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Zigmanuir [339]

2 years ago

10

The volumes of two similar prisms are 891 cm3 and 33 cm3. The surface area of the larger prism is 153 cm2. What is the surface a

rea of the smaller prism?

1 answer:

Kitty [74]2 years ago

7 0

I think it would be 271

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Use inverse operations to find the inverse of y = 27x?<br> The inverse of y = 27x3 is y =

VLD [36.1K]

Answer:

y = (∛x)/3

Step-by-step explanation:

To undo the multiplication by 27, you multiply by its inverse:

(1/27)y = (1/27)(27x^3)

y/27 = x^3 . . . . . . . . . . . simplify

To undo the cube, you take the cube root:

(∛y)/(∛27) = ∛(x^3)

(∛y)/3 = x

Apparently, you want the inverse function, so you swap the variables:

y = (∛x)/3

_____

You can swap the variables at the beginning or end. It doesn't matter. If you do it at the beginning, you have ...

x = 27y^3

and you're solving for y. You use the same inverse operations that we used above.

7 0

2 years ago

Find the dicontinuities of the function. f(x) = x2 + 12x + 27 x2 + 4x + 3 . There is a removable discontinuity at (, ).

ValentinkaMS [17]

Solution-

$f(x)=\frac{x^{2}+12x+27 }{x^{2} +4x+3} =\frac{x^{2}+9x+3x+27 }{x^{2} +3x+x+3}=\frac{(x+3)(x+9)}{(x+3)(x+1)}$

Equating the denominator to 0,

⇒ (x+3)(x+1) = 0

⇒ x= -3, -1

∴ Discontinuities of the function f(x) is at -3, -1

As (x+3) is also a factor of the numerator, you will have what's called a removable discontinuity. That will show up as a hole in the graph.

∴ At x = -3, you will have removable discontinuity.

4 0

2 years ago

Read 2 more answers

Mr. and Mrs. Smith's children all play different sports. They want to buy a combination of soccer training sets,which each cost

8090 [49]

C is the answer because a number of children are unknown.

5 0

2 years ago

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Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised t

zimovet [89]

Answer:

(b) 1/792

Step-by-step explanation:

The complete question is;

<em>Counting: Grading One professor grades homework by randomly choosing 5 out of 12 homework problems to grade.</em>

<em>(a) How many different groups of 5 problems can be chosen from the 12</em>

<em>problems?</em>

<em>(b)Probability extension: Jerry did only 5 problems of one assignment. What is the probability that the problems he did comprised the group that was selected to be graded?</em>

<u>In (a)</u>

<u></u>

Apply the formula

$\frac{n!}{(n-r)!(r!)}$

where n=12 and r=5

substitute values

$=\frac{12!}{(12-5)!(5!)} \\\\\\=\frac{12!}{(7!)(5!)} \\\\\\=\frac{12*11*10*9*8}{5*4*3*2*1} \\\\\\=\frac{95040}{120} \\\\\\=792$

In (b)

If Jerry did only 5 problems of one assignment then the probability will be

$\frac{5}{12} *\frac{4}{11} *\frac{3}{10} *\frac{2}{9} *\frac{1}{8} =\frac{1}{792}$

<em />

<em />

4 0

2 years ago

Shishir bought 4000 orange at 70 paisa each. But 400 of them were rotten. He sold 2000 oranges at 90 paisa each.If he plans to m

Ipatiy [6.2K]

Answer:

He needs to sell the rest of the oranges at 75 paisa each.

Step-by-step explanation:

Consider the given information that, Shishir bought 4000 orange at 70 paisa each.

Note: 1 rupees = 100 paisa

Thus, 70 paisa = 70/100 rupees = 0.70 rupees

Therefore, the cost price of 4000 oranges is:

4000×0.70 rupees = 2800 rupees

The selling price of 2000 oranges is:

2000×0.90 rupees = 1800 rupees

The number of oranges now Shishir have:

4000 - 2000 - 400 = 1600

He wants to make a profit of RS 200. Thus the selling price of 4000 oranges should be:

2800 rupees + 200 rupees = 3000 rupees

He earned 1800 rupees by selling 2000 oranges at 90 paisa. So, the remaining amount that he needs to make with 1600 oranges is:

3000 rupees - 1800 rupees = 1200 rupees

Therefore, the cost of one orange is:

1600 oranges = 1200 rupees

1 orange = 1200/1600 rupees

1 orange = 0.75 rupees

Hence, he needs to sell the rest of the oranges at 75 paisa each.

4 0

2 years ago