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2 years ago

Which equation is correct for ∆FON?

Mathematics

1 answer:

lbvjy [14]2 years ago

7 0

the picture in the attached figure

we know that

in the right triangle FON

sin F=opposite side angle F/hypotenuse

in this problem

opposite side angle F=ON--------> 9 units

Hypotenuse=FN--------> 18 units

sin F=ON /FN-------> sin F=9 /18

therefore

the answer is the option

B.) sin F 9/18

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Find the distance between the points given. (-1, -1) and (1, 3) √5 √(17) 2√5

Pani-rosa [81]

Answer:

2√5

Step-by-step explanation:

d = √(x2 - x1)² + (y2 - y1)²

= √[1 - (-1)]² + [3 - (-1)]

= √(2)² + (4)²

= √(4) + (16)

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2 years ago

Given the first three terms of a linear sequence: 3x – 2;x+9; 2x + 5 Determine the value of x.

Yuliya22 [10]

Answer:

x = 5

Step-by-step explanation:

The difference between consecutive terms will be equal , then

a₂ - a₁ = a₃ - a₂ , that is

x + 9 - (3x - 2) = 2x + 5 - (x + 9) ← distribute parenthesis on both sides

x + 9 - 3x + 2 = 2x + 5 - x - 9 , simplify both sides

- 2x + 11 = x - 4 ( subtract x from both sides )

- 3x + 11 = - 4 ( subtract 11 from both sides )

- 3x = - 15 ( divide both sides by - 3 )

x = 5

5 0

2 years ago

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What is the mean of variable a piz help me i have one more question

timama [110]

I think the mean of variable a is 4

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2 years ago

Find the inverse of the given matrix, if it exists.Aequals=left bracket Start 3 By 3 Matrix 1st Row 1st Column 1 2nd Column 0 3

BabaBlast [244]

Answer:

$A^{-1}=\left[ \begin{array}{ccc} \frac{1}{9} & \frac{4}{27} & - \frac{2}{27} \\\\ \frac{8}{9} & \frac{5}{27} & \frac{11}{27} \\\\ - \frac{4}{9} & \frac{2}{27} & - \frac{1}{27} \end{array} \right]$

Step-by-step explanation:

We want to find the inverse of $A=\left[ \begin{array}{ccc} 1 & 0 & -2 \\\\ 4 & 1 & 3 \\\\ -4 & 2 & 3 \end{array} \right]$

To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 4&1&3&0&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Subtract row 1 multiplied by 4 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ -4&2&3&0&0&1\end{array}\right]$

Add row 1 multiplied by 4 to row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&2&-5&4&0&1\end{array}\right]$

Subtract row 2 multiplied by 2 from row 3

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&-27&12&-2&1\end{array}\right]$

Divide row 3 by −27

$\left[ \begin{array}{ccc|ccc}1&0&-2&1&0&0 \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Add row 3 multiplied by 2 to row 1

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&11&-4&1&0 \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

Subtract row 3 multiplied by 11 from row 2

$\left[ \begin{array}{ccc|ccc}1&0&0&\frac{1}{9}&\frac{4}{27}&- \frac{2}{27} \\\\ 0&1&0&\frac{8}{9}&\frac{5}{27}&\frac{11}{27} \\\\ 0&0&1&- \frac{4}{9}&\frac{2}{27}&- \frac{1}{27}\end{array}\right]$

As can be seen, we have obtained the identity matrix to the left. So, we are done.

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2 years ago

Number 14 please and thank you

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You can do a notebook a door , a frame or a mirror, mostly anything that has angles

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2 years ago