How am I suppose to answer this question

Mathematics

1 answer:

lukranit [14]2 years ago

6 0

I think C i don’t sure sorry

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Given matrices S and T below, which statement is true?

harina [27]

The matrices are
S =(4 11 T= ( -8 11
-3 -8) 3 4 )
Inverse of a matrix is a matrix derived from another matrix such that if you pre- multiply it with the original matrix you get a unit matrix.
if we multiply S and T
ST will be
( 4 11 × (-8 11 = ( 1 0
-3 -8) -3 -4) 0 1)
and also TS
( -8 11 × (4 11 = ( 1 0
-3 -4) -3 -8) 0 1)
therefore, matrices S and T are inverses of each other because ST = TS= I

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8 0

3 years ago

Read 2 more answers

Marie wants to know the most popular car for parents. She plans to survey the people entering the mall from 10:00 a.m. to 11:00

Andreyy89

Answer:

C. Marie should make sure she surveys both the fathers and mothers.

3 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.