:D PLZ HELP WITH MI MATH

Mathematics

2 answers:

Bas_tet [7]2 years ago

5 0

18cm square

I think domt know

STALIN [3.7K]2 years ago

3 0

It’s 18 so you are correct

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Can some one plz help I'm comfuzzled

scZoUnD [109]

14+ 13y - 13y = 20y - 13y -21 (taking away the smallest y value from both sides)

14 = 7y - 21

14 + 21 = 7y - 21 +21 (+21 on both sides to remove -21 if we do -14 then we will have nothing on other side)

35 = 7y

35/7 = 7y/7 (dividing 7 on both sides)

5 = y

answer is 100% correct no doubts. hope it helped you.

5 0

3 years ago

PLEASE HELP I GIVE THANKS

ololo11 [35]

The answer would be B because it crosses the X axis closer to the 0,0 than the original line

5 0

3 years ago

*HIGHER POINTS* Solve for m<A

Musya8 [376]

Answer:

The answer should be about 114.

Step-by-step explanation:

Of you look at the shape of the other angles. B and D seem to be about 90 and the shape of A is just a little bit bigger than ninety. I started by adding 78, 90, and 90 together which gave me 246. Then I subtracted 246 from 360 which gave me 114. I had estimated the answer to be around 110.

6 0

2 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.