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2 years ago

A football player runs 10.7 m/s toward the end zone. If he runs for 3.7

Mathematics

1 answer:

sukhopar [10]2 years ago

4 0

Answer:

39.59

Step-by-step explanation:

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A radio station had 141 tickets to a concert. They gave away 2 times as many tickets to listeners as to employees. How many tick

erica [24]

Answer:

Option C: 47 tickets

Step-by-step explanation:

Given that:

Tickets given to employees = 1 part

Tickets given to listeners = 2 parts

This means 141 tickets are totally divided into 3 parts so:

1 part = 141/3

1 part = 47

So now,

Tickets given to employees = 1 part = 47 tickets

Tickets given to listeners = 2 parts = 47 *2 tickets = 94 tickets

I hope it will help you!

4 0

3 years ago

The equation r=3c +5 represents the values shown in the table below C- 6, 8, 12, 18 R- 23, 29, ?, 59 What is the missing value i

almond37 [142]

The missing value is 53.

3 0

3 years ago

Read 2 more answers

A television normally sells for $320. This week, it's on sale at 60% of the normal price. What is the sale price?

KIM [24]

60%=60/100=0.60

320*.60=192

Answer=$192

8 0

2 years ago

Read 2 more answers

How will adding the value 90 affect the mean and median of the data set 1, 5, 5, 6, 9, 10?

Soloha48 [4]

I believe the answer is B for this question

6 0

2 years ago

Read 2 more answers

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.

3 0

3 years ago