help what is this 8y to the second power - 3y to the second power

(\mathrm{A}_{4}\) is larger than \(\mathrm{B}_{2}\)

Yes, \(\mathrm{A}_{4}\) is larger than \(\mathrm{B}_{2}\). To show this, we can use the following equation:

\[\mathrm{A}_{4} = a \cdot c \cdot \mathrm{H}\]

and

\[\mathrm{B}_{2} = b_{1} \cdot c_{1}\]

Given that \(a = 34\), \(c = 45\), \(\mathrm{H} = 194\), \(b_{1} = 7\), and \(c_{1} = 3\), then \(\mathrm{A}_{4} = 34 \cdot 45 \cdot 194 = 291930\) and \(\mathrm{B}_{2} = 7 \cdot 3 = 21\). As \(291930 > 21\), \(\mathrm{A}_{4}\) is larger than \(\mathrm{B}_{2}\).

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The water level after 4 weeks will be 29.2 feet.

The water level of the lake will increase by 10% each week during the winter rainstorms. This means that each week, the water level will increase by 0.10 × 20 = <<0.10*20=2>>2 feet.

After the first week, the water level will be 20 + 2 = <<20+2=22>>22 feet.
After the second week, the water level will be 22 + 2 = <<22+2=24>>24 feet.
After the third week, the water level will be 24 + 2 = <<24+2=26>>26 feet.
And so on.

We can use the formula A = P(1 + r)^n to find the water level after n weeks, where A is the final amount, P is the initial amount, r is the rate of increase, and n is the number of weeks.

In this case, P = 20, r = 0.10, and n is the number of weeks.

So, the water level after n weeks will be:

A = 20(1 + 0.10)^n

We can plug in different values of n to find the water level after a certain number of weeks.

For example, to find the water level after 4 weeks, we can plug in n = 4:

A = 20(1 + 0.10)^4 = 29.2 feet

So, the water level after 4 weeks will be 29.2 feet.

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The solution for x is 1/(3^16).

To solve for x, we need to convert the logarithmic equation to exponential form. The general formula for converting a logarithmic equation to an exponential equation is:

log_b(x) = y => b^y = x

In this case, the base is 3, the exponent is -2 - 6 - 8 + (1/9) - (1/9), and x is the value we are trying to find. So, we can write the exponential equation as:

3^(-2 - 6 - 8 + (1/9) - (1/9)) = x

Simplifying the exponent gives us:

3^(-16) = x

Now, we can solve for x by taking the inverse of both sides:

x = 1/(3^16)

Therefore, the solution for x is 1/(3^16).

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Answer:

I think the answer is 57.14

Answer:

57.14

Step-by-step explanation:

57.14 is the result of rounding 57.14 to the nearest 0.01

By assuming and solving with the rectangle measurements, we get the equation, 2x^2 + 6x - 33 = 0.

Follow the steps below to reach the answer:

A region of 11 cm2 divides the two rectangles A and B.

Rectangle A has a length of x cm.

Rectangle B has a length of (x+3) cm.

Form an equation in x and demonstrate that it simplifies when given the fact that rectangle A's width is 2 cm bigger than rectangle B's width:

2x^2 + 6x - 33 = 0

Rectangle A's length and area being x and 11, respectively, results in the following for rectangle A's width: 11 / x

Rectangle B's length is equal to x + 3 and its area is equal to 11, hence its width is as follows: 11 / (x + 3)

Given that width of the rectangle, A is 2 cm Larger than the width of rectangle B, we obtain: 11/x =  (11/ ( x + 3 ) ) + 2

In addition, based on the situation, we get

11/x - 11/ (x+3) = 2 cm

Let this be equation (1)

The main challenge at hand is to change this equation into the quadratic equation's standard form.

It is obtained by multiplying both sides of equation (1) by x*(x+3). As a result, you will get,

11*(x+3) - 11x = 2x*(x+3)

11x + 33 - 11x = 2x^2 + 6x

2x^2 + 6x - 33 = 0.

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Answer:

2

Step-by-step explanation:

The transformation of the equation y = -2 I x - 3 I + 3 from the parent function y = I x I includes a horizontal shift of 3 units to the right, a vertical stretch by a factor of 2, a reflection across the x-axis, and a vertical shift of 3 units up.

The transformation of the equation y = -2 I x - 3 I + 3 from the parent function y = I x I can be described as follows:

The parent function is shifted 3 units to the right. This is indicated by the "-3" inside the absolute value bars in the equation.
The parent function is vertically stretched by a factor of 2. This is indicated by the "-2" in front of the absolute value bars in the equation.
The parent function is reflected across the x-axis. This is indicated by the negative sign in front of the "2" in the equation.
The parent function is shifted 3 units up. This is indicated by the "+3" outside the absolute value bars in the equation.

In summary, the transformation of the equation y = -2 I x - 3 I + 3 from the parent function y = I x I includes a horizontal shift of 3 units to the right, a vertical stretch by a factor of 2, a reflection across the x-axis, and a vertical shift of 3 units up.

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Roland made 16 sales this week, while Vickie made 8 more sales than Roland.

An equation is an expression that shows the relationship between two or more numbers and variables. Equations can either be linear, quadratic, cubic and so on depending on the degree.

Let x represent the number of sales Roland has.

Roland earns a base of $49 plus $15 per sale. HenceL

Total earning by Roland = 15x + 49

Vickie made 8 more sales than Roland. Vickie earns a base of $145 plus $6 per sale.

Total earning by Vickie = 6(x + 8) + 145 = 6x + 193

Roland and Vickie are salespeople who earned the same amount this week, Therefore:

15x + 49 = 6x + 193

9x = 144

x = 16

Roland made 16 sales this week.

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Answer:

2. 29 3. 9:11, [tex]\frac{9}{11}[/tex], 55 4(a) 10.7

Step-by-step explanation:

2. 29

3. 9:11

   [tex]\frac{9}{11}[/tex]

   [tex]\frac{11}{20} *100%\\= 55%[/tex]

4(a)10.7

Answer:  The total number of milkshakes sold is 7 + 18 = 25.

The number of milkshakes sold with whipped cream is 7.

To find the percentage of milkshakes with whipped cream, we can use the formula:

percentage = (part/whole) x 100

Substituting the values, we get:

percentage = (7/25) x 100

percentage = 28

Therefore, 28% of the milkshakes had whipped cream.

Step-by-step explanation:

The minima and minima of the function y=x^4-(32/3)x^3-2x^2+2x in the interval [0,3] are:
Maxima: x=2

Minima: None

The minima and maxima of a function are the lowest and highest points on the function within a given interval. To find these points, we need to take the derivative of the function and set it equal to zero to find the critical points. The critical points are where the function changes direction, and are potential minima or maxima. We can then use a conditional statement to determine if the critical points are within the given interval and if they are minima or maxima.

The derivative of the function is:
y'=4x^3-3(32/3)x^2-4x+2

Setting the derivative equal to zero, we get:
4x^3-32x^2-4x+2=0

We can use the Rational Root Theorem to find the potential rational roots of this equation. The potential rational roots are ±1, ±2, ±1/2, and ±1/4. Using synthetic division, we find that x=2 is a root. This gives us the factor (x-2), and we can use synthetic division again to find the other factors. The factored form of the equation is:
(x-2)(4x^2-12x+1)=0

Using the quadratic formula, we can find the other two roots:
x=3±√(3^2-4(4)(1))/2(4)
x=3±√(9-16)/8
x=3±√(-7)/8
x=3±i√7/8

The only real root is x=2, so this is the only critical point. We can use a conditional statement to determine if this critical point is within the given interval and if it is a minima or maxima. The critical point x=2 is within the interval [0,3], so we need to determine if it is a minima or maxima. We can do this by taking the second derivative of the function and plugging in the critical point:
y''=12x^2-6(32/3)x-4
y''(2)=12(2^2)-6(32/3)(2)-4
y''(2)=48-64-4
y''(2)=-20

Since the second derivative is negative at the critical point, this means that the critical point is a maxima. Therefore, the maxima of the function is at x=2.

In conclusion, the minima and maxima of the function y=x^4-(32/3)x^3-2x^2+2x in the interval [0,3] are:
- Maxima: x=2
- Minima: None

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The perimeter of the inside of the track is approximately 354.36 meters.

What is the circumference?

In geometry, the circumference is the perimeter of a circle or ellipse.

To find the perimeter of the inside of the track, we need to find the length of the inside edge of the track and add it to twice the length of the semicircles at each end.

The inside edge of the track is formed by two parallel lines, each offset by a distance of 56 m from the outer edge. The length of this edge is equal to the length of the long straight edge minus twice the offset distance:

length of inside edge = 130 m - 2(56 m)

length of inside edge = 18 m

The length of each semicircle at each end is equal to half the circumference of a full circle with a radius equal to the width of the track (56 m). Therefore, the length of both semicircles is:

2 x (1/2 x 2π x 56 m) = 2π x 56 m

Adding the length of the inside edge to twice the length of the semicircles, we get:

perimeter of inside track = 18 m + 2π x 56 m

perimeter of inside track = 18 m + 112π m

the perimeter of the inside track ≈ 354.36 m (rounded to two decimal places)

Hence, the perimeter of the inside of the track is approximately 354.36 meters.

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Given:-

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Solution:-

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hope it helps! :)

The temperature at 7 AM is approximately 87 degrees.

The outside temperature over a day can be modeled as a sinusoidal function. In this case, we know that the high temperature of 98 degrees occurs at 6 PM and the average temperature for the day is 85 degrees. To find the temperature at 7 AM, we can use the equation:

T(t) = A sin(B(t - C)) + D

where T(t) is the temperature at time t, A is the amplitude, B is the frequency, C is the horizontal shift, and D is the vertical shift.

We know that the average temperature is 85 degrees, so D = 85. We also know that the high temperature of 98 degrees occurs at 6 PM, so A = (98 - 85)/2 = 6.5 and C = 6.

Now we can plug in the values and solve for the temperature at 7 AM:

T(7) = 6.5 sin(B(7 - 6)) + 85

Since we want to find the temperature to the nearest degree, we can round the answer to the nearest whole number:

T(7) = 6.5 sin(B) + 85 ≈ 87 degrees

Therefore, the temperature at 7 AM is approximately 87 degrees.

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The probability that the product operates is 0.6055.

The probability can be meant as the chance something will be happened or not. The probability that the product operates is the probability that all 47 integrated circuits are not defective. Since the probability that any integrated circuit is defective is 0.01, the probability that it is not defective is 1 - 0.01 = 0.99. Since the integrated circuits are independent, the probability that all 47 are not defective is the product of the probability that each one is not defective. This is given by:

P(product operates) = 0.99⁴⁷ ≈ 0.6055

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Applying Trigonometric ratios, the missing sides, expressed as radicals in simplest form are:

26. u = 3√2; v = 3

27. a = 3/2; b = 3/2

28. x = 7√2; y = 7

29. u = 5√2; v = 5√2

Radicals are mathematical expressions that involve the square root or nth root of a number or a variable.

We can express each of the missing sides in radicals using the appropriate Trigonometry ratio in each case as follows:

26. Use sine ratio to find u:

sin 45 = 3/u

u = 3/sin 45

u = 3 / 1/√2 [sin 45 = 1/√2]

u = 3 * √2/1

u = 3√2

Use tangent ratio to find v:

tan 45 = 3/v

v = 3/tan 45

v = 3/1 [tan 45 = 1]

v = 3

27. Use sine ratio to find a:

sin 45 = a / 3√2

a = 3√2 * sin 45

a = 3√2 * 1/√2 [sin 45 = 1/√2]

a = 3/2

Use cosine ratio to find b:

cos 45 = b / 3√2

b = 3√2 * cos 45

b = 3√2 * 1/√2 [cos 45 = 1/√2]

b = 3/2

28. Use cosine ratio to find x:

cos 45 = 7 / x

x = 7/ cos 45

x = 7 / 1/√2 [cos 45 = 1/√2]

x = 7 * √2/1

x = 7√2

Use tangent ratio to find y:

tan 45 = y/7

y = 7 * tan 45

y = 7 * 1 [tan 45 = 1]

y = 7

29. Use sine ratio to find u:

sin 45 = u/10

u = 10 * sin 45

u = 10 * 1/√2 [sin 45 = 1/√2]

u = 10/√2

Rationalize

u = 5√2

Use cosine ratio to find v:

cos 45 = v/10

v = 10 * cos 45

v = 10 * 1/√2 [sin 45 = 1/√2]

v = 10/√2

Rationalize

v = 5√2

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We use standard units when measuring an object because :

A. Standard units are used because they are relatively permanent.

B. Standard units enable communication over time.

C. Standard units enable communication over distance.

D. Standard units are exact.

We use standard units when measuring an object because they are relatively permanent and enable communication over time, distance, and exactness.

Standard units are used because they are relatively permanent. This means that they do not change over time and are consistent. Standard units enable communication over time. Since they are consistent, they allow for accurate communication of measurements even if they are taken at different times.

Standard units enable communication over distance. They allow for accurate communication of measurements even if the people communicating are in different locations. Standard units are exact. This means that they are precise and accurate, allowing for consistent measurements.

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The answers to each part about the scale factor is given above.

Scale factor is the ratio of final dimensions to the initial dimensions.

Mathematically, we can write -

K = {final dimension/initial dimension}

Given are the dimensions of the drawing as -

Original                        1.5 in.          3 yd

Enlargement               2.5 in.          5 yd

{ 1 } -

We can write the scale factor as -

K = 3/1.5

K = 2

{ 2 } -

Area of the scaled drawing -

A = 2.5 x 5

A = 2.5 x 180

A = 450 square inches

{ 3 } -

To find the area of the actual drawing, multiply the area of the scale drawing by the multiple of the scale factor.

Therefore, the answers to each part about the scale factor is given above.

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Using the lengths of the time spent we can plot the histogram by the values in the table.

A histogram is a visual depiction of a frequency distribution with continuous classes that has been categorised.

To create a histogram, follow the steps shown below.

Mark the class intervals on the X-axis and the frequencies on the Y-axis to get started.

Both axes must have the exact same scales. Intervals between classes must be exclusive. Create rectangles with appropriate frequencies as the heights and bases acting as class intervals.

As the class limits are shown on the horizontal axis and the frequencies are shown on the vertical axis, a rectangle is constructed on each class interval.

If the intervals are identical, the height of each rectangle is proportional to the associated class frequency.

From the given information, we have the following:

Length of time (t)     Frequency

0 ≤ t < 10                    5

10 ≤ t < 25                  24

25 ≤ t < 30                 12

30 ≤ t < 50                 8

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The complete question is:

A \times B \times C = \left[\begin{array}{ll} -40 & -60 \\ -68 & 70 \\ 6 & 0 \end{array}\right]

To find the product of the matrices A, B and C, we can use the following equation:



$$A \times B \times C = \left[\begin{array}{lll} (A \times B)_{11} & (A \times B)_{12} & (A \times B)_{13} \\ (A \times B)_{21} & (A \times B)_{22} & (A \times B)_{23} \\ (A \times B)_{31} & (A \times B)_{32} & (A \times B)_{33} \end{array}\right] \times C = \left[\begin{array}{ll} (A \times B \times C)_{11} & (A \times B \times C)_{12} \\ (A \times B \times C)_{21} & (A \times B \times C)_{22} \\ (A \times B \times C)_{31} & (A \times B \times C)_{32} \end{array}\right]$$



