Easy question, decent amount of points! 25 points, if you're curious!
Be quick or it is gone.

Answer:

[tex]40\ m^{3}[/tex]

Step-by-step explanation:

we know that

If two figures are similar then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z-----> scale factor

x-----> the volume of the dilated cone

y-----> the volume of the original cone

[tex]z^{3}=\frac{x}{y}[/tex]

In this problem we have

[tex]z=1/5[/tex]

[tex]y=5,000\ m^{3}[/tex]

substitute and solve for x

[tex](1/5)^{3}=\frac{x}{5,000}[/tex]

[tex](1/125)=\frac{x}{5,000}[/tex]

[tex]x=5,000/125=40\ m^{3}[/tex]

ANSWER

[tex] \boxed { \sqrt{} }30 \degree[/tex]

[tex] \boxed { \sqrt{} }210 \degree[/tex]

EXPLANATION

We want to solve

[tex] \cot( \theta) = \sqrt{3} [/tex]

where

[tex]0 \degree \: \leqslant x \leqslant 360 \degree[/tex]

We reciprocate both sides of this trigonometric equation to obtain:

[tex] \tan( \theta) = \frac{1}{ \sqrt{3} } [/tex]

We take arctangent of both sides to get;

[tex] \theta = \tan ^{ - 1} ( \frac{1}{ \sqrt{3} } ) [/tex]

[tex] \theta = 30 \degree[/tex]

This is the principal solution.

The tangent ratio is also positive in the third quadrant.

The solution in the third quadrant is

[tex]180 + \theta = 180 + 30 = 210 \degree[/tex]

Answer: 1. C)  5/18, 25/90

2.  A) 6/10

Step-by-step explanation: 1) We need to find the ratios those makes a proportion:

Let us check given options one by one.

3/15, 12/55  

Converting them into simplest fractions.

3÷3/15÷3 = 1/5

12/55 can't be reduce more.

1/5 ≠ 12/55

So, 3/15, 12/55  don't form a proportion.

8/24, 12/35

Converting them into simplest fractions.

8÷8/24÷8 = 1/3

12/35 can't be reduce more.

8/24 ≠ 12/35

So, 8/24, 12/35 don't form a proportion.

So, 5/18, 25/90 form a proportion.

4/11, 16/25

Converting them into simplest fractions.

4/11 and 16/25 both can't be reduce more.

4/11≠16/25

So, 4/11, 16/25 don't form a proportion.

___________________________________________________

Let us reduce 9/15 into simplest fraction.

Now, let us convert each and every option in simplest fractions.

16÷1/21÷1 = 16/21

36÷2/50÷2 = 18/25

45÷5/70÷5 = 9/14

We can see 6/10 gives lowest fraction 3/5 as 9/15 gives.

Therefore, 6/10 form a proportion with 9/15.

The weight of a can of soup can be determined using the variation formula. By substituting the given values into the equation, we can find the weight of a can that is 4 inches high with a diameter of 2 inches.

To find the weight of the can that is 4 inches high with a diameter of 2 inches, we can use the given information that the weight of a can of soup varies jointly with the height and the square of the diameter. We are given that a can 8 inches high with a diameter of 3 inches weighs 28.8 ounces. This gives us enough information to set up a proportion to solve for the weight of the can we are looking for.

Let's assign variables to the height (h), diameter (d), and weight (w) of the can. From the given information, we have the following equation:

w = khd^2

Substituting the given values, we have:

28.8 = k(8)(3^2)

Here, we have one equation with one unknown. We can now solve for the constant k by dividing both sides of the equation by (8)(3^2).

After finding the value of k, we can substitute it back into the equation to find the weight of the can that is 4 inches high with a diameter of 2 inches.

w = k(4)(2^2)

Solving for w will give us the weight of the can.

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The simplified expression is √10 - √14 + √15 - √21.

