Determineeeeeeeeeeeeeeee m < h

Mathematics

2 answers:

Elden [556K]3 years ago

7 0

Answer:

Angle supplementary to angle with measure of 120 is 60.

(120+x=180)

Triangle: 60+32+h = 180

h=88

Let me know if this helps!

Lilit [14]3 years ago

3 0

I can confirm that the answer above is correct :)

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Order number from least to greates <br> 9/5 -2.5 -1.1 -4/5 0.8

Brut [27]

Answer:

Step-by-step explanation:

Start by writing these five numbers in just one format (improper fractions with denominator 10):

9/5 = 18/10, and

-2.5 = -25/10, and

-1.1 = -11/10, and

-4/5 = -8/10, and, finally

0.8 = 8/10

The smallest of these (the least) is -25/10, and

the largest is 18/10.

Arranged in ascending order, we have:

-25/10, -11/10, -8/10, 8/10, 18/10, or

-2.5, -1.1, -4/5, 0.8, 9/5

5 0

3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.