MATHEMATICS MIDDLE SCHOOL

ABC has vertices A(-1,6), B(2,10) and C(7,-2). Find the measure of each angle of the triangle. Round decimal to the nearest tenth

Answers

Answer 1

Answer:

Angle A = 98.1°

Angle B = 59.5°

Angle C = 22.4°

Step-by-step explanation:

When given vertices (x1,y1) ,(x2,y2), we would be using the formula:

√(x2 - x1)² + (y2 - y1)²

ABC has vertices A(-1,6), B(2,10) and C(7,-2)

Side AB = A(-1,6), B(2,10)

√(x2 - x1)² + (y2 - y1)²

= √(2-(-1))² + (10 - 6)²

= √(3² + 4²)

= √9+16

= √25

= 5

Side BC = B(2,10), C(7,-2)

√(x2 - x1)² + (y2 - y1)²

√(7-2)² + (-2 - 10)²

= √5² +(-12)²

= √(25 + 144)

= √169

= 13

Side AC = A(-1,6), C(7,-2)

√(x2 - x1)² + (y2 - y1)²

√(7-(-1))² + (-2 - 6)²

= √6² +(-8)²

= √(64 + 64)

= √128

= 11.314

To find the measure of each angle we would be using the cosine rule

Where

a² = b² + c² - 2bc × Cos A

b² = a² + c² - 2ac × Cos B

c² = a² + b² - 2ab × Cos C

Sketching a triangle we find out that

Side AB = c = 5

Side AC = b = 11.314

Side BC = a = 13

Hence

1) for Side BC = a

a² = b² + c² - 2bc × Cos A

Cos A = b² + c² - a²/ 2bc

Cos A = 11.314² + 5² -13²/(2 × 11.313 × 5)

A = arc cos[(11.314² + 5² -13²)/(2 × 11.313 × 5)]

A = 98.13°

To the nearest tenth = 98.1°

2) for Side AC = b

b² = a² + c² - 2ac × Cos B

Cos B = a² + c² - b²/ 2ac

Cos B = 13² + 5² - 11.314/2 × 13 × 5

B = arc cos[ (13² + 5² - 11.314)/(2 × 13 × 5)]

B = 59.49°

To the nearest tenth = 59.5°

3) for Side AB = c

c² = a² + b² - 2ab × Cos C

Cos C = a² + b² - c²/ 2ab

Cos C = 13² + 11.314² - 5²/ 2 × 13 × 11.314

C = arc cos[(13² + 11.314² - 5²)/ (2 × 13 × 11.314)]

C = 22.38°

To the nearest tenth = 22.4°

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explain how u got the answer

Answers

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Answers

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Answers

The number is = 10x+y
if reverse the digits, the number is: 10y+x
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I suggest this system of equations:
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The number is: 10x+y=10(1)+4=14

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Answers

A boat travels 6 miles in 24 minutes. What is the average speed of the boat?

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Help please! this is due tomorrow so please help out! will mark as brainiest!!

Answers

Answer:

$\large\boxed{\bold{Q1}\ -(128)/(9)}\n\boxed{\bold{Q2}\ 2^(25)}$

Step-by-step explanation:

$\bold{Q1}\n\n(-2(3^2\cdot3^3)^2)/(3^(12)\cdot2^(-6))\qquad\text{use}\ a^n\cdot a^m=a^(n+m)\n\n=-(2(3^(2+3))^2)/(3^(12)\cdot2^(-6))=-(2(3^5)^2)/(3^(12)\cdot2^(-6))\qquad\text{use}\ (a^n)^m=a^(nm)\n\n=-(2\cdot3^(5\cdot2))/(3^(12)\cdot2^(-6))=-(2\cdot3^(10))/(3^(12)\cdot2^(-6))\qquad\text{use}\ (a^n)/(a^m)=a^(n-m)\n\n=-2^(1-(-6))\cdot3^(10-12)=-2^(7)\cdot3^(-2)\qquad\text{use}\ a^(-n)=(1)/(a^n)\n\n=-2^7\cdot(1)/(3^2)=-(2^7)/(3^2)=-(128)/(9)$

$\bold{Q2}\n\n(4^2)/(4^(-7))\cdot(2^4\cdot2^3)\cdot(3^(-2))^0\qquad\text{use}\ (a^n)/(a^m)=a^(n-m),\ a^n\cdot a^m=a^(n+m),\ a^0=1\n\n=4^(2-(-7))\cdot2^(4+3)\cdot1=4^(9)\cdot2^7=(2^2)^(9)\cdot2^7\qquad\text{use}\ (a^n)^m=a^(nm)\n\n=2^(2\cdot9)\cdot2^7=2^(18)\cdot2^7\qquad\text{use}\ a^n\cdot a^m=a^(n+m)\n\n=2^(18+7)=2^(25)$

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Answers

You can solve this problem in multiple ways, but you'll have to divide the total volume with the amount per servings to find the number of servings (makes sense logically if you think about it).

The first way is to just use the fractions given. This will usually be your method in handling this if you're not familiar with it.

Convert 6 3/4 into an improper fraction. To do this, we need to put the 6 into the numerator somehow. We can approach this by multiplying the whole number (6) with the denominator (4) and adding it to (3). Basically, 6 x 4 + 3 = 27. This will be our new numerator. If you are not familiar with this concept, please let me know.

Our new fraction for the total amount of juice is $(27)/(4)$ . Divide this with the amount per serving, $(3)/(8)$

$(27)/(4) / (3)/(8)$
We flip the second fraction over and multiply it.

$(27)/(4) * (8)/(3) = (9)/(1) * (2)/(1) = 18 servings$ (I crossed out common numbers here, unfortunately I don't know how to do that here so let me know if my explanation was not clear enough. I skipped a few steps.)

18 servings would be your answer.

Or you can convert 6 3/4 into decimal, which would be 6.75. 3/8 in decimal is .375 and you can do long division from there. That way is a bit more tedious in my opinion, but sometimes more convenient.

Cheers!

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