To find each element of the product, we use the following equation:



$$(A \times B \times C)_{ij} = \sum_{k=1}^{3} A_{ik} \times B_{kj} \times C_{ij}$$



Where $i$ and $j$ represent the row and column numbers respectively. For example, to find the element $(A \times B \times C)_{11}$, we have:



$$(A \times B \times C)_{11} = \sum_{k=1}^{3} A_{1k} \times B_{k1} \times C_{11} = (-5 \times -8 \times -4) + (1 \times 7 \times -4) + (-7 \times 5 \times -4) = -40$$



Therefore, the product of A, B and C is:



$$A \times B \times C = \left[\begin{array}{ll} -40 & -60 \\ -68 & 70 \\ 6 & 0 \end{array}\right]$$

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To solve for r, we can divide both sides of the equation by Pt: r = (A-P)/Pt

The formula A=P+Prt is used to calculate the total amount of money (A) after a certain period of time when a principal amount (P) is invested at a certain interest rate (r) for a certain amount of time (t). To solve for the specified variable r, we need to rearrange the formula and isolate r on one side of the equation. Here are the steps to do so:

Step 1: Subtract P from both sides of the equation to get:
A - P = P + Prt - P

Step 2: Simplify the right side of the equation to get:
A - P = Prt

Step 3: Divide both sides of the equation by Pt to get:
(A - P) / Pt = r

Step 4: Simplify the left side of the equation to get:
r = (A - P) / Pt

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The composition of the permutation (3 5 ) three times is (3 5) and the permutation (1 2 4) to the power of 2023 is (1 2 4)

The composition of the permutation (3 5 ) three times is (3 5) and the permutation (1 2 4) to the power of 2023 is (1 2 4).For part (c), we can find the composition of the permutation (3 5) three times by applying the permutation to itself three times. The first time we apply it, we get (5 3), the second time we get (3 5), and the third time we get (5 3) again. Since the permutation (5 3) is the same as the original permutation (3 5), the composition of the permutation (3 5) three times is (3 5).For part (d), we can find the permutation (1 2 4) to the power of 2023 by applying the permutation to itself 2023 times. Since the permutation (1 2 4) is a cycle of length 3, applying it to itself 3 times will give us the identity permutation (1 2 4)(1 2 4)(1 2 4) = (1)(2)(4). Therefore, we can reduce the exponent of 2023 to 2023 mod 3 = 1, and the permutation (1 2 4) to the power of 2023 is the same as the permutation (1 2 4) to the power of 1, which is (1 2 4).

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Hence, a 19-gram sample would contain 2.5909 x 10⁵ germs.

How come we use measure?

We can compare unknown amounts to known values with the use of measurement. We can quantify the size, length, and speed of objects with the use of measurement. The final result won't be accurate without measurement.

We may start by calculating the bacterium to dirt sample weight ratio:

bacteria per gram = 1.5 x 10⁵ / 11 = 13636.36...

We are able to employ this ratio to determine how many bacteria are present in a 19-gram specimen:

bacteria = bacteria per gram x weight of sample

bacteria = 13636.36... x 19

bacteria = 259090.91...

We must write the number in scientific notation because it is more than 10⁵. This can be achieved by adding a factor of 10⁵ and shifting the decimal place five places to a left:

bacteria = 2.5909 x 10^5

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The numerator is \((25x^{2}+15x+9)\) and the denominator is \(2(x+2)\).

To simplify the given rational expression, we need to factor the numerator and denominator and then cancel out any common factors.

