The negative sign can be read as "the opposite of." true or false

The required, Tom can arrange the 3 books in the combination of 35 different ways on his bookshelf.

The number of ways to choose a subset of 3 books from a set of 7 books can be calculated using the combination formula:

C(n, k) = n! / (k!(n-k)!)

Where n is the total number of items and k is the number of items we want to choose.

In this case, Tom wants to arrange 3 books out of a set of 7 books. Using the combination formula, we have:

C(7, 3) = 7! / (3!(7-3)!)

= 7! / (3!4!)

Calculating the factorials:

7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040

3! = 3 * 2 * 1 = 6

4! = 4 * 3 * 2 * 1 = 24

Substituting these values into the formula:

C(7, 3) = 5040 / (6 * 24)

= 5040 / 144

= 35

Therefore, Tom can arrange the 3 books in 35 different ways on his bookshelf.

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The cylinder has base diameter of 6 centimeters and the height, [tex] h [/tex], of 8 centimeters.

Since the base diameter of the cylinder is 6 cm, therefore, it's radius, [tex] r [/tex], will be 3 cm.

Now, we know that the volume, [tex] V [/tex], of such a cylinder will be:

[tex] V=\pi r^2h [/tex]

Plugging in the values given we get the volume to be:

[tex] V=\pi(3)^2\times 8=3.14\times9\times8=226.08 [/tex] [tex] cm^3 [/tex]

Since Rene needs to make 2000 [tex] cm^3 [/tex] of ice and one cylinder's volume is 226.08 [tex] cm^3 [/tex], Rene would need...

[tex] \frac{2000}{226.08}\approx8.846\approx\boldsymbol{9} [/tex] such identical cylindrical containers.

Final answer:

To make 2000 cm3 of ice, Rene would need a minimum of 9 identical containers, given the volume of one container is 226.08 cm3.

Explanation:

To determine the number of identical cylindrical containers needed to make 2000 cm3 of ice, we must first calculate the volume of one container. Given that the diameter of the base is 6 centimeters, the radius (r) is half of that, which is 3 centimeters. We're also given the height (h) of 8 centimeters. We can use the formula for the volume of a cylinder, V = πr2h.

Plugging in the values, we get:

V = 3.14 × (3 cm)2 × 8 cm

which simplifies to V = 3.14 × 9 cm2 × 8 cm = 226.08 cm3.

Now we can divide the total volume of ice Rene wants to make by the volume of one container:

2000 cm3 ÷ 226.08 cm3 = approx 8.85.

Since Rene cannot use a fraction of a container, she would need a minimum of 9 identical containers to make at least 2000 cm3 of ice.

Answer:

16 units

Step-by-step explanation:

RST was dilated by a scale factor of 1/2.  This means R'S'T' has sides that are 1/2 the length of the sides of RST.

Since the legs of R'S'T' are 8 units, this gives us the equation

8 = 1/2x

Divide both sides by 1/2:

8÷1/2 = x

8 × 2/1 = x

16 = x

Each leg in RST is 16 units.

Stella divided 12/8.

If we divide 12 by 8, we get exactly 1.5.

Let us check given options one by one .

A) Stella made an error : Stella didn't made an error because it's just division of two numbers.

B) The answer is exactly 1.5.  : On dividing 12 by 8, we get exactly 1.5, so this option is correct.

C) The calculator rounded the answer. : On dividing 12 by 8, we get exactly 1.5. So, no rounding is required.

D) The calculator truncated the answer. : On dividing 12 by 8, we get exactly 1.5. So, no truncation in the answer.

Therefore, correct option is B) The answer is exactly 1.5.

Answer:

[tex]x=10.5[/tex]

Step-by-step explanation:

Use Thales' Theorem which says: If two lines, not necessarily parallel, are cut by a system of parallel lines, then the segments that result on one of the two lines are proportional to the corresponding segments obtained on the other.

Take a look at the picture I attached you, from it, It is true that:

[tex]\frac{AB}{A'B'} =\frac{BC}{B'C'}[/tex]

Hence:

[tex]\frac{12}{6} =\frac{21}{x}[/tex]

Solving for x:

[tex]x=\frac{21}{2} =10.5[/tex]

Proving that ΔWXZ is congruent to ΔYZX does not directly prove any of the provided statements about a parallelogram without additional context. Likely, it could imply that diagonals of a parallelogram bisect each other, but more information is needed.

If you prove that ΔWXZ is congruent to ΔYZX, you have demonstrated the congruence of triangles within a geometric figure, likely referencing elements of a parallelogram. However, the specific elements such as sides or diagonals are not indicated in the proof of congruence between the triangles in question.

From the options you have provided and general geometric principles, proving congruent triangles within a figure doesn't necessarily directly prove any of those statements without additional context. If we assume ΔWXZ and ΔYZX are created by a diagonal of a parallelogram, congruent triangles might imply that diagonals of a parallelogram bisect each other (Option A). However, it could also relate to the congruence of opposite sides depending on the specifics of the figure, not the diagonals being congruent (Option B), because congruence of triangles does not guarantee equal length of diagonals, just segments within the triangles.

The solution is that 27x^3+64 factors into (3x+4)(9x^2-12x+16).  

The 2nd factor is the quadratic factor that you wanted.

Polynomials are sums of k-xⁿ terms, where k can be any number and n can be any positive integer.

3x+2x-5, for example, is a polynomial.

here, we have,

we know that,

In general, a^3+b^3 = (a+b)(a^2-ab+b^2).

given that,

27x^3+64

now, we have to factorize the given expression.

using the above formula, we get,

27x^3+64

= (3x+4)(9x^2-12x+16)

Hence, The solution is that 27x^3+64 factors into (3x+4)(9x^2-12x+16).  

The 2nd factor is the quadratic factor that you wanted.

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Final answer:

Switching the axes of two hyperbolas affects the equations of the asymptotes.

Explanation:

When the transverse axis and conjugate axis are switched in two hyperbolas, it affects the equations of the asymptotes. The equations of the asymptotes are determined by the ratio of the coefficients of x² and y² in the hyperbola equation. When the axes are switched, this ratio changes, resulting in different equations for the asymptotes.

For example, let's consider two hyperbolas with equations:

1) 8x² + 10xy - 3y² - 2x - 4y - 2 = 0

2) 8y² + 10xy - 3x² - 2x - 4y - 2 = 0

In the first hyperbola, the ratio of the coefficients of x² and y² is 8/(-3), which determines the equation of the asymptotes. In the second hyperbola, the ratio is (-3)/8, leading to different equations for the asymptotes.

Answer:

Simple Random answer

Step-by-step explanation:

It says that he divided town into eight regions and randomly chose 10. The sample she formed was the Simple Random Answer.

Answer:

upper quartile is 93

lower quartile is 80.5

Step-by-step explanation:

I got a 100 on quiz

According to the information we can infer that the Lower Quartile is 23.5, the upper quartile is 32,5, and the interquartile range is 9.

Arrange the data in ascending order: 20, 22, 25, 28, 29, 30, 32, 33, 34.

Calculate the position of the lower quartile using the formula: (n + 1) / 4, where n is the number of data points.

Since 2.5 falls between the 2nd and 3rd data points, you need to find the average of these two values:

Upper Quartile (Q3):

Calculate the position of the upper quartile using the formula: 3 * (n + 1) / 4.

Since 7.5 falls between the 7th and 8th data points, find the average of these two values:

Interquartile Range (IQR):

Calculate the IQR by subtracting Q1 from Q3:

So, the lower quartile (Q1) is 23.5, the upper quartile (Q3) is 32.5, and the interquartile range (IQR) is 9.

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The probability that a randomly selected score is greater than 334 is 0.0228

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have [tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] )

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write

[tex]P(Z \leq z) = P(Z < z) )[/tex]

Also, know that if we look for Z = z in z tables, the p value we get is

[tex]P(Z \leq z) = \rm p \: value[/tex]

For this case, we're given that:

If we take:
X = a randomly selected score obtained in the GRE exam.

Then, by the given data, we have:

[tex]X \sim N(\mu = 310, \sigma = 12)[/tex]

And therefore, we can write:

Probability that a randomly selected score is greater than 334 = P(X > 334)

Converting X to standard normal distribution:

[tex]Z = \dfrac{X - \mu}{\sigma} = \dfrac{X - 310}{12}[/tex]

The probability we need can be calculated as:

[tex]P(X > 334) = 1 - P(X \leq 334)\\P(X > 334) = 1 - P(Z \leq \dfrac{334 - 310}{12})\\\\P(X > 334) = 1 - P(Z \leq 2)[/tex]

From the z-tables, the p-value for Z = 2 is 0.9772

Thus, we get [tex]P(Z \leq 2) = 0.9772[/tex], and therefore,

[tex]P(X > 334) = 1 - P(Z \leq 2) = 1- 0.9772 = 0.0228[/tex]

Thus, the probability that a randomly selected score is greater than 334 is 0.0228

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Answer:

A

Step-by-step explanation:

n1=8. 7+1=8

n2=9. 7+2=9

n3=10. 7+3=10

n4=11. 7+4=11

n5=12. 7+5=12