To simplify the expression (√2 + √3)(√5 - √7), we can use the distributive property of multiplication.

Expanding the expression, we get:

(√2 + √3)(√5 - √7) = √2 x √5 + √2 x (-√7) + √3 x √5 + √3 x (-√7)

Now, simplifying each term with FOIL method we have:

√2 * √5 = √(2 x 5) = √10

√2 * (-√7) = -√(2 x 7) = -√14

√3 * √5 = √(3 x 5) = √15

√3 * (-√7) = -√(3 x 7) = -√21

Combining the simplified terms, we get:

√10 - √14 + √15 - √21

Therefore, the simplified expression is √10 - √14 + √15 - √21.

In this process, we applied the distributive property to expand the expression and then simplified each term by multiplying the square roots together. Finally, we combined the like terms to obtain the simplified expression.

For more such answers on the FOIL method

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Answer:

√10 + √15 - √14 - √21

Step-by-step explanation:

(√2 + √3)(√5 - √7)

= √2(√5 - √7) + √3(√5 - √7)

= √10 - √14 + √15 - √21

= √10 + √15 - √14 - √21

Is there a viable solution when time is 30 minutes? Answer: No, the tub will be empty by then. 

A tub filled with 50 quarts of water empties at a rate of 2.5 quarts per minute.

Let w be the quarts of water left in the tub.

Let t be the time in minutes.

So, the equation to model this situation is :

Modelling equation: [tex]w=50-2.5t[/tex]

This equation is viable only up to when t is 20 minutes. That will give w = 0. More than 20 minutes is not possible.

The probability that at least one of the three children gets the disease from their mother is 97.3%.

To find the probability that at least one child gets the disease from their mother when a female carrier of the gene has three children, we can follow these steps:

1. Define the probabilities:

  - Probability that a child gets the disease from their mother (given): [tex]\( P(D) = 0.70 \)[/tex].

  - Probability that a child does not get the disease from their mother: [tex]\( P(D^c) = 1 - P(D) = 0.30 \)[/tex].

2. Identify the total number of children:

  - Number of children: [tex]\( n = 3 \)[/tex].

3. Calculate the probability that none of the children get the disease:

  - For all three children to not get the disease, the probability is [tex]\( (P(D^c))^n = (0.30)^3 \)[/tex].

4. Find the probability that at least one child gets the disease:

  - The probability that at least one child gets the disease is the complement of the probability that none of the children get the disease.

  - Thus, [tex]\( P(\text{at least one child gets the disease}) = 1 - P(\text{none of the children get the disease}) \)[/tex].

Let's do the calculations step by step.

Step-by-Step Calculations:

1. Calculate the probability that none of the children get the disease:

  [tex]\[ P(\text{none of the children get the disease}) = (P(D^c))^3 = (0.30)^3 \][/tex]

2. Perform the exponentiation:

  [tex]\[ (0.30)^3 = 0.30 \times 0.30 \times 0.30 = 0.027 \][/tex]

3. Calculate the probability that at least one child gets the disease:

  [tex]\[ P(\text{at least one child gets the disease}) = 1 - P(\text{none of the children get the disease}) = 1 - 0.027 = 0.973 \][/tex]

Conclusion:

The probability that at least one child gets the disease from their mother is [tex]\( 0.973 \)[/tex], or 97.3%.

In mathematics, discontinuities and restrictions in rational expressions are caused by values that make the denominator zero, as division by zero is undefined. Graphically, these are represented as vertical asymptotes or holes. They can be accounted for by noting the x-values where the denominator equals zero.

In mathematics, discontinuities and restrictions in rational expressions are caused by values that make the denominator of the expression equal to zero, as division by zero is undefined. A rational expression is a ratio of two polynomials, typically written in the form P(x)/Q(x), where P(x) and Q(x) are polynomial expressions and x is a variable.

