The mystery number is a three-digit number. Its digits are 3, 6, and
2.
• The 6 has a digit on its left and on its right.
• The 2 is not the last digit.
What is the mystery number?​

The ratio of the depth of the liquid to the depth of the cup is 4/5.

Geometry and Ratio

The formula for the volume of a cone is:

\(V=\frac{1}{3}\pi r^{2}h\)

where r is the radius of the base of the cone and h is the height of the cone.

When the conical cup is filled with water, there will be a difference in the radius of the water and the cup and a difference in the height of the water and the height of the cup (can be seen in the picture).

From the difference in radius and height of the water with the conical cup, a ratio can be made of similarity rule of triangle (see picture).

If the radius of water is \(r_{1}\) and the height of water is \(h_{1}\), radius of cup is \(r_{2}\) and the height of cup is \(h_{2}\), so:

\(\frac{r_{1} }{r_{2} } =\frac{h_{1} }{h_{2} }\)

\(r_{1}\) with \(r_{2}\) and  \(h_{1}\) with  \(h_{2}\) have the same certain ratio. Let's assume the ratio is x, then it can be written:

\(r_{2}=x.r_{1} \\h_{2}=x.h_{1}\)

The ratio of the volume of water to the volume of the conical cup is 64/125, meaning that 64 of the water in the conical cup with a volume of 125.

From here we can substitute it in the volume ratio:

\(\frac{V_{1} }{V_{2}}=\frac{\frac{1}{3}\pi r_{1}^{2}h_{1} }{\frac{1}{3} \pi r_{2}^{2}h_{2}} =\frac{64}{125} \\\frac{V_{1} }{V_{2}}=\frac{\frac{1}{3}\pi r_{1}^{2}h_{1} }{\frac{1}{3} \pi (xr_{1})^{2}(x.h_{1})} =\frac{64}{125} \\\frac{V_{1} }{V_{2}}=\frac{\frac{1}{3}\pi r_{1}^{2}h_{1} }{\frac{1}{3}\pi r_{1}^{2}h_{1}x^{3} } =\frac{64}{125} \\\frac{V_{1} }{V_{2}}=\frac{1}{x^{3}}=\frac{64}{125}\\ \frac{1}{x^{3}}=\frac{64}{125}\\\\x^{3}=\frac{125}{64} \\x=\frac{5}{4}\)

The ratio between the radius and height of the cup and the water is 5/4. This means that the ratio between the radius and height of the water and the cup is 4/5.

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Answer:

The equation of the line in standard form is -8x + y = 20

Step-by-step explanation:

The standard form of the linear equation is Ax + By = C, where

The slope-intercept form of the linear equation is y = m x + b, where

Parallel lines have equal slopes

∵ The line is parallel to the line y = 8x - 2

→ Compare it with the slope-intercept form above

∴ m = 8

∵ The parallel lines have the same slopes

∴ The slope of the line = 8

→ Substitute it in the slope-intercept form above

∴ y = 8x + b

∵ The line passes through the point (-2, 4)

∴ x = -2 and y = 4

→ Substitute x by -2 and y by 4 in the equation to find b

∵ 4 = 8(-2) + b

∴ 4 = -16 + b

→ Add 16 to both sides

∴ 4 + 16 = -16 + 16 + b

∴ 20 = b

→ Substitute the value of b in the equation above

∴ y = 8x + 20

→ Subtract both sides by 8x

∵ y - 8x = 8x - 8x + 20

∴ y - 8x = 20

→ Switch x and y

∴ -8x + y = 20

∴ The equation of the line in standard form is -8x + y = 20

Answer:

Answer is A

Step-by-step explanation:

\(m = y2 - y1 \div x2 - x1 \\ y2 = m(x2 - x1) + y1\)

Answer:

Step-by-step explanation:

If -3/2 and 5 are solutions to the quadratic equation, hence the expression will give;

(x-(-3/2))(x-5) = 0

(x+3/2)(x-5) = 0

Compare the resulting expression with (ax+b)(x+c)=0, we can see that;

ax = x

divide both sides by x

ax/x = x/x

a = 1

b = 3/2

-5 = +c

c = -5

Hence a = 1,  = 3/2 and c = -5

Congruent quadrilateral related parts are congruent.

The true statements are:

Statements 2, 4, and 6 are Not Congruent

Statements 1, 3, and 5 are Congruent

From the given figure:

Quadrilaterals 1, 2 and 4 are congruent

Quadrilateral 3 is not congruent to any of the other quadrilaterals.

The true statements are:

Quadrilateral 1 and quadrilateral 2 Congruent

Quadrilateral 1 and quadrilateral 3 Not Congruent

Quadrilateral 1 and quadrilateral 4 Congruent

Quadrilateral 2 and quadrilateral 3 Not Congruent

Quadrilateral 2 and quadrilateral 4 Congruent

Quadrilateral 3 and quadrilateral 4 Not Congruent

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Answer:

-17 9/10

Step-by-step explanation:

Answer:

When the input value (x) is 0, the output value (y) is also 0. In this exponential function, any value raised to the power of 0 equals 1, and 2.3 raised to the power of 0 equals 1, so 2.3x (0) = 2.3 * 1 = 2.3, and the output value is 0.

Answer:

→100ml.

Step-by-step explanation:

If 1 litre = 1000ml, then

\(\small\sf{\quad\dfrac{2}{10} \: of \: 500 \: ml}\)

\(: \implies\small\sf{\dfrac{2}{10} \: of \: 500 \: ml}\)

\(: \implies\small\sf{\dfrac{2}{10} \times 500 \: ml}\)

\(: \implies\small\sf{\dfrac{2 \times 500 \: ml}{10}}\)

\(: \implies\small\sf{\dfrac{1000 \: ml}{10}}\)

\(: \implies\small\sf{\cancel{\dfrac{1000 \: ml}{10}}}\)

\(: \implies\small{\sf{100 \:ml}}\)

∴ The answer is 100 ml.

Answer: sorry i have not learnd that yet

Step-by-step explanation:

Answer:I need the answer please

Step-by-step explanation:

Answer:

1. 9/10

2. 15/16

4. 3/5

5. 3/5

Step-by-step explanation:

1. 7/10 + 2/10 = 9/10

2. 13/16 + 2/16 = 15/16

4. 7/15 + 2/15 = 9/15 = 3/5

5. 9/20 + 3/20 = 12/20 = 3/5

Answer: 1 r1 for second

Step-by-step explanation: 13/16+2/16=15/16 divide 15/16 gets 1 r1

We can change t = 4 intο the fοllοwing mathematical functiοn tο get the car's wοrth after 4 years V(4) = 18,000(0.75)⁴ and V(4) = 5,695.31.

A mathematical fοrmula, rule, οr law describes the relatiοnship between twο independent variables and οne dependent variable (the dependent variable). The creatiοn οf physical cοnnectiοns relies οn functiοns in bοth mathematics and physics.

Tο mοdel the value οf the car after t years, we can use the fοllοwing expοnential decay functiοn:

\(V(t) = V0(1 - r)^t\)

where V0 is the initial value οf the car (in this case, $18,000), r is the annual depreciatiοn rate (in this case, 25% οr 0.25), and t is the number οf years since the car was purchased.

Substituting the values, we get:

\(V(t) = 18,000(1 - 0.25)^t\)

\(V(t) = 18,000(0.75)^t\)

Tο find the value οf the car after 4 years, we can substitute t = 4 intο the equatiοn:

V(4) = 18,000(0.75)⁴

V(4) = 5,695.31.

Therefοre, the value οf the car after 4 years is $5,695.31.

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Answer:

b

Step-by-step explanation:

you need to round 2.51 to 3 because it was the correct answer

The present age of Kevin is 6 years.

Using the provided data, we can create two equations that specify the ages of Kevin and Daniel.

Let Kevin's present age be k and Daniel's present age be d.

As per the data given:

Kevin is 3 years older than Daniel. This can be written as:

k = d + 3

Two years ago, Kevin was 4 times as old as Daniel

Two years ago, Kevin was k - 2 years old, and Daniel was d - 2 years old.

k - 2 = 4(d - 2)

Now we have two independent equations, and we can solve for our two unknowns.

Solving our first equation for d.
We get:

d = k - 3

Substituting this into our second equation, we get the equation:

k - 2 = 4((k - 3) -2)

k - 2 = 4k - 12 - 8

k - 2 = 4k - 20

4k - k = 20 - 2

3k = 18

k = 6

Therefore the answer is 6 years.

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If we sample the function x(t) = cos(75t) at the Nyquist frequency, the resulting discrete frequency would be half of the Nyquist frequency, which is equal to half of the highest frequency component in the continuous signal.

In this case, the highest frequency component in x(t) is 75 Hz, as determined by the coefficient of t in the cosine function. According to the Nyquist-Shannon sampling theorem, to accurately represent a signal, the sampling frequency must be at least twice the highest frequency component. Therefore, the Nyquist frequency in this scenario would be 2 * 75 Hz = 150 Hz.

Since we are sampling at the Nyquist frequency, the resulting discrete frequency would be half of the Nyquist frequency, which is 150 Hz / 2 = 75 Hz. Hence, when sampling x(t) at the Nyquist frequency, the resulting discrete frequency would be 75 Hz.

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Get y alone on one side first in order to graph a linear inequality with two variables, like x and y. Then, take a look at the equation that results when you replace the inequality sign with an equality sign. This equation's graph is a line. A dashed line should be drawn if the inequality is strict ( < or >).

The inequality symbols  ‘<’, ‘>’, ‘≤’ or ‘≥’ can be used to compare any two values in an expression that is a linear inequality. These values may take the form of numbers, algebraic expressions, or a combination of the two. For instance, numerical inequalities like 10<11 and 20>17 and algebraic inequalities like x>y, y<19-x, x ≥ z > 11 are examples (also called literal inequalities). 

An ordinary linear function's graph can be seen when we plot the graph for inequalities. The graph, however, is the region of the coordinate plane that satisfies the inequality for an inequality and a line for a linear function, respectively.

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Answer:

Which part of the story's plot structure does this sentence illustrate? I think its B. Climax

The range of scores between the upper and lower quartiles of a distribution is called the interquartile range. The median is the score that divides a distribution into two equal halves, while quartiles divide a distribution into quarters.

The range of scores between the upper and lower quartiles of a distribution is called the interquartile range. The interquartile range (IQR) is the difference between the 75th percentile (upper quartile) and the 25th percentile (lower quartile). It is used to measure the spread of the middle 50% of the data, providing a sense of the distribution's variability. Percentiles are a way of dividing a distribution into hundredths, often used to describe a student's performance relative to their peers.
Quartiles are three values ​​that divide the statistical data into four parts, each containing the same observation. A quarter is a type of quantity. First quartile: Also called Q1 or lower quartile. Second quartile: Also called Q2 or median. Third quarter: Also called Q3 or upper quarter.

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(D) the floor of a classroom is a perfect example of a line segment.

Therefore, (D) the floor of a classroom is a perfect example of a line segment.

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The correct form of the question is given below;
Which is an example of a line segment?

(A) the edge of a book

(B) the corner of a box

(C) a beam of light

(D) the floor of a classroom

I. Simplify the following fully:

1.1 (2x-1)(x^(2)-3x+1)

= 2x^(3) - 6x^(2) + 2x - x^(2) + 3x - 1

= 2x^(3) - 7x^(2) + 5x - 1

1.2 12(2x+3)(5-x)

= 12(-2x^(2) + 7x + 15)

= -24x^(2) + 84x + 180

1.3 (x^(3)-1)/((x+2)+x(x+2))+(x-1)/(2x+4)

= (x^(3)-1)/(x^(2)+3x+2)+(x-1)/(2(x+2))

= (x-1)(x^(2)+x+1)/(x+2)(x+1)+(x-1)/(2(x+2))

= (2(x-1)(x^(2)+x+1)+(x-1)(x+1))/(2(x+2)(x+1))

= (2x^(3)+x^(2)-x-1+x^(2)-1)/(2(x+2)(x+1))

= (2x^(3)+2x^(2)-x-2)/(2(x+2)(x+1))

1.4 (3)/(x^(2)-9)+(2)/((x-3)^(2))

= (3)/((x+3)(x-3))+(2)/((x-3)^(2))

= (3(x-3)+2(x+3))/(x+3)(x-3)^(2)

= (3x-9+2x+6)/(x+3)(x-3)^(2)

= (5x-3)/(x+3)(x-3)^(2)

2. Factorise:

2.1 6x^(2)+7x-20

= (2x-4)(3x+5)

2.2 x^(3)+x^(2)-x-1

= (x+1)(x^(2)-1)

= (x+1)(x+1)(x-1)

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(a) The probability that it loses money on a single policy is 0.025

(b) The probability that the company makes a profit on these policies is 3.75%

Here we have given that that an insurance company found that 2.5% of male drivers between the ages of 18 and 25 are involved in serious accidents annually.

Now for a single policy as we are given the probability of accident here is calculated as,

=> 25/1000 = 0.025

Where as suppose that the company writes 1,000 such policies to a collection of drivers, then the probability that the company makes a profit on these policies is calculated as,

=> 2000 - [65,000 x 25/1000]/100

=> [2000 - 1625]/100

=> 375/100

=> 3.75%

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Answer:

Carlos=1.5041

Mykala=2.6991

William=4.1350

Emily=4.1773

The function f (x) is discontinuous at x = 2 because there is a vertical asymptote there. Therefore, the y-values of f as x approaches 2 from the right will approach negative infinity.

