Apair of parallel lines is cut by a transversal. one of the angles formed measures 58°. which statements about the other seven angles formed are true? (multiple answers) there are three more angles with the same measure. all the other angles have the same measure. the rest of the angles measure 122°. only three of the angles measure 122°. four of the angles measure 122°.

Correct options are 1 and 5.

Step-by-step explanation:

If parallel lines m and n are cut by a transeversal t, then corresponding angles are equal in measure. Therefore,

∠2≅∠6;

∠1≅∠5;

∠3≅∠7;

∠4≅∠8.

If parallel lines m and n are cut by a transeversal t, then interior angles on the same side are supplementary. Thus,

∠3+∠5=180°;

∠4+∠6=180°.

If one of the angles ( let it be angle 3) formed measures 58°, then m∠5=180°-58°=122°. Then

m∠2=m∠3=m∠6=m∠7=58°;

m∠1=m∠4=m∠5=m∠8=122°.

You can conclude that there are three more angles with the same measure of 58° and four of the angles measure 122°.

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2