Luke is going to invest $7,600 and leave it in an account for 5 years. Assuming the
interest is compounded daily, what interest rate, to the nearest tenth of a percent,
would be required in order for Luke to end up with $8,900?

Answer:

The expression -7x^0 cannot be a part of a rational expression because any non-zero number raised to the power of 0 is equal to 1. Therefore, -7x^0 is equal to -7(1) = -7, which is a constant and not a variable expression. In a rational expression, the numerator and denominator should be polynomial expressions with variables.

The other options 4, 3x^2-4rootx, and 3x^2-4x+5 can be part of a rational expression as they are polynomial expressions with variables.

The Surface area of the composite figure is calculated as approximately: 234 sq. cm

Here, we have,

to Find the Surface Area of a Composite Figure:

The surface area of the composite figure is the area surrounding the faces of the solid as a whole. Therefore, we have:

Surface area (SA) = Surface area of the square prism + surface area of the square pyramid - 2(area of base)

Area of base = area of square = 6 * 6 = 36 sq. cm.

Surface area of the square prism = 2a² + 4ah

a = 6 cm

h = 4 cm

Plug in the values:

Surface area of the square prism = 2(6²) + 4*6*4

= 72 + 96

= 168 sq. cm.

Surface area of the square pyramid = 2bs + b²

b = side length = 6 cm

s = slant height = √(8² + 3²) = 8.5 cm

Plug in the values:

Surface area of the square pyramid = 2 * 6 * 8.5 + 6² = 138 sq. cm.

Surface area of the composite figure = 168 + 138 - 2(36) = 234 sq. cm

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complete question:

Find the surface area of each composite figure. Use 3.14 for π. Round to the nearest tenth. 12m. 6cm. 6cm. 4cm.​

Answer:

a. The probability that all 5 selected transistors are defective is 0.0024 or 0.24%.

To calculate this probability, we use the formula for the probability of the intersection of independent events:

P(all 5 defective) = P(defective) x P(defective) x P(defective) x P(defective) x P(defective)

where P(defective) = 5/12 for each transistor.

Plugging in the values, we get:

P(all 5 defective) = (5/12) x (5/12) x (5/12) x (5/12) x (5/12) = 0.0024

b. The probability that none of the selected transistors are defective is 0.163 or 16.3%.

To calculate this probability, we use the complement rule:

P(none defective) = 1 - P(at least one defective)

To calculate P(at least one defective), we can use the complement rule again:

P(at least one defective) = 1 - P(none defective)

So, we need to find P(none defective) first:

P(none defective) = P(not defective) x P(not defective) x P(not defective) x P(not defective) x P(not defective)

where P(not defective) = 7/12 for each transistor.

Plugging in the values, we get:

P(none defective) = (7/12) x (7/12) x (7/12) x (7/12) x (7/12) = 0.163

Then, we can find P(at least one defective):

P(at least one defective) = 1 - P(none defective) = 1 - 0.163 = 0.837

Finally, we can find P(none are defective) using the complement rule:

P(none defective) = 1 - P(at least one defective) = 1 - 0.837 = 0.163.

The inequality 2|x - 3| + 6 > 10 has the solution set:

x < 1 or x > 5

To solve the inequality 2|x - 3| + 6 > 10, we need to isolate the absolute value expression |x - 3|.

We can do this by subtracting 6 from both sides of the inequality:

2|x - 3| > 4

Then we can divide both sides by 2:

|x - 3| > 2

Now we need to consider two cases, one where x - 3 is positive, and another where it is negative:

When x - 3 is positive, the inequality |x - 3| > 2 becomes:

x - 3 > 2

Solving for x gives:

x > 5

When x - 3 is negative, the inequality |x - 3| > 2 becomes:

-(x - 3) > 2

Solving for x gives:

x < 1

Therefore, the solution set for the inequality 2|x - 3| + 6 > 10 is:

x < 1 or x > 5

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The complete question:

Match each absolute value inequality on the left side with its solution set on the right