) The value of f (-10), f (-100), f (-1000000) are f (-10) = 0.1818181818,f (-100) = 0.1818181818, and f (-1000000) = 0.1818181818.b) The values of f (2.01), f (2.0001), and f (2.000001) are f (2.01) = -197.5099502, f (2.0001) = -19999.5000333, and f (2.000001) = -1999999.50000017.    c) As x approaches negative infinity, the denominator of the expression approaches negative infinity and the numerator approaches negative infinity. The fraction is approaching zero from the negative direction, which is a horizontal asymptote.    d) As x approaches 2 from the right side, the denominator approaches zero, and the numerator approaches -4. The fraction is approaching negative infinity from the right direction, so there is a vertical asymptote at x = 2. The function f (x) is discontinuous at x = 2 because there is a vertical asymptote there. Therefore, the y-values of f as x approaches 2 from the right will approach negative infinity.Answer:The values of f (-10), f (-100), and f (-1000000) are f (-10) = 0.1818181818, f (-100) = 0.1818181818, and f (-1000000) = 0.1818181818.The values of f (2.01), f (2.0001), and f (2.000001) are f (2.01) = -197.5099502, f (2.0001) = -19999.5000333, and f (2.000001) = -1999999.50000017.As x approaches negative infinity, the denominator of the expression approaches negative infinity and the numerator approaches negative infinity. The fraction is approaching zero from the negative direction, which is a horizontal asymptote.As x approaches 2 from the right side, the denominator approaches zero, and the numerator approaches -4. The fraction is approaching negative infinity from the right direction, so there is a vertical asymptote at x = 2. The function f (x) is discontinuous at x = 2 because there is a vertical asymptote there. Therefore, the y-values of f as x approaches 2 from the right will approach negative infinity.

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Answer:

4

Step-by-step explanation:

Answer:

2 and 3

Step-by-step explanation:

consider perfect cubes either side of 23

8 < 23 < 27 , then

\(\sqrt[3]{8}\) < \(\sqrt[3]{23}\) < \(\sqrt[3]{27}\) , that is

2 < \(\sqrt[3]{23}\) < 3

Heyy there! I am AncientEnigma29 and I will be answering your question step by step

\( \sf \: f(x) = ln(x)\)

Replace f(x) with y

\( \sf \: y = ln(x)\)

Swap x and y with each other

\( \tt \: x = ln(y)\)

Solve for y...

\( \bf \: ln(y) = log e (y)\)

Rule of log used: \( \boxed{ \rm \:c = log_{a}b \: then \: {a}^{c} = b }\)

By the rule we have, x=ln(y) is equal to y=e^x

Replace y with f^-1 (x)

\( \mathcal{ {f}^{ - 1}(x) = {e}^{x} }\)

To receive a score on this portion of the final assessment, students should form groups with 1 to 3 members.

The question specifies that groups of 4 members or larger will receive a zero score on this portion of the final assessment. This requirement is set by the Committee on the Status of Endang.

The purpose of this restriction may be to encourage collaboration and ensure fair evaluation by limiting the group size to a manageable number. By restricting group sizes to 1-3 members, it promotes individual and small group participation, allowing each student to actively contribute to the assessment.

The Committee on the Status of Endang likely established this rule to maintain the integrity of the assessment process and prevent potential issues that may arise from larger groups, such as unequal distribution of work, lack of participation, or excessive collaboration. By setting a maximum group size, the committee aims to ensure fairness and maintain the academic standards of the assessment.

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Answer: 3 ^12

Step-by-step explanation: PLEASE TRUST ME I AM DOING THE DIAGNOSTIC

Answer:

3^12

Step-by-step explanation:

I got it right

Answer:

If you mean by density then the density of hydrogen is approximately 20 times the density of helium.

The type of transformation in this problem is given as follows:

Vertical translation.

The four translation rules are defined as follows:

For this problem, we have a translation of 2 units up, which is called a vertical translation.

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The required Matlab code to find the greatest common divisor of a number using the Euclidean algorithm is shown.

To find the greatest common divisor (GCD) of two numbers using the Euclidean algorithm in MATLAB, you can use the following code:

% Replace '12345678' with your actual student number

studentNumber = '12345678';

% Extract the first 4 digits as the first number

firstNumber = str2double(studentNumber(1:4));

% Extract the last 3 digits as the second number

secondNumber = str2double(studentNumber(end-2:end));

% Find the GCD using the Euclidean algorithm

gcdValue = gcd(firstNumber, secondNumber);

% Display the result

disp(['The GCD of ' num2str(firstNumber) ' and ' num2str(secondNumber) ' is ' num2str(gcdValue) '.']);

Make sure to replace '12345678' with your actual student number. The code extracts the first 4 digits as the first number and the last 3 digits as the second number using string indexing. Then, the gcd function in MATLAB is used to calculate the GCD of the two numbers. Finally, the result is displayed using the disp function.

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Answer:

B is the correct answer.

Step-by-step explanation:

Pi=3.14

So, 7+3.14=10.14

So yes, B is your correct answer.

Mark brainliest please!