For example, in the rational expression (x+2)/(x-3), x-3 is the denominator. The restriction is determined by setting the denominator equal to zero. If x-3 equals zero, then the value of x that makes this true is 3. Therefore, x cannot be 3 as this would make the denominator zero, creating a discontinuity in the rational expression.

Graphically, discontinuities in rational expressions are represented as vertical asymptotes or holes. A vertical asymptote occurs when the denominator of the fraction is zero but the numerator is not. A hole occurs when both the numerator and the denominator are zero at the same point. These can be accounted for by noting the x-values where the denominator equals zero.

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The polynomial representing the amount of paper wasted when cutting the largest possible circle out of a square of side length l is (4-π)/4 * l².

The question at hand is concerned with finding the polynomial that represents the amount of paper wasted when cutting out the largest possible circles from square pieces of paper. The side length of each square is given as l. The area of each square is l², while the area of the circle that can be cut from the square is calculated using the formula πr², where r is the radius of the circle. Because the largest circle that fits in the square touches all four sides, the diameter of the circle equals the side length l, making the radius r equal to l/2.

To find the polynomial for the wasted paper, first we calculate the area of the circle: π(l/2)², which simplifies to πl²/4. To find the wasted area, subtract the area of the circle from the area of the square: l² - πl²/4. This difference represents the wasted paper and can be further simplified to a single polynomial: (4/4)l² - (π/4)l², which simplifies to (4-π)/4 * l². This is the polynomial representing the amount of paper wasted for each square piece of paper.

Answer:

The land should be recorded in the purchaser's books at : $153,000.

Step-by-step explanation:

Since the parcel of land is purchased by the company for $153,000, hence the land should be recorded in the purchaser's books at : $153,000.

Answer:

Step-by-step explanation:

Pepe is putting a fence in his backyard to enclose the garden in the form of a triangle.

In the garden already has sides enclosed with 8 feet and 5 feet.

We know a triangle is possible when sum of length of two sides > third side

so third side < 8 + 5

or third side should be less than 13.

Final answer:

Ben incorrectly assumed that ¼ of two pizzas combined would still be ¼; however, the correct sum is ¼ + ¼ which equals ½ of a pizza. To add fractions, only the numerators are added when denominators match. Mastery of fractions is critical in both academics and day-to-day life.

Explanation:

Ben's error lies in the misunderstanding of fractions and how they add up. If Ben has two pizzas and each has ¼ left, combining the two portions does not result in having ½ of a pizza, but instead, he would have ¼ + ¼ which equals ½. This is because when adding fractions, you only add the numerators (the top numbers) if the denominators (the bottom numbers) are the same. Therefore, ¼ of one pizza plus ¼ of another pizza would equal ½ of a pizza. It's important to understand that fractions represent parts of a whole and that these parts must be added correctly to find the total amount.

Being comfortable with fractions is important not only for academic success but also for everyday life, such as understanding discounts, following recipes, or splitting bills. Utilizing real-life examples like these can help strengthen one's intuitive sense of fractions and make mathematical concepts more relatable.

Consider the coordinate plane:

1. The origin is the point where Sharon and Jacob started - (0,0).

2. North - positive y-direction, south - negetive y-direction.

3. East - positive x-direction, west - negative x-direction.

Then,

The distance between two points with coordinates [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex] can be calculated using formula

[tex]D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.[/tex]

Therefore, the distance between  Jacob and Sharon is

[tex]D=\sqrt{(12-(-4))^2+(-5-3)^2}=\sqrt{16^2+8^2}=\sqrt{256+64}=\sqrt{320}=8\sqrt{5}\approx 11.18\ m.[/tex]

Final Answer:

The point (0,0) does not satisfy the inequality [tex]y ≥ x^2[/tex] + x - 4, as 0 is not greater than or equal to -4. Additionally, it does not satisfy [tex]y > x^2 + 2x + 1[/tex], as 0 is not greater than 1. Therefore, (0,0) does not meet the conditions of the system of inequalities.

