If isosceles triangle ABC has a 130° angle at vertex B, which statement must be true? O mZA= 15° and mzC= 35
O mZA+ mZB = 155°
O mZA+ mzC = 60
OMZA= 20° and m2C=30

The true statement about the given angles of the triangle is m∠A + m∠B = 155⁰.

The sum of angles of Isosceles triangle must add up to 180 degrees.

Let the base angles = x and x

x + x + 130 = 180 (sum of angles in a triangle)

2x + 130 = 180

2x = 180 - 130

2x = 50

x = 25

Given angle B = 130°

B + A = B + C = 130 + 25 = 155°

Thus, the true statement about the given angles of the triangle is m∠A + m∠B = 155⁰.

Learn more about angles of Isosceles triangle here: link

Option B m<ZA+m<ZB = 155° is correct option.

Step-by-step explanation:

We know that sum of angles of triangle is equal to 180°

And Isosceles triangle, 2 angles are same

So, let  Angle 1 = x

Angle 2 = x

Angle 3 = 130°

Finding values of angle x

x+x+130=180

2x=180-130

2x=50

x=25

So, Angle 1 = 25 °

Angle 2 = 25°

As, angle B is greater than 90°, the sum of angles added must be greater than 90°

So, Option A, C and D can't be correct, as their sum is less than 90°

Option B m<ZA+m<ZB = 155° is correct option.

Verify:

m<ZA = 25° (as found earlier)

m<ZB = 130° (given)

So, m<ZA+m<ZB = 155°

25°+130° = 155°

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.