All categories

2 years ago

PLEASE HELP ASAP

Mathematics

1 answer:

Delicious77 [7]2 years ago

5 0

Answer: EH = 12

Step-by-step explanation:

CD : GH = 2 : 3 ⇒ AD : EH = 2 : 3

AD = 8

AD : EH

2 : 3

×4 ×4

8 : 12

You might be interested in

A bag contains balls in a radio of 3 5 4 if there are twelve balls how many balls are in the bag!!

Elena-2011 [213]

Twelve balls are in the bag

4 0

3 years ago

Read 2 more answers

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.

6 0

3 years ago

After running 14 laps the runner had compleated 25% of the race. How many laps does he still have to compleate?

miss Akunina [59]

Answer:

42 laps

Step-by-step explanation:

7 0

2 years ago

2. Find the area of the region bounded by the graphs of y = -x2 + 3x + 4 and y = 4. (5 points)

Kipish [7]

Answer:

4.5 sq. units

Step-by-step explanation:

area = -⅓x³+3/2 x² ] 0 to 3

= -⅓(3)³+ 3/2 (3)² -0

= -9 + 27/2

= -9 + 13.5

= 4.5 sq. units

8 0

3 years ago

Write two ratios that are equivalent to 3/11

julsineya [31]

The answer is 3:11 and 3 to 11

5 0

3 years ago