2|x - 3| + 6 > 10 - 2

[tex]x=\textit{lbs of aluminum at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{lbs of aluminum at 54\%}\\\\ ~~~~~~ 54\%~of~y\implies \cfrac{54}{100}(y)\implies 0.54 (y) \\\\\\ \textit{48 lbs of aluminum at 52\%}\\\\ ~~~~~~ 52\%~of~48\implies \cfrac{52}{100}(48)\implies 0.52 (48)\implies 24.96 \\\\[-0.35em] ~\dotfill[/tex]

[tex]\begin{array}{lcccl} &\stackrel{lbs}{quantity}&\stackrel{\textit{\% of lbs that is}}{\textit{AL only}}&\stackrel{\textit{lbs of}}{\textit{AL only}}\\ \cline{2-4}&\\ \textit{1st alloy}&x&0.3&0.3x\\ \textit{2nd alloy}&y&0.54&0.54y\\ \cline{2-4}&\\ mixture&48&0.52&24.96 \end{array}~\hfill \begin{cases} x + y = 48\\\\ 0.3x+0.54y=24.96 \end{cases} \\\\[-0.35em] ~\dotfill[/tex]

[tex]\stackrel{\textit{using the 1st equation}}{x+y=48\implies y=48-x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation from above}}{0.3x~~ + ~~0.54(48-x)~~ = ~~24.96}\implies 0.3x+25.92-0.54x=24.96 \\\\\\ 0.30x-0.54x=-0.96\implies -0.24x=-0.96\implies x=\cfrac{-0.96}{-0.24} \\\\\\ \boxed{x=4}\hspace{9em}y=48-x\implies \boxed{y=44}[/tex]

The probability of randomly choosing two boys is 0.174

If the selection is done at random, then the probability of choosing a boy is equal to the quotient between the number of boys and the total number of people.

There are 12 boys and 16 + 12 = 28 students in total, then the probability is:

P = 12/28

Now, for the second selection the probability is computed in the same way, but now there are 11 boys and 27 students (one was already choosen), then:

P = 11/27

the joint probability (for the two selections) is given by the product between the individual probabilities, we will get:

P = (12/28)*(11/27)

P = 0.174

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18 will be the result of the given problem.

Applying the order of operations (PEMDAS), we first calculate the value inside the parentheses, which is 3 + 6 = 9.

Then we multiply 4 by 9, which gives 36.

Finally, we divide 36 by 2, which gives the final answer of 18. Therefore:

4 x (3 + 6) ÷ 2 = 18

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Part A: The equation of the circle in standard form is x² + (y - 7)² = 2².

Part B: Center: (h, k) = (0, 7), Radius: 2

Part C: Please check the attachment.

In this problem we find the definition of the equation of a circle in general form, whose standard form must be found:

(x - h)² + (y - k)² = r²

Where:

This can be done by completing the square. First, write the equation of the circle:

x² + y² = 14 · y - 45

Second, complete the square by algebra properties:

x² + y² - 14 · y + 45 = 0

x² + (y² - 14 · y + 49) = 4

x² + (y - 7)² = 2² (PART A)

Third, write the coordinates of the center and the radius of the circle:

Center: (h, k) = (0, 7), Radius: 2 (PART B)

Finally, we proceed to graph the resulting expression.

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Answer:

Set your calculator to Degree mode.

10² = 6.78² + 4² - 2(6.78)(4)(cos x)

100 = 61.9684 - 54.24(cos x)

38.0316 = -54.24(cos x)

cos x = -.701173

[tex]x = {cos}^{ - 1} ( - .701173)[/tex]

x = 134.52°

Answer:

15

Step-by-step explanation:

I'm going to be very honest with you; one of my kids had this problem recently and I remember the answer is 15.

99=15+14+13+12+11+10+9+8+7

So the largest number is 15.

