in the question, we can see two triangles, ΔABC and ΔA'B'C' where the second triangle is the result of transformation from the first one.

(A) If we rotate ΔABC 180° counterclockwise about the origin, then the image will coincide with ΔA'B'C'. So, this transformation can take place here.

(B) If we reflect ΔABC across the origin, then also the image will coincide with ΔA'B'C' and so this transformation can also take place.

(C) If we rotate ΔABC through 180° clockwise about the origin, the we will see the image will be same as ΔA'B'C'. Hence, this transformation can also take place.

(D) Finally, if we reflect ΔABC across the line y = -x, the the image formed will be different from ΔA'B'C', in fact, it is ΔA'D'E', as shown in the attached figure. So, this transformation can not take place here.

Thus, the correct option is (D).


6 0
3 years ago
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If g(x)=1/x were shifted 3 units to the left and 3 units up, what would the new equation be
emmainna [20.7K]

Answer:

g(x) 1/3-4

Step-by-step explanation:

4 0
3 years ago
The joint probability density function of X and Y is given by fX,Y (x, y) = ( 6 7 x 2 + xy 2 if 0 &lt; x &lt; 1, 0 &lt; y &lt; 2
fredd [130]

I'm going to assume the joint density function is

f_{X,Y}(x,y)=\begin{cases}\frac67(x^2+\frac{xy}2\right)&\text{for }0

a. In order for f_{X,Y} to be a proper probability density function, the integral over its support must be 1.

\displaystyle\int_0^2\int_0^1\frac67\left(x^2+\frac{xy}2\right)\,\mathrm dx\,\mathrm dy=\frac67\int_0^2\left(\frac13+\frac y4\right)\,\mathrm dy=1



b. You get the marginal density f_X by integrating the joint density over all possible values of Y:

f_X(x)=\displaystyle\int_0^2f_{X,Y}(x,y)\,\mathrm dy=\boxed{\begin{cases}\frac67(2x^2+x)&\text{for }0

c. We have

P(X>Y)=\displaystyle\int_0^1\int_0^xf_{X,Y}(x,y)\,\mathrm dy\,\mathrm dx=\int_0^1\frac{15}{14}x^3\,\mathrm dx=\boxed{\frac{15}{56}}

d. We have

\displaystyle P\left(X

and by definition of conditional probability,

P\left(Y>\dfrac12\mid X\frac12\text{ and }X

\displaystyle=\dfrac{28}5\int_{1/2}^2\int_0^{1/2}f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\boxed{\frac{69}{80}}

e. We can find the expectation of X using the marginal distribution found earlier.

E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}

f. This part is cut off, but if you're supposed to find the expectation of Y, there are several ways to do so.

  • Compute the marginal density of Y, then directly compute the expected value.

f_Y(y)=\displaystyle\int_0^1f_{X,Y}(x,y)\,\mathrm dx=\begin{cases}\frac1{14}(4+3y)&\text{for }0

\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87

  • Compute the conditional density of Y given X=x, then use the law of total expectation.

f_{Y\mid X}(y\mid x)=\dfrac{f_{X,Y}(x,y)}{f_X(x)}=\begin{cases}\frac{2x+y}{4x+2}&\text{for }0

The law of total expectation says

E[Y]=E[E[Y\mid X]]

We have

E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}

\implies E[Y\mid X]=1+\dfrac1{6X+3}

This random variable is undefined only when X=-\frac12 which is outside the support of f_X, so we have

E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87

5 0
3 years ago
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