First, let's factor the numerator:
\(500x^{3}-108=4(125x^{3}-27)=4(5x-3)(25x^{2}+15x+9)\)

Next, let's factor the denominator:
\(40x^{2}+56x-48=8(5x^{2}+7x-6)=8(5x-3)(x+2)\)

Now, we can cancel out the common factor of \(4(5x-3)\) from the numerator and denominator:
\(\frac{4(5x-3)(25x^{2}+15x+9)}{8(5x-3)(x+2)}=\frac{(25x^{2}+15x+9)}{2(x+2)}\)

Therefore, the simplified rational expression is \(\frac{(25x^{2}+15x+9)}{2(x+2)}\).

In this simplified form, the numerator is \((25x^{2}+15x+9)\) and the denominator is \(2(x+2)\).

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Isabel had 10 kilogram of oranges when she started. She ate 2 kilogram of the oranges, leaving her with 8 kilogram. This means that she has 8/10 kilogram of oranges left.

It is important to understand how to calculate fractions in order to answer questions like this. In this case, the question was asking how much of a kilogram remains after Isabel ate 2 kilogram. To answer this, we had to figure out what fraction of the 10 kilogram Isabel ate, which was 2/10. We then subtracted this fraction from 1 in order to figure out how much of a kilogram remains, which was 8/10.

This kind of calculation is useful when trying to figure out how much of something remains after a certain amount has been taken away. It can be used in a wide variety of contexts, such as when trying to figure out how much of a product or item remains after some has been sold or used. Understanding how to calculate fractions can be very helpful in these kinds of scenarios.

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If we consider the number line with rational number marked on it, then the number of parts to be present between 1 and 2 and -1 and -2 will be one part each.

A number line is a pictorial representation or drawing of numbers in which there are equal intervals between each number and is used to represent the real numbers on it. Real numbers are those numbers which may be positive or negative integers, rational numbers or irrational numbers. In general, a quantity which can be expressed as an infinite decimal expansion is called as a real number.

If we draw a number line and mark the points as 1, 2, 3,...,∞ and -∞,..., -3, -2, -1 on the right and left hand side of 0, then the intervals between each real number is equal to 1. Hence the number of parts between 1 and 2 is one part and -1 and -2 is one part. However, these parts may be subdivided into more parts to get more fine value (smaller value).

This value will be a fractional value between 1 and 2 or -1 and -2. Hence parts between two numbers on a number line can be infinity, but for sake of simplicity, one is taken in general case.

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Answer:

x ∈ { ( 6 + [tex]\sqrt{23}[/tex] * i) , { ( 6 - [tex]\sqrt{23}[/tex] * i) }

Step-by-step explanation:

Given: [tex]x^2-12x+59=0[/tex]

First, collect like terms. [tex]x^2[/tex] and -12x, 59 and 0:

[tex]x^2-12x=0-59[/tex]

+59 is changed to -59 when transferred. Then subtract:

( x - 12 ) x = -59

Finally divide both sides by -59:

x ∈ { ( 6 + [tex]\sqrt{23}[/tex] * i) , { ( 6 - [tex]\sqrt{23}[/tex] * i) }

Answer:

Step-by-step explanation:

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The distance across the lake is 120 feet.

Right angled triangle are those triangle for which one of the angle is 90 degrees.

Given is a right angled triangle.

Pythagoras theorem states that,

Square of the hypotenuse of a right angled triangle is equal to the sum of the squares of the two legs of the triangle.

The distance across the lake is one of the leg of the right triangle.

Length of hypotenuse = 208 feet + 104 feet

                                     = 312 feet

Length of other leg = 192 feet + 96 feet

                                = 288 feet

Distance across the lake = √(312² - 288²)

                                         = √14,400

                                         = 120 feet

Hence 120 feet is the distance across the lake.

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Answer:The father's position relative to the surface of the water is 4.2 meters below the surface of the water.

Step-by-step explanation:

What is an expression?

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

The height relative to the water surface is calculated as:-

Height from iceberg to water surface = 0.5 meters

The height from iceberg to deep inside the water is = 4.7 meters

Father's height relative to the water surface is,

H= 4.7 - 0.5 = 4.2 meters.

The image of the condition is also attached with the answer below.

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