Plz help me solve it step by step (find the the area)

21. In the figure given below, AC is parallel to DE. Find the valuesof xy and z and hence find the 2DBE.

algol13

Answer:

X= 50°

Y= 70°

Z= 30°

BDE= 30°

2BDE= 60°

Step-by-step explanation:

(2x -70 )+z+(2x+20)=180...(sum of angle on a straight line)

2x -70 = BDE... alternate angles

Y + (2x-70)+(50+x-20) = 180...(sum of angles in a triangle)

X-20 = z ... alternate and opposite angles

(2x -70 )+z+(2x-+20)=180

2x-70 + x-20 +2x +20= 180

5x -70= 180

5x = 250

X= 50°

X-20 = z

50-20= z

30° = z

2x -70 = BDE

2(50) -70 = BDE

100-70 = BDE

30°= BDE

Y + (2x-70)+(50+x-20)

Y + 100-70 +50 +50 -20 = 180

Y + 200-90=180

Y= 70°

2BDE = 2*30

2BDE= 60°

8 0

3 years ago

What is 5x5x3x5x3x2 simplified using exponents

8090 [49]

Answer: 5 to the 3rd times 3 to the second times 2

Step-by-step explanation:

8 0

4 years ago

Can you help me solve systems of equations by elimination step by step 6x+y=15 -7x-2y=-10

Anna71 [15]

Answer: (4, -9)

Step-by-step explanation:

Use elimination method. Manipulate one (or both) equations to eliminate one of the variables and solve for the remaining variable. I will be eliminating y

6x + y = 15 → 2(6x + y = 15) → 12x + 2y = 30

-7x - 2y = -10 → 1(-7x + 2y = -10) → -7x - 2y = -10

5x = 20

x = 4

Next, replace "x" with "4" into either equation and solve for y.

6(4) + y = 15

24 + y = 15

y = -9

Check:

Plug in x = 4 and y = -9 into the other equation to verify it makes a true statement.

-7x - 2y = -10

-7(4) - 2(-9) = -10

-28 - -18 = -10

-28 + 18 = -10

-10 = -10 $\checkmark$

6 0

4 years ago

Express the system as AX = B; then solve using matrix inverses

Wittaler [7]

Answer with explanation:

1. The given equations are

3x -5 y=2

-x+2 y= 0

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{cc}3&-5\\-1&2\end{array}\right] ,\\\\ X=\left[\begin{array}{c}x&y\end{array}\right],\\\\B=\left[\begin{array}{c}2&0\end{array}\right]$

$\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B$

Adj.A=Transpose of cofactor of Matrix A

$Adj.A=\left[\begin{array}{cc}2&1\\5&3\end{array}\right] ,\\\\ |A|=6-5\\\\|A|=1\\\\\left[\begin{array}{c}x&y\end{array}\right]=\left[\begin{array}{cc}2&5\\1&3\end{array}\right] \times \left[\begin{array}{c}2&0\end{array}\right]\\\\x=4, y=2$

2.

The given equations are

x+y-z=2

x+z=7

2 x +y+z=13

⇒The matrix in the form of , AX=B, is

$A=\left[\begin{array}{ccc}1&1&-1\\1&0&1\\2&1&1\end{array}\right]\\\\ X=\left[\begin{array}{ccc}x\\y\\z\end{array}\right]\\\\B= \left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\\rightarrow X=A^{-1}B\\\\\rightarrow X=\frac{Adj.A}{|A|}\times B\\\\a_{11}=-1,a_{12}=1,a_{13}=1,a_{21}=-2,a_{22}=3,a_{23}=1,a_{31}=1,a_{32}=-2,a_{33}=-1\\\\|A|=1\times(0-1)-1\times(1-2)-1\times(1-0)\\\\=-1+1-1\\\\|A|=-1\\\\Adj.A=\left[\begin{array}{ccc}-1&-2&1\\1&3&-2\\1&1&-1\end{array}\right]$

$\frac{Adj.A}{|A|}=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\\\\X=A^{-1}B\\\\\left[\begin{array}{ccc}x\\y\\z\end{array}\right]=\left[\begin{array}{ccc}1&2&-1\\-1&-3&2\\-1&-1&1\end{array}\right]\times\left[\begin{array}{ccc}2\\7\\13\end{array}\right]\\\\x=3,y=3,z=4$

7 0

3 years ago

What ratios are equivalent to 4:3

Brums [2.3K]

Answer:

here are a few that are equivalent

8:6

28:21

48:36

68:51

4 0

3 years ago

Read 2 more answers