Step-by-step explanation:

The point (0,0) is not a solution to the given system of inequalities. First, considering the inequality [tex]y ≥ x^2 + x - 4[/tex], when substituting x=0 and y=0 into the equation, we find that 0 is not greater than or equal to -4. Therefore, (0,0) does not satisfy the conditions of the first inequality. Moving on to the second inequality, [tex]y > x^2 + 2x + 1[/tex], substituting x=0 and y=0 results in 0 not being greater than 1.

Consequently, (0,0) fails to meet the requirements of the second inequality as well. In summary, the point (0,0) does not simultaneously satisfy both inequalities, rendering it unsuitable as a solution to the system.

Analyzing solutions to systems of inequalities involves evaluating each inequality independently to ensure the chosen point satisfies all conditions. In this instance, the failure of (0,0) to satisfy either inequality demonstrates that it does not conform to the system's criteria. When dealing with systems of inequalities, it is essential to carefully assess each component to accurately determine the solution set.

No, (0,0) is not a solution to the given system.

The system consists of two inequalities: y ≥ x^2 + x - 4 and y > x^2 + 2x + 1. To check whether (0,0) is a solution, substitute x = 0 and y = 0 into both inequalities. For the first inequality, x^2 + x - 4 becomes -4, and 0 is not greater than or equal to -4. Therefore, (0,0) does not satisfy the first inequality. Moving on to the second inequality, x^2 + 2x + 1 becomes 1, and 0 is not greater than 1. Hence, (0,0) fails to satisfy the second inequality as well. As a result, (0,0) is not a solution to the system of inequalities.

In summary, by substituting the coordinates of (0,0) into both inequalities, we find that the point does not meet the conditions set by either inequality. Therefore, (0,0) is not a solution to the system. This conclusion is based on the specific values obtained through substitution, demonstrating that the coordinates do not satisfy the given inequalities.

Final Answer:

The inverse function is [tex]\( f^{-1}(x) = x^2 + 5 \)[/tex] and the range of [tex]\( f^{-1} \)[/tex] is [tex]\( y \geq 5 \)[/tex] or [5, ∞].

Explanation:

To find the inverse function, [tex]\( f^{-1} \)[/tex], for the function [tex]\( f(x) = \sqrt{x-5} \)[/tex], we'll need to follow these steps:

1. Write the function as an equation: [tex]\( y = \sqrt{x-5} \)[/tex].
2. To find the inverse, we exchange the roles of x and y. The equation now reads [tex]\( x = \sqrt{y-5} \)[/tex].
3. Our next task is to solve this equation for y. To do so, we need to eliminate the square root by squaring both sides of the equation:
[tex]\[ x^2 = (\sqrt{y-5})^2 \\\\\[ x^2 = y - 5 \][/tex]
Now, we add 5 to both sides in order to isolate y:
[tex]\[ y = x^2 + 5 \][/tex]
This is our inverse function: [tex]\( f^{-1}(x) = x^2 + 5 \)[/tex].

Regarding the range of [tex]\( f^{-1} \)[/tex], we need to consider the domain of the original function f(x). The original function [tex]\( f(x) = \sqrt{x-5} \)[/tex] is only defined for [tex]\( x \geq 5 \)[/tex], because you cannot take the square root of a negative number in real numbers.

Since the domain of f(x) becomes the range of [tex]\( f^{-1}(x) \)[/tex], the range of the inverse function must be [tex]\( y \geq 5 \)[/tex], because the smallest value of x is 5, which when inputted into the inverse gives us [tex]\( 5^2 + 5 = 25 + 5 = 30 \)[/tex], and it only grows larger for larger values of x.

So the inverse function is [tex]\( f^{-1}(x) = x^2 + 5 \)[/tex] and the range of [tex]\( f^{-1} \)[/tex] is [tex]\( y \geq 5 \)[/tex].