What is the area of AABC with ZA = 20, zB = 55 and c = 114 cm (round up to 2 decimal places)?

  

dy
Find an expression for y given
dx
= 7×5
Find the area of AABC whose c = 6.5cm, b = 13cm and ZA = 100 (Round up your
answer to 2 decimal places).
Find the area of AABC whose ZA = 20, zB = 55 and c = 14.14 cm (Round up your
answer to 2 decimal places).

Introduction:
In the field of mathematics, solving equations and geometric problems has always been an important aspect. One such equation is finding an expression for y given dx = 7×5. On the other hand, calculating the area of a triangle is one of the basic geometric problems that students come across. Two questions have been presented below related to finding the area of a triangle AABC with different parameters.

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Description:
The first question presented is an equation that requires finding the value of y when dx equals 7 times x to the power of 5. This equation falls under the category of solving linear equations where one variable is unknown. The method of isolating the unknown variable and solving the equation is used to find the result.

The second and third questions presented require calculating the area of a triangle AABC using different parameters such as side lengths and angle measures. The formula to calculate the area of a triangle is 1/2 times base times height. In both questions, either base or height is unknown, and using the given measurements, the formula is applied to find the area. In the second question, the height of the triangle is unknown, whereas in the third question, the base length is unknown. Using the given angle measures and side lengths, the missing parameters are calculated and substituted in the formula to get the area of the triangle.

Objectives:

– To derive an expression for y given dx = 7x^5
– To calculate the area of a triangle (AABC) given the length of two sides and an angle or the length of three sides

Learning Outcomes:

After completing this content, learners should be able to:

– Use calculus to find the equation for y in terms of x given dx = 7x^5
– Apply trigonometric formulas to find the area of a triangle given its side lengths and angle measures
– Round up answers to two decimal places to ensure proper accuracy in numerical calculations

Header 1: Deriving Equation for y

– Define basic calculus terminology and principles
– Apply calculus principles to solve for an equation in terms of given variables
– Demonstrate an understanding of how to check work for accuracy.

Header 2: Finding the Area of AABC

– Define and apply the sine formula for finding the area of a triangle
– Identify and explain the different parameters involved in calculating the area of a triangle
– Round answers to two decimal places to ensure proper accuracy in numerical calculations.

Solution 1:

To find an expression for y given dx = 7x^5, we can start by taking the derivative of both sides with respect to x. Using the power rule, we get:

dy/dx = 35x^4

Now, we can integrate both sides with respect to x to get an expression for y. Integrating the left side gives us:

y = ∫ dy = ∫ 35x^4dx

y = 35/5 x^5 + C

where C is the constant of integration. Therefore, the expression for y is:

y = 7x^5 + C

Solution 2:

To find the area of AABC whose c = 6.5cm, b = 13cm and ZA = 100, we can use the formula for the area of a triangle given two sides and the included angle:

A = (1/2) bc sin(A)

where A is the included angle between sides b and c. We have c = 6.5cm, b = 13cm, and ZA = 100, so we can use the sine formula to find sin(A):

sin(A) = sin(100) = 0.9848

Therefore, the area of AABC is:

A = (1/2)(6.5)(13)(0.9848) = 49.11 cm^2

To find the area of AABC whose ZA = 20, zB = 55 and c = 14.14 cm, we can use the formula for the area of a triangle given three sides:

A = √(s(s-a)(s-b)(s-c))

where s is the semi-perimeter of the triangle, given by:

s = (a + b + c)/2

We have ZA = 20, zB = 55, and c = 14.14 cm, so we can use the law of cosines to find side a:

a^2 = c^2 + b^2 – 2bc cos(A)

cos(A) = (b^2 + c^2 – a^2)/(2bc) = (55^2 + 14.14^2 – 20^2)/(2*55*14.14) = 0.8848

A = cos^-1(0.8848) = 29.16

Therefore, s = (20 + 55 + 14.14)/2 = 44.57, and the area of AABC is:

A = √(44.57(44.57-14.14)(44.57-55)(44.57-20)) = 295.79 cm^2 (rounded up to 2 decimal places)

Suggested Resources/Books:
1. “Calculus: Early Transcendentals” by James Stewart
2. “Geometry Essentials For Dummies” by Mark Ryan
3. “Introduction to Trigonometry” by Geoffrey Boothroyd

Similar Asked Questions:
1. What is the formula for finding the area of a triangle?
2. How do you solve for a missing angle in a triangle?
3. What is the Pythagorean Theorem?
4. How do you find the perimeter of a triangle?
5. What is the relationship between the sides and angles of a right triangle?dy
Find an expression for y given
dx
= 7×5
Find the area of AABC whose c = 6.5cm, b = 13cm and ZA = 100 (Round up your
answer to 2 decimal places).
Find the area of AABC whose ZA = 20, zB = 55 and c = 14.14 cm (Round up your
answer to 2 decimal places).

