Given: \overline{AB} \parallel \overline{DC}AB∥DCand \overline{AB} \cong \overline{DC}.AB≅DC. Prove: \triangle ABE \cong \triangle CDE△ABE≅△CDE.

45

step-by-step explanation:

9.55 degrees.

step-by-step explanation:

the given angle is

1 cycle is equal to 2π radians. so 4π will be equal to two cycles.

5.166667π-4π = 1.166667π

hence, this angle goes a little more than negative x axis.

1.166667π-π= 0.166667π

hence, the reference angle is 0.166667π radians.

1 radian = 57.2958 degrees

now, in degrees, this should be equals to 9.55 degrees.

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

  -3         -3  

y=-4x+5

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15