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3 years ago

What is the midpoint between -9 and 15 on the number line? 3 12 6 2

Mathematics

2 answers:

Elena L [17]3 years ago

5 0

Answer:

Step-by-step explanation:

The midpoint of two numbers on a number line can be calculated by finding the sum of the two given numbers and dividing that sum by 2. The resulting answer will be the midpoint of the given numbers.

In other words we can say that midpoint is the average of the given numbers. The given numbers are -9 and 15. The sum of these two numbers is:

-9 + 15 = 6

Half of 6 is 3, so 3 is the midpoint of -9 and 15 on the number line.

In order to verify you can calculate the difference from 3 to 15 and from 3 to -9, the result will be the same.

Alborosie3 years ago

5 0

ANSWER

The midpoint of -6 and 15 is 3.

EXPLANATION

The midpoint of any two given numbers on the real number line is given by,

$N_m= \frac{ n_{1} +n_{2}}{2}$

Let

$n_1 = - 9$

and

$n_2 = 15$

We substitute the given numbers into the formula to obtain;

$N_m= \frac{ - 9+15}{2}$

Simplify;

$N_m= \frac{ 6}{2}$

$N_m= 3$

You might be interested in

Identify the theorem or postulate that is related to the measures of the angles in the pair, and find the unknown angle measures

aleksley [76]

Angles formed when two parallel are intersected by a common transversal include, corresponding, alternate interior and exterior, same-side interior, and vertically opposite angles

The correct option is Same–Side Int. ∠s Thm. m∠1 = 30°, m∠5 = 150°

The reason the selection is correct is as follows:

The given angles are;

m∠1 = (30·x - 30)°. m∠5 = (40·x + 70)°

∠1, and ∠5, are located on the same side of the transversal and are both formed on the interior side of the parallel lines

Therefore, they are same–side interior angles

Same side interior angles theorem states that same side interior angles formed between parallel lines are supplementary

Therefore, we have;

m∠1 + m∠5 = 180°

Which gives;

(30·x - 30)° + (40·x + 70)° = 180°

(70·x + 40)° = 180°

70·x = 180° - 40° = 140°

$x = \dfrac{140^{\circ}}{70} = 2^{\circ}$

m∠1 = (30·x - 30)° = (30 × 2 - 30)° = 30°

m∠1 = 30°

m∠5 = (40·x + 70)° = (40 × 2 + 70)° = 150°

m∠5 = 150°

The correct option is therefore;

Same-Side Int. ∠s Thm. m∠1 = 30°, m∠5 = 150°

Learn more about parallel lines cut by a transversal here:

brainly.com/question/533025

8 0

2 years ago

A 51-foot wire running from the top of a tent pole to the ground makes an angle of 58° with the ground. If the length of the ten

AfilCa [17]

Answer:

35.11 ft

Step-by-step explanation:

This given situation can be thought of as triangle $\triangle PQR$ where PQ is the length of pole.

PR is the length of rope.

and QR is the distance of bottom of pole to the point of fastening of rope to the ground.

And $\angle Q \neq 90^\circ$

Given that:

PQ = 44 ft

PR = 51 ft

$\angle R = 58^\circ$

To find:

Side QR = ?

Solution:

We can apply Sine Rule here to find the unknown side.

Sine Rule:

$\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}$

Where

a is the side opposite to $\angle A$

b is the side opposite to $\angle B$

c is the side opposite to $\angle C$

$\dfrac{PR}{sinQ}=\dfrac{PQ}{sinR}\\\Rightarrow sin Q =\dfrac{PR}{PQ}\times sinR\\\Rightarrow sin Q =\dfrac{51}{44}\times sin58^\circ\\\Rightarrow \angle Q =79.41^\circ$

Now,

$\angle P +\angle Q +\angle R =180^\circ\\\Rightarrow \angle P +58^\circ+79.41^\circ=180^\circ\\\Rightarrow \angle P = 42.59^\circ$

Let us use the Sine rule again:

$\dfrac{QR}{sinP}=\dfrac{PQ}{sinR}\\\Rightarrow QR =\dfrac{sinP}{sinR}\times PQ\\\Rightarrow QR =\dfrac{sin42.59}{sin58}\times 44\\\Rightarrow QR = 35.11\ ft$