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Gemiola [76]
3 years ago
10

A cellphone weighs about 2.8x10^n pounds. Which value of n is most reasonable?

Mathematics
2 answers:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

Approximate weight of cell phone is between 100 gm to 200 gm.

1 Pound = 0.45359237 Kilograms

           = 0.4545 kg

2.8 pound = 2.8 × 0.4545

               = 1.2727 Kg

It is given that weight of cell phone is 2.8 \times 10^n pounds, means 1.2727 \times 10^n Kilograms.

Which is equal to, 1272.7 \times 10^n grams.

Because, 1 Kilogram = 1,000 gram

If, n=0

Weight of cell phone = 1272.7 grams=1.2727 Kg

when, n=1

Weight of cell phone = 12727 grams=12.727 Kg

n= -2,

Weight of cell phone = 12.727 grams

n=- 3

Weight of cell phone = 1.2727 grams

None, of the weights are appropriate weight of cell phone.Neither of four options provided here is true.

Sidana [21]3 years ago
5 0

Answer:

The correct answer is B.) -1

Step-by-step explanation:

I think we had different answer choices but for me It was B which is -1

Hope this helps!!

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Hector is visiting a cousin who lives 350 miles away he has driven 90 miles how many more miles does it take to drive to reach h
bija089 [108]
260 more miles,since 350-90=260
8 0
3 years ago
( 32.45 - 4.8) - 2.06
Igoryamba

Answer:

25.59

Step-by-step explanation:

The first thing that you will do is solve parenthesis (remember PEMDAS).

(32.45-4.8) - 2.06

(27.65) - 2.06

Then do the subtraction.

(27.65) - 2.06=

25.59

Hope this helps!

5 0
3 years ago
Given the quadriatic equation
vazorg [7]

Answer:

⇒  The given quadratic equation is x2−kx+9=0, comparing it with ax2+bx+c=0

∴  We get, a=1b=−k,c=9

⇒  It is given that roots are real and distinct.

∴  b2−4ac>0

⇒  (−k)2−4(1)(9)>0

⇒  k2−36>0

⇒  k2>36

⇒  k>6 or k<−6

∴  We can see values of k given in question are correct.

8 0
2 years ago
The slipe of the line passing through the points (7, 5) and (21, 15) is
SOVA2 [1]

Answer:

5/7

Step-by-step explanation:

You find a slope by using the equation y2-y1/x2-x1.

In this case: 15-5/21-7

10/14

5/7

6 0
2 years ago
I don't even know where to start on this problem?
velikii [3]
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7 0
3 years ago
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