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2 years ago

10

PLEASE HELP ASAP!! Geometry:

1 answer:

AlexFokin [52]2 years ago

7 0

Answer:

2.87b

Step-by-step explanation:

done done done donendone

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frozen [14]

$\sqrt{-4}$ can also be written as 2i
where i = √-1

$\frac{2i}{4-2-3i} = \frac{2i}{2-3i}$

Rationalizing denominator,

$\frac{2i(2+3i)}{(2-3i)(2+3i)} = \frac{4i-6}{4+9} = \frac{4i-6}{13}$

8 0

3 years ago

The product of 11/13 and 4 is:

kirza4 [7]

Answer:

11/13 x 4 = 3 5/13

Step-by-step explanation:

4 0

3 years ago

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An environment engineer measures the amount ( by weight) of particulate pollution in air samples ( of a certain volume ) collect

Serggg [28]

Answer:

$k = 1$

$P(x > 3y) = \frac{2}{3}$

Step-by-step explanation:

Given

$f \left(x,y \right) = \left{ \begin{array} { l l } { k , } & { 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x } & { \text 0, { elsewhere. } } \end{array} \right.$

Solving (a):

Find k

To solve for k, we use the definition of joint probability function:

$\int\limits^a_b \int\limits^a_b {f(x,y)} \, = 1$

Where

${ 0 \leq x} \leq 2,0 \leq y \leq 1,2 y \leq x }$

Substitute values for the interval of x and y respectively

So, we have:

$\int\limits^2_{0} \int\limits^{x/2}_{0} {k\ dy\ dx} \, = 1$

Isolate k

$k \int\limits^2_{0} \int\limits^{x/2}_{0} {dy\ dx} \, = 1$

Integrate y, leave x:

$k \int\limits^2_{0} y {dx} \, [0,x/2]= 1$

Substitute 0 and x/2 for y

$k \int\limits^2_{0} (x/2 - 0) {dx} \,= 1$

$k \int\limits^2_{0} \frac{x}{2} {dx} \,= 1$

Integrate x

$k * \frac{x^2}{2*2} [0,2]= 1$

$k * \frac{x^2}{4} [0,2]= 1$

Substitute 0 and 2 for x

$k *[ \frac{2^2}{4} - \frac{0^2}{4} ]= 1$

$k *[ \frac{4}{4} - \frac{0}{4} ]= 1$

$k *[ 1-0 ]= 1$

$k *[ 1]= 1$

$k = 1$

Solving (b): $P(x > 3y)$

We have:

$f(x,y) = k$

Where $k = 1$

$f(x,y) = 1$

To find $P(x > 3y)$ , we use:

$\int\limits^a_b \int\limits^a_b {f(x,y)}$

So, we have:

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {f(x,y)} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 {1} dxdy$

$P(x > 3y) = \int\limits^2_0 \int\limits^{y/3}_0 dxdy$

Integrate x leave y

$P(x > 3y) = \int\limits^2_0 x [0,y/3]dy$

Substitute 0 and y/3 for x

$P(x > 3y) = \int\limits^2_0 [y/3 - 0]dy$

$P(x > 3y) = \int\limits^2_0 y/3\ dy$

Integrate

$P(x > 3y) = \frac{y^2}{2*3} [0,2]$

$P(x > 3y) = \frac{y^2}{6} [0,2]\\$

Substitute 0 and 2 for y

$P(x > 3y) = \frac{2^2}{6} -\frac{0^2}{6}$

$P(x > 3y) = \frac{4}{6} -\frac{0}{6}$

$P(x > 3y) = \frac{4}{6}$

$P(x > 3y) = \frac{2}{3}$

8 0

2 years ago

So I don’t really understand this so yeh pls help *pic included* by the way this is volume of cubes and cuboids and the question

faust18 [17]

Answer:

A2: 4*4*4 = 64 cm³; 6*6*6 = 216 cm³; etc...

B2: 7*4.5*4 = 126 cm³; 10.5*6*2 = 126 cm³; etc...

Step-by-step explanation:

So in A2 you have cubes, so l=b=h, and only one has to be given, called the "length of side".

In B2 they are different, hence there are 3 rows with l, b and h. But still you multiply them l x b x h

3 0

2 years ago

2+2(560)+5638+163936

dmitriy555 [2]

170696 is the answer fellow user

8 0

3 years ago

Read 2 more answers