Help me find the area please .

I need help asap pls

sp2606 [1]

9514 1404 393

Answer:

a) ∠DAE = 33°; ∠ABD = 57°

b) ∠CEB = 90°

c) ∠ABE = 22°

d) ∠ADE = 15°

Step-by-step explanation:

The diagonals of a rhombus meet at right angles. Each bisects the corner angles at its ends. Adjacent angles are supplementary, opposite angles are congruent, and each diagonal creates two isosceles triangles.

a) ∠DAE = 90° - ∠ADE = 90° -57°

∠DAE = 33°

∠ADE = ∠ABD = 57°

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b) ∠CEB = 90° . . . . . the diagonals meet at right angles

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c) ∠ABE = 44°/2

∠ABE = 22°

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d) ∠ADE = 90° -(1/2)∠DAB = 90° -150°/2

∠ADE = 15°

8 0

3 years ago

Given g(x)=-x+4g(x)=−x+4, find g(5)g(5) whats the answer?

Bas_tet [7]

Answer:

g(x) = -4g(x) = -x+4

= g(5) = -4(5) = -(5)+4

= g(5) = -20 = -1

5 0

2 years ago

Read 2 more answers

If Jay had 5 cards and Jill had 10 cards. What is the product?

cupoosta [38]

Hello there,

The word "product" mean's to multiply.

So we are going to do 5 x 10.

This will give us 50 because when we do
10+10+10+10+10 is give's us 50. That is
what multiplication mean's.

Hope this helps

Your correct answer will be 50 card's

~Jurgen

5 0

3 years ago

I WILL GIVE BRAINLIEST JUST ANSWER (DUE TODAY AT 12:00am)

Alexus [3.1K]

Look up flash cards they helped me cheat on my test

4 0

3 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago