if you didnt have a potractor,how could you reconstruct the quarter or it from 0 degrees to 90degrees

1 answer:

4 0

43. Without having a protractor, how can we construct the quarter or it from 0 degrees to 90 degrees. This question is very simple. There are a lot of ways of how constructing a right angle without using a protractor. One of the easiest way is creating a straight vertical line then connect another horizontal from from vertical’s end on the bottom to the right or left. By that you can have a right angle or 90 degrees.

You might be interested in

Find the value of a in the equation 6a-7=3a+14

aleksandr82 [10.1K]

Answer:

a=7

Step-by-step explanation:

Given,

6a-7=3a+14

or,6a-3a=14+7

or,3a=21

or,a=7 [ANSWER]

6 0

4 years ago

Read 2 more answers

Please help me !! D is also an option but I couldn't get it in the picture

LiRa [457]

Answer: A) (-1, 0) to (1, 2)

<u>Step-by-step explanation:</u>

Complex numbers are written in the form of ai + b ; where "a" represents the x-coordinate and "b" represents the y-coordinate --> (a, b)

-i → -1i + 0 → (-1, 0)

2 + i → 1i + 2 → (1, 2)

Which graph connects those two coordinates? OPTION A

5 0

4 years ago

In △ABC , line AB is extended up to point E and line BC is extended up to D . The measure of some of the angles is given as; ∠BA

Over [174]

Answer:

$\angle ACD = 130 ^\circ$ .

Step-by-step explanation:

$\angle ABC$ and $\angle EBC$ are supplementary angles. Their sum should be $180^\circ$ . $\angle EBC = 120^\circ$ $\implies \angle ABC = 180^\circ - \angle EBC = 180^\circ - 120^\circ = 60^\circ$ .

The sum of the three angles in triangle $\triangle ABC$ : $(\angle ABC + \angle BAC + \angle ACB)$ should be equal to $180^\circ$ .

$\angle ABC = 60^\circ$ and $\angle BAC = 70^\circ$ $\implies \angle ACB = 180^\circ - \angle ABC - \angle ACB = 180^\circ - 70^\circ - 60^\circ = 50^\circ$ .

$\angle ACB$ and $\angle ACD$ form another pair of supplementary angles. Their sum should also be $180^\circ$ . $\angle ACB = 50^\circ$ $\implies \angle ACD = 180^\circ - \angle ACB = 180^\circ - 50^\circ = 130^\circ$ .