All categories

3 years ago

HELP ME PLZ

Mathematics

2 answers:

mr Goodwill [35]3 years ago

7 0

You are asked to use the volume formula V = lwh which basically means to find the volume by multiplying the length, width and height of the prism. These are all given to you as expression so you are asked to multiply the expressions. That is, find: $( \frac{ x^{2} -x}{4x+1} )( \frac{2x-5}{1} )( \frac{24 x^{2} +6x}{x-1} )$

First factor each of the expressions in the numerators and then multiply those in the numerators putting them over those in the denominator. It turns out those in the denominators (in this problem) cannot be factored. When you do this you are left with:
$\frac{(x)(x-1)(2x-5)(6x)(4x+1)}{(4x+1)(x-1)}$

You can divide out (sometimes called cancelling out) those terms expressions that are in both the numerator and the denominator. That leaves you with:
$\frac{(x)(2x-5)(6x)} {1}$ which we can clean up by multiplying x and 6x in the numerator.

Thus, the volume of the prism is: $6 x^{2} (2x-5)$
We do not multiply this out because it is now written as a product and thus factored (as you were asked).

g100num [7]3 years ago

7 0

These are the answers!!

Write 2x – 5 over 1. Factor numerators and denominators, when possible . Divide out common factors to simplify. Determine that the volume is (x)(6x)(2x – 5).

You might be interested in

Find the area of the sector

sergij07 [2.7K]

Answer:

$\boxed {\boxed {\sf a \approx 190.9 \ yd^2}}$

Step-by-step explanation:

There are 2 formulas for the area of a sector, but since we are given the central angle in degrees (not radians), we will use this formula:

$a= \frac {\theta}{360} \times \pi r^2$

Where θ is the central angle and r is the radius.

For this circle, the radius is 9 yards and the central angle is 270 degrees. We can substitute these values into the formula.

$a= \frac {270}{360} \times \pi \times (9 \ yd )^2$

Solve the fraction.

$a=0.75 \times \pi \times (9 \ yd)^2$

Solve the exponent.

(9 yd)²= 9 yd* 9 yd=81 yd²

$a= 0.75 \times \pi \times 81 \ yd^2$

Multiply all three numbers together.

$a= 190.851753706 \ yd^2$

The question asks us to round to the nearest tenth.

190.8<u>5</u>1753706

The 5 in the hundredth place tells us to round the 8 up to a 9.

$a \approx 190.9 \ yd^2$

The area of the sector is approximately <u>190.9 square yards. </u>

8 0

3 years ago

Read 2 more answers

An experiment consists of tossing 4 coins successively. The number of sample points in this experiment is

IgorLugansk [536]

Answer:

C. 4

Step-by-step explanation:

Hi, the number of sample points is the sum of all attempts you executed during an experiment. Because the experiment proposes tossing a coin 4 times, then you have 4 sample points.

Attempt 1 = coin toss 1 = sample point 1

Attempt 2 = coin toss 2 = sample point 2

Attempt 3 = coin toss 3 = sample point 3

Attempt 4 = coin toss 4 = sample point 4

6 0

4 years ago

A sample of 18 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is n

ki77a [65]

Answer:

i. x=- - - - - -

ii. o=- - - - - -

iii. Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.

c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.

N(X, o/ $\sqrt{n}$ )

So, the distribution is

N(2, 0.12/ $\sqrt{16}$ )

i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411

iii. The error bound is 0.0411

i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582

iii. The error bound is 0.0582

g) The difference in the confidence intervals in part (f) and part (e) is of the level of confidence. The change is, the change in the area being calculated for the normal distribution. Therefore, the larger confidence level results in larger area and larger interval.

Hence the interval in part is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

Step-by-step explanation:

i. x=- - - - - -

ii. o=- - - - - -

iii. Sx=- - - - -

b) The random variable X signifies the weights of the bags of candies which are selected at random.

c) The statistics variable X is a measure on a sample that is used as an estimate of the population mean.X is the mean weight of the 16 bags of candies that were selected.

d) The distribution needed for this problem is normal distribution with parameters because the population standard deviation is known.

N(X, o/ $\sqrt{n}$ )

So, the distribution is

N(2, 0.12/ $\sqrt{16}$ )

i. The 90% confidence interval for the population mean weight of the candies is 1.9589 , 2.0411

iii. The error bound is 0.0411

i. The 98% confidence interval for the population mean weight of the candies is 1.9418 , 2.0582

iii. The error bound is 0.0582

Hence the interval in part is larger than the one in part (e).

h) The interval in part (f) signifies that with 98% confidence that the population mean weight of the bag of candies lies between 1.942 ounce and 2.058 ounce.

5 0

3 years ago

List two things that affect the end behavior of a polynomial function

Thepotemich [5.8K]

Step-by-step explanation:

The end behavior of a polynomial function is the behavior of the graph of f(x) as x approaches positive infinity or negative infinity. The degree and the leading coefficient of a polynomial function determine the end behavior of the graph.

4 0

3 years ago

Simplify the question

Ilia_Sergeevich [38]

The answer is x^6 y^3

7 0

3 years ago

Read 2 more answers