The way we solved it:

Call the 1st number x. Each number after x is x+(a number) all the way up to x+8 (the 9th number in the series of 9 consecutive whole numbers):

  x, x + 1, x + 2, x + 3, x + 4, x + 5, x + 6, x + 7 and x + 8

Add up those consecutive numbers (I know, it's kind of messy):

             x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 + x + 7 + x + 8

Combine the like terms:

           9x + 36 = 99

Subtract 36 from both sides:

            9x = 63

Divide by 9 on both sides:

            x = 7

So the largest number will be: x + 8

            7+8 = 15

And then check that it works:

99=15+14+13+12+11+10+9+8+7

This shaded region represents the solution to the system of inequalities.

We have,

The inequalities:

y ≤ -1 and x ≥ 1

To graph y ≤ -1, we first draw the horizontal line y = -1.

Since we want y to be less than or equal to -1, we shade the area below the line.

To graph x ≥ 1, we draw the vertical line x = 1.

Since we want x to be greater than or equal to 1, we shade the area to the right of the line.

The shaded regions of both inequalities overlap in the area to the right of x = 1 and below y = -1.

Thus,

This shaded region represents the solution to the system of inequalities.

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Comparing the median of each box-and-whisker plot, the type of transportation that is LEAST likely to take more than 30 minutes is: d. train.

In order to determine the transportation that is LEAST likely to take more than 30 minutes, we have to compare the median of each data set represented on the box-and-whisker plot for each transportation.

The box-and-whisker plot that has the lowest median would definitely represent the the transportation that is LEAST likely to take more than 30 minutes, since median represents the typical minutes or center of the data.

Therefore, from the box-and-whisker plots given, the one for train has the lowest median. Therefore train would LEAST likely take more than 30 minutes.

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The function of the graph is f(x) = x²-3x -10

A quadratic function is a function in which the highest power of it's variable is 2. quadratic equation is given in the form f(x) = ax²+ bx +c

To form the quadratic function from a graph, we need to find the roots of the function from the graph. The root is the point where the curve intercept the x axis.

Therefore the roots are -2 and 5.

f(x) = x² -( z+y)x +zy

where z and y are the roots

z+y = -2 +5 = 3

zy = -2 × 5 = -10

f(x) = x²-3x -10

therefore the function of the graph is f(x) = x²-3x -10

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We can simplify the square roots:

√2 = √(2 × 1) = √2 × √1

√18 = √(9 × 2) = √9 × √2 = 3√2

√50 = √(25 × 2) = √25 × √2 = 5√2

Now we can find the mean:

(mean) = (sum of numbers) / (number of numbers)

(mean) = (√2 + 3√2 + 5√2) / 3

(mean) = (9√2) / 3

(mean) = 3√2

Therefore, the mean of the numbers √2, √18, and √50 is 3√2.

We can use the exchange rate of 1 Euro = R18.32. Therefore,400 € = 400 x R18.32 = R7,328.

1.1.1 According to the table, the weakest currency is the yen with 14 yen being equivalent to only R0.14.1.1.2 R292 318 460 in words is Two hundred and ninety-two million, three hundred and eighteen thousand, four hundred and sixty Rand. 1.1.3

To determine the amount in rand that needs to be spent on ventilators in Spain, we need to find out the total amount of money raised, then use the ratio provided (1:2:4) to determine the amount to be spent on ventilators.

The ratio indicates that for every 1 unit of money, 2 units are to be spent on staff payment and 4 units on purchasing ventilators.

Therefore, if the total amount of money raised is €14 million, which is equivalent to R292 313 460, then the amount to be spent on ventilators is:4/(1+2+4) × 292 313 460 = R175 388 076.1.1.4 To find out how much 400€ would be in rand,

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Cameron and Rebecca are organizing a lunch for their friends. Within the dishes to be served, they will prepare a tomato, carrot and broccoli salad. The number of tomatoes and carrots are 10-partners and the total amount of vegetables is 12. If we increase the number of broccoli by three and decrease the number of tomatoes by two, the amount of tomatoes and broccoli will also be 10-partners. What is the number of tomatoes, carrots and broccoli bought by Cameron and Rebecca?

Cameron and Rebecca purchased 7 tomatoes, 3 carrots and 2 broccoli.

