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sesenic [268]
4 years ago
7

You are traveling down the interstate averaging 50 mph. How much time has passed when you have traveled 120 miles?

Mathematics
2 answers:
Llana [10]4 years ago
7 0
All you need to do is use the formular distance over speed which gives an answer of
2.4
frutty [35]4 years ago
6 0
You traveled 120 miles

You are traveling at 50mph

So divide 120 by 50

120 ÷ 50=  2\frac{20}{50} ⇒ 2 \frac{2}{5}

Answer: You traveled 2  \frac{2}{5} hours or 2.4 hours 
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A square classroom has a wall that is 21 feet long. What is the area of the classroom?
Rasek [7]

<u>Answer and Step-by-Step explaination:</u>

Area =Length time Width

Area= 21*21

Area=441

6 0
3 years ago
Help please! Find the Surface area. Whoever gets it right ill give Brainliest!
11Alexandr11 [23.1K]

Answer:

284.6 square yards

284.6 yd^2

Step-by-step explanation:

Base: 7 × 13 = 91

Rectangle Sides: 6.1 × 13 = 79.3 (x2 = 158.6)

Triangle Sides: 5 × 7 = 35 (÷2 ×2 = 35)

Add: 91 + 158.6 + 35 = 284.6

8 0
3 years ago
Find the equation of a line given the point and slope below.
Levart [38]

Step-by-step explanation:

\frac{y - 6}{x - 3}  =  - 3

- 3(x - 3) = y - 6

- 3x + 9 = y - 6

y =  - 3x + 9  + 6 =  - 3x + 15

Option C

4 0
4 years ago
Let f (x) = 3x − 1 and ε &gt; 0. Find a δ &gt; 0 such that 0 &lt; ∣x − 5∣ &lt; δ implies ∣f (x) − 14∣ &lt; ε. (Find the largest
s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and \varepsilon>0, we want to find a \delta such that |f(x)-14|

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that |x-5|. Then

|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta.

Then, if |x-5|, then |f(x)-14|. So, in this case, if 3\delta \leq \varepsilon we get that |f(x)-14|. The maximum value of delta is \frac{\varepsilon}{3}.

By definition, this procedure proves that \lim_{x\to 5}f(x) = 14. Note that f(5)=14, so this proves that f is continous at x=5.

3 0
3 years ago
A fair die is rolled once. Let A be the event of rolling an even number, and let B be the event of rolling a number greater than
IRISSAK [1]

Answer:

A ∩ B = {4, 6}

Step-by-step explanation:

A die had 6 faces

S = {1, 2, 3, 4, 5, 6}

If A  be the event of rolling an even number, then;

A = {2, 4, 6}

If B be the event of rolling a number greater than 3, then;

B = {4, 5, 6}

A ∩ B are the values that are common to both sets

A ∩ B = {4, 6}

3 0
3 years ago
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