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jarptica [38.1K]

2 years ago

12

A rectangular prism with a volume of four cubic units is filled with cubes and side links of 1/3 unit. How many 1/3 unit cubes d

oes it take to fill the prism?

1 answer:

Artist 52 [7]2 years ago

7 0

Answer:

Number of box that fills the prism=36.36 cubes

Step-by-step explanation:

Number of cubes that can fill the rectangular prism= Volume of the rectangular prism / Volume of the cube

Volume of the rectangular prism= $4 units ^3$

Side length of the cube= $1/3 unit$

Volume of the cube :

$Side*side*side\\\\=1/3*1/3*1/3\\\\=1/9$

$=0.11units^3$

Number of box that fills the prism= 4/0.11= 36.36 cubes

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a bag contains eleven counters. five are white. a counter is taken out the bag and is not replaced. a second counter is taken ou

Fed [463]

Answer:

$\displaystyle P(A)=\frac{6}{11}$

Step-by-step explanation:

<u>Probabilities</u>

The question describes an event where two counters are taken out of a bag that originally contains 11 counters, 5 of which are white.

Let's call W to the event of picking a white counter in any of the two extractions, and N when the counter is not white. The sample space of the random experience is

$\Omega=\{WW,WN,NW,NN\}$

We are required to compute the probability that only one of the counters is white. It means that the favorable options are

$A=\{WN,NW\}$

Let's calculate both probabilities separately. At first, there are 11 counters, and 5 of them are white. Thus the probability of picking a white counter is

$\displaystyle \frac{5}{11}$

Once a white counter is out, there are only 4 of them and 10 counters in total. The probability to pick a non-white counter is now

$\displaystyle \frac{6}{10}$

Thus the option WN has the probability

$\displaystyle P(WN)=\frac{5}{11}\cdot \frac{6}{10}=\frac{30}{110}=\frac{3}{11}$

Now for the second option NW. The initial probability to pick a non-white counter is

$\displaystyle \frac{6}{11}$

The probability to pick a white counter is

$\displaystyle \frac{5}{10}$

Thus the option NW has the probability

$\displaystyle P(NW)=\frac{6}{11}\cdot \frac{5}{10}=\frac{30}{110}=\frac{3}{11}$

The total probability of event A is the sum of both

$\displaystyle P(A)=\frac{3}{11}+\frac{3}{11}=\frac{6}{11}$

$\boxed{\displaystyle P(A)=\frac{6}{11}}$

7 0

2 years ago

Solve:<br><br> a.x3+x2-8x - 12x+2<br> b. x3-4x2-3x + 18x-3<br> c. x2 + 4x + 4<br> d. x2 - 6x + 9

NeTakaya

<h2>Steps:</h2>

So firstly, this function is asking us to divide f(x) by g(x) so let's set that up as such:

$(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\div \frac{x-3}{x+2}$

Next, remember that <u>dividing by a number is the same as multiplying by its reciprocal.</u> To find the reciprocal of a number, flip the numerator and denominator around. With this info, flip the second fraction to it's reciprocal and change the sign to multiplication:

$(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\times \frac{x+2}{x-3}$

Next, we are going to factor x² - x - 6. Firstly, what two terms have a product of -6x² and a sum of -x? That would be -3x and 2x. Replace -x with 2x - 3x:

$(\frac{f}{g})(x)=\frac{x^2+2x-3x-6}{1}\times \frac{x+2}{x-3}$

Next, factor x² + 2x and -3x - 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

$(\frac{f}{g})(x)=\frac{x(x+2)-3(x+2)}{1}\times \frac{x+2}{x-3}$

Now you can rewrite it as:

$(\frac{f}{g})(x)=\frac{(x-3)(x+2)}{1}\times \frac{x+2}{x-3}$

Next, multiply:

$(\frac{f}{g})(x)=\frac{(x-3)(x+2)^2}{x-3}$

Next, divide:

$(\frac{f}{g})(x)=(x+2)^2$

And lastly, simplify:

A good tip: $(x+y)^2=x^2+2xy+y^2$

$(\frac{f}{g})(x)=x^2+4x+4$

<h2>Answer:</h2>

<u>The correct option is C. x² + 4x + 4.</u>

7 0

3 years ago