Number of tomatoes = x

Number of carrots = 10 - x

Number of broccoli = y

System of Equations:

x + 10 - x + y = 12

y = 12 - 10

y = 2

x - 2 + y + 3 = 10 (If we increase the number of broccoli by three and decrease the number of tomatoes by two, the amount of tomatoes and broccoli will also be 10-partners)

x + 1 + y = 10

x + y = 9

y = 9 - x

Substitution of y:

x + 10 - x + y = 12

x + 10 - x + 9 - x = 12

-x + 19 = 12

-x = 12 - 19

x = 7 ⇒ 10 - x = 3

Cameron and Rebecca purchased 7 tomatoes, 3 carrots and 2 broccoli.

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For the 4-hour infusion, 1500ml of the 20% acetylcysteine solution must be made up and combined with 500ml of glucose.

To calculate the volume of acetylcysteine that needs to be added to the glucose, we can use the following formula:

The volume of acetylcysteine (in ml) = (Total amount of acetylcysteine needed in mg / Concentration of acetylcysteine in mg/ml) x 100

We have been asked to add 3000mg of acetylcysteine to 500 ml of glucose. If the concentration of the ampoule containing acetylcysteine is 20%, this means that there are 20g of acetylcysteine in 100 ml of solution, or 200mg/ml.

Using the formula above, we can calculate the volume of acetylcysteine that needs to be added as follows:

Volume of acetylcysteine = (3000mg / 200mg/ml) x 100 = 1500ml

Therefore, 1500ml of the 20% acetylcysteine solution needs to be drawn up and added to the 500ml of glucose to make the 4-hour infusion.

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Answer: £234

Step-by-step explanation:

First, we need to subtract the deposit (£350) from the total cost of the trip (£1754):

£1754 - £350 = £1404

Then we need to divide the remaining amount of money (£1404) by the number of months Janice will be on holiday (6 months):

£1404 ÷ 6 months = £234 each month

Hope this helps!

Answer:6

Step-by-step explanation:

The simplified expression of [tex]\sqrt[4]{567(x^9)(y^{11})}[/tex] is [tex]3x^2y^2\sqrt[4]{7xy^3}[/tex].

Given expression:

[tex]\sqrt[4]{567(x^9)(y^{11})}[/tex].

We can simplify the expression using the property that says for any non-negative numbers a and b, and any integer n ≠ 0,

[tex](a\times b)^{n} = a^n \times {b^n}[/tex]

Using this property, we can simplify the expression as follows:

[tex](567x^9y^{11})^{1/4} = (567)^{1/4} \times (x^9)^{1/4} \times (y^{11})^{1/4}[/tex]

Now we need to simplify,

[tex](81\times 7)^{1/4}x^{9/4}y^{11/4}\\=(81)^{1/4}(7)^{1/4}(x^{\frac{8}{4}+\frac{1}{4})(y^{\frac{8}{4}+\frac{3}{4})[/tex]

Also,

[tex]{3^4}^{1/4}(7)^{1/4}(x^{2+\frac{1}{4}})(y^{2+\frac{3}{4}})\\=3^1 7^{1/4}x^2x^{1/4}y^2y^{1/4}\\=3x^2y^2(7^{1/4}x^{1/4}y^{3/4})\\=3x^2y^2(7xy^3)^{1/4}\\=3x^2y^2\sqrt[4]{7xy^3}[/tex]

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The end behavior of the quadratic function is f(x) = (x + 5)² - 2

We have the table of values which represents a quadratic function f(x).

A quadratic equation is represented as;

f(x) = a(x - h)² + k

Where, Vertex = (h, k)

If we plot the graph according to the given table, the vertex will be,

(h, k) = (-5, -2)

Substitute, (h, k) = (-5, -2) in f(x) = a(x - h)² + k

we have, f(x) = a(x + 5)² - 2

Also, the graph, we have the point (0, 23)

This means that

a(0 + 5)² - 2 = 23

25a = 25

a = 1

Substitute a = 1 in f(x) = a(x + 5)² - 2

f(x) = (x + 5)² - 2

Hence, the equation is polynomial f(x) = (x + 5)² - 2

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The area of triangle UVW to the nearest square inch is approximately 2 square inches.

The area of triangle UVW can use the Heron's formula:

Area = √(s(s-a)(s-b)(s-c))

s is the semiperimeter of the triangle and a, b and c are the lengths of its sides.

First, we can calculate the semiperimeter s:

s = (u + v + w)/2

s = (5.5 + 4.4 + 9.6)/2

s = 9.75

The length of each of the three sides of the triangle:

a = v = 4.4 inches

b = u = 5.5 inches

c = w = 9.6 inches

Then, we can plug these values into the Heron's formula and simplify:

Area = √(s(s-a)(s-b)(s-c))

Area = √(9.75(9.75-4.4)(9.75-5.5)(9.75-9.6))

Area = √(9.75(5.35)(4.25)(0.15))

Area ≈ √(3.296)

Area ≈ 1.81

Rounding this to the nearest square inch, we get:

Area ≈ 2 square inches

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The value of inverse of function is,

⇒ f ⁻¹ (x) = x² / 36

We have to given that;

Function is,

⇒ f (x) = 6√x

Now, We can find the inverse of the function is,

⇒ f (x) = 6√x

⇒ y = 6√x

Solve for x;

⇒ y/6 = √x

⇒ (y/6)² = x

⇒ x = y² / 36

⇒ f ⁻¹ (x) = x² / 36

Thus, The value of inverse of function is,

⇒ f ⁻¹ (x) = x² / 36

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After simplification,

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

Given,

(x/4)+(1/8)/(x^2/4)

Simplifying,

Firstly take LCM of 4 and 8,

LCM of 4 and 8 will be 8.

Then,

(2x +1) /(8) / x^2 /4

Further,

Take reciprocal of the denominator,

(2x +1) /(8) (4/x^2)

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

Hence after simplification the expression becomes (2x + 1)/2(x^2).

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right isosceles triangle.

Right Scalene Triangle

Angles

Acute angle:  An angle whose measure is less than 90 degrees.

Right angle:  An angle whose measure IS 90 degrees.  As a consequence, the two rays that form the angle are perpendicular.

Obtuse angle:  An angle whose measure is greater than 90 degrees.

Triangles (defined by side lengths)

Equilateral Triangle:  A triangle with all three sides congruent (all three sides have the same length)

Isosceles Triangle:  A triangle that contains two congruent sides (two sides with the same length).

Scalene Triangle: A triangle with three different length sides.

Triangles (defined by angle measures)

Acute Triangle:  A triangle whose angles are all acute angles

Right Triangle:  A triangle that contains one right angle.

Obtuse Triangle:  A triangle that contains one obtuse angle.

By angle

Our triangle has an angle that appears to be a right angle (the two sides look perpendicular).  This would be a Right Triangle (when classified by angles).

By side length

All three sides of this triangle are different in length.  This would be a Scalene Triangle (when classified by side length).

Combining the attributes:

This is both a Right Triangle and a Scalene Triangle.  Together, it is a Right Scalene Triangle.

Right Isosceles

To be a Right Isosceles Triangle, the triangle needs to be both a Right Triangle, and an Isosceles Triangle.

A Right triangle contains one right angle, and an Isosceles Triangle contains a pair of sides that are the same length.

While this triangle does contain a right angle, all three sides appear to be different length (scalene).

This is not a Right Isosceles Triangle.

Acute Isosceles

To be an Acute Isosceles Triangle, the triangle needs to be both an Acute Triangle, and an Isosceles Triangle.

An Acute Triangle contains all angles that are acute angles (less than 90 degrees), and an Isosceles Triangle contains a pair of sides that are the same length.

However, this triangle contains a right angle (equal to 90 degrees, not less than), so it cannot be an Acute Triangle, and this triangle does not contain a pair of sides that are the same length, so it cannot be an Isosceles Triangle.

This is not an Acute Isosceles Triangle.

Acute Scalene

To be an Acute Scalene Triangle, the triangle needs to be both an Acute Triangle, and a Scalene Triangle.

An Acute Triangle contains all angles that are acute angles (less than 90 degrees), and a Scalene Triangle has sides that are all different length.

