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°.

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet