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2 years ago

What is the area of this triangle?

Mathematics

1 answer:

Diano4ka-milaya [45]2 years ago

6 0

ANSWER

56.1 square inches.

EXPLANATION

The area of this triangle can be calculated using the formula,

$Area = \frac{1}{2} ab \sin( C)$

where a=8in.

b=14.2 in.

and C=99° is the included angle.

We plug in these values to obtain,

$Area = \frac{1}{2} \times 8 \times 14.2 \sin( 99 \degree)$

$Area = 56.1007$

We round to the nearest tenth to get,

56.1 sq. inch

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Here's what I mean when I write x/2 JUST IN CASE.

а.

Let the total number of pencils be x, then this implies that Maggie received x/2 pencils from Katie. Maggie keeps 5 and give the rest to Jamil, this means

that Jamil receives

x/2-5 pencils.

b. Expression of pencils given to Jamil is: x/2-5

Substitute x=16 in this expression: x/2-5

Simplify and evaluate: = 8-5= 3

Maggie gives 3 pencils to Jamil.

c. Expression of pencils given to Jamil is: x/2-5 to

Equate this expression with 1 to form an equation and solve for x all, the total

number of pencils if Jamil received 1 pencil:

x/2 - 5 = 1

Solve for x:

x = 2(1+ 5)

Evaluate:

x = 2(6) = 12

Katie had 12 pencils if Jamil received 1 pencil.

RESULT

a. Expression of number of pencils given to Jamil is x/2-5, where x is total number of pencils.

b. Maggie gives 3 pencils to Jamil.

c. Katie had 12 pencils if Jamil received 1 pencil.

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2 years ago

Find two unit vectos that are orthogonal to both [0,1,2] and [1,-2,3]

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Answer:

Let the vectors be

a = [0, 1, 2] and

b = [1, -2, 3]

( 1 ) The cross product of a and b (a x b) is the vector that is perpendicular (orthogonal) to a and b.

Let the cross product be another vector c.

To find the cross product (c) of a and b, we have

$\left[\begin{array}{ccc}i&j&k\\0&1&2\\1&-2&3\end{array}\right]$

c = i(3 + 4) - j(0 - 2) + k(0 - 1)

c = 7i + 2j - k

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( 2 ) Convert the orthogonal vector (c) to a unit vector using the formula:

c / | c |

Where | c | = √ (7)² + (2)² + (-1)² = 3√6

Therefore, the unit vector is

$\frac{[7,2,-1]}{3\sqrt{6} }$

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

The other unit vector which is also orthogonal to a and b is calculated by multiplying the first unit vector by -1. The result is as follows:

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

In conclusion, the two unit vectors are;

[ $\frac{7}{3\sqrt{6} }$ , $\frac{2}{3\sqrt{6} }$ , $\frac{-1}{3\sqrt{6} }$ ]

and

[ $\frac{-7}{3\sqrt{6} }$ , $\frac{-2}{3\sqrt{6} }$ , $\frac{1}{3\sqrt{6} }$ ]

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