While this triangle does have sides that are all different length, it contains a right angle, whose measure IS 90 degrees (not less than 90 degrees).  So it cannot be an Acute Triangle

This is not an Acute Scalene Triangle.

Answer:

Answer is -2/21r + (-19)

Step-by-step explanation:

(-1/7r - 9 -2/3r) - (-5/7r + 10)

= -1/7r - 9 - 2/3r + 5/7r - 10  (distributing the negative sign)

= (-1/7r + 5/7r) + (-2/3r) - 9 - 10  (grouping the like terms)

= 4/7r - 2/3r - 19  (combining the like terms)

= (12r - 14r)/21r - 38r/21r  (getting a common denominator)

= -2r/21r - 38r/21r

= (-2r - 38r)/21r

= -40r/21r

= -40/21

Therefore, the simplified expression is B. -2/21r + (-19).

C. The total profit, if all shares for both stocks are sold at the close of business on day 7, is $587.62

D. The best day to sell TechCo shares would be day 7, as it had the highest closing price. For HealthCo, day 7 would also be the best day to sell.

E. There is no difference in profit or loss compared to part C.

Part 2

A. Diversity can be evaluated based on the portfolio in question

B. By analyzing the types of assets present in the portfolio, it is possible to determine where potential risks may originate.

C. The individual carrying a particular portfolio can be characterized based on their risk tolerance, financial goals, and stage of life.

Part 1:

A. TechCo (TCO) and HealthCo (HCO) have their shares priced at $100 and $50 respectively.

Name of Stock Symbol Quantity of Shares Purchased Purchase Price per Share Total Purchase Value

TechCo TCO 75 $100 $7,500

HealthCo HCO 150 $50 $7,500

B. Here is an illustration of the potential ending values:

Name of Stock Closing Price Day 1 Closing Price Day 2 Closing Price Day 3 Closing Price Day 4 Closing Price Day 5 Closing Price Day 6 Closing Price Day 7

TechCo $101 $99 $100 $98 $102 $101 $103

HealthCo $51 $50 $52 $49 $51 $52 $54

C. If we sell all shares on day 7, we can calculate the profit/loss and the brokerage fee:

TechCo: 75 shares * $103/share = $7,725

HealthCo: 150 shares * $54/share = $8,100

Total = $7,725 + $8,100 = $15,825

Brokerage fee = $15,825 * 1.5% = $237.38

Profit = $15,825 - $15,000 (initial investment) - $237.38 (brokerage) = $587.62

D. The best day to sell TechCo shares would be day 7, as it had the highest closing price. For HealthCo, day 7 would also be the best day to sell.

E. Since we've already assumed selling on the best day, there is no difference in profit or loss compared to part C.

Part 2:

A. Diversity can be evaluated based on the portfolio in question. A portfolio that encompasses investments in a variety of sectors, asset classes, and regions would exhibit a high level of diversity.

B. By analyzing the types of assets present in the portfolio, it is possible to determine where potential risks may originate.

An investment portfolio that primarily consists of stocks or other assets with high risks would be considered as an aggressive portfolio. A portfolio that primarily consists of bonds or low-risk assets is considered to be a conservative one.

A balance between the two would be a moderate portfolio.

C. The individual carrying a particular portfolio can be characterized based on their risk tolerance, financial goals, and stage of life.

For example, a young professional with a high-risk tolerance might carry an aggressive portfolio to maximize growth, while a retired individual might have a conservative portfolio to preserve their capital.

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To find the volume of the figure, we use the formula: (2cm x 8cm x 7cm) + (1cm x 6cm x 8cm). The correct answer is (B)

To find the volume of a joined shape, you have to solve for the volume of each component of the shape and sum them together.

In this case, we have 2 shapes joined and both are cuboids.

Volume of a cuboid = length x breath x height

Volume₁ = 8 x 7 x 2 = 2 x 8 x 7

Volume₂ = 8 x 1 x 6 = 1 x 6 x 8

Total volume = (2cm x 8cm x 7cm) + (1cm x 6cm x 8cm)

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