Introduction:
In the field of mathematics, solving equations and geometric problems has always been an important aspect. One such equation is finding an expression for y given dx = 7×5. On the other hand, calculating the area of a triangle is one of the basic geometric problems that students come across. Two questions have been presented below related to finding the area of a triangle AABC with different parameters.

Description:
The first question presented is an equation that requires finding the value of y when dx equals 7 times x to the power of 5. This equation falls under the category of solving linear equations where one variable is unknown. The method of isolating the unknown variable and solving the equation is used to find the result.

The second and third questions presented require calculating the area of a triangle AABC using different parameters such as side lengths and angle measures. The formula to calculate the area of a triangle is 1/2 times base times height. In both questions, either base or height is unknown, and using the given measurements, the formula is applied to find the area. In the second question, the height of the triangle is unknown, whereas in the third question, the base length is unknown. Using the given angle measures and side lengths, the missing parameters are calculated and substituted in the formula to get the area of the triangle.

Objectives:

– To derive an expression for y given dx = 7x^5
– To calculate the area of a triangle (AABC) given the length of two sides and an angle or the length of three sides

Learning Outcomes:

After completing this content, learners should be able to:

– Use calculus to find the equation for y in terms of x given dx = 7x^5
– Apply trigonometric formulas to find the area of a triangle given its side lengths and angle measures
– Round up answers to two decimal places to ensure proper accuracy in numerical calculations

Header 1: Deriving Equation for y

– Define basic calculus terminology and principles
– Apply calculus principles to solve for an equation in terms of given variables
– Demonstrate an understanding of how to check work for accuracy.

Header 2: Finding the Area of AABC

– Define and apply the sine formula for finding the area of a triangle
– Identify and explain the different parameters involved in calculating the area of a triangle
– Round answers to two decimal places to ensure proper accuracy in numerical calculations.

Solution 1:

To find an expression for y given dx = 7x^5, we can start by taking the derivative of both sides with respect to x. Using the power rule, we get:

dy/dx = 35x^4

Now, we can integrate both sides with respect to x to get an expression for y. Integrating the left side gives us:

y = ∫ dy = ∫ 35x^4dx

y = 35/5 x^5 + C

where C is the constant of integration. Therefore, the expression for y is:

y = 7x^5 + C

Solution 2:

To find the area of AABC whose c = 6.5cm, b = 13cm and ZA = 100, we can use the formula for the area of a triangle given two sides and the included angle:

A = (1/2) bc sin(A)

where A is the included angle between sides b and c. We have c = 6.5cm, b = 13cm, and ZA = 100, so we can use the sine formula to find sin(A):

sin(A) = sin(100) = 0.9848

Therefore, the area of AABC is:

A = (1/2)(6.5)(13)(0.9848) = 49.11 cm^2

To find the area of AABC whose ZA = 20, zB = 55 and c = 14.14 cm, we can use the formula for the area of a triangle given three sides:

A = √(s(s-a)(s-b)(s-c))

where s is the semi-perimeter of the triangle, given by:

s = (a + b + c)/2

We have ZA = 20, zB = 55, and c = 14.14 cm, so we can use the law of cosines to find side a:

a^2 = c^2 + b^2 – 2bc cos(A)

cos(A) = (b^2 + c^2 – a^2)/(2bc) = (55^2 + 14.14^2 – 20^2)/(2*55*14.14) = 0.8848

A = cos^-1(0.8848) = 29.16

Therefore, s = (20 + 55 + 14.14)/2 = 44.57, and the area of AABC is:

A = √(44.57(44.57-14.14)(44.57-55)(44.57-20)) = 295.79 cm^2 (rounded up to 2 decimal places)

Suggested Resources/Books:
1. “Calculus: Early Transcendentals” by James Stewart
2. “Geometry Essentials For Dummies” by Mark Ryan
3. “Introduction to Trigonometry” by Geoffrey Boothroyd

Similar Asked Questions:
1. What is the formula for finding the area of a triangle?
2. How do you solve for a missing angle in a triangle?
3. What is the Pythagorean Theorem?
4. How do you find the perimeter of a triangle?
5. What is the relationship between the sides and angles of a right triangle?

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