All categories

3 years ago

What is 8.038×108 in standard form?

Mathematics

2 answers:

Angelina_Jolie [31]3 years ago

6 0

Answer:

803,800,000

Step-by-step explanation:

8.038×10^8 in standard form

To get standard form we multiply the decimal number by 10^8

10^8 is 100000000

Now we multiply

8.038×100000000

Move the decimal number 3 places to the right and add 5 more zeros

803800000

So the standard form of the given expression is

803,800,000

skad [1K]3 years ago

3 0

Okay. I think what you mean by this is 8.038 * 10^8. 10^8 = 100,000,000. When you multiply that by 8.038, you get 803,800,000. The answer is D.

You might be interested in

The least common denominator of 1/4 and 3/8?

castortr0y [4]

8 so it would be 2/8 plus 3/8

3 0

3 years ago

Read 2 more answers

Find the volume of the cylinder in terms of pi. Height 11m Radius 3.4m

Katen [24]

The volume is V=399.48 m to the power of three

5 0

3 years ago

Read 2 more answers

An excellent swimmer is confident he can judge his speed in the water exactly. He bets he can swim two lengths of a 50-meter Oly

Zolol [24]

(1.5 + x) / 2 = 2 .... multiply both sides by 2
1.5 + x = 2 * 2
1.5 + x = 4
x = 4 - 1.5
x = 2.5 meters per second

7 0

3 years ago

What is 5.8×10^6 mg/s in grams per minute? A. 9.7×10^9 g/min B. 9.7 × 10^7 g/min C. 3.48 × 10^5 g/min D. 3.48 × 10^3 g/min.

Luda [366]

1 second = 5.8x10^6 mg
60 seconds = 5.8x10^6 x 60 = 3.48x10^8 mg
(i.e 3.48x10^8 mg/minute)

1000 mg = 1 g
3.48x10^8 mg = 3.48x10^8/1000 = 3.48x10^5 g

Therefore, 5.8x10^6 mg/s = 3.48x10^5 g/min

6 0

2 years ago

Read 2 more answers

A thermometer is taken from an inside room to the outside, where the air temperature is 20° f. after 1 minute the thermometer re

fomenos

The change in the temperature is proportional to the difference in temperatures. Let's call the initial temperature of the room and the thermometer $E_0$ and the current temperature $E(t)$ where $t$ is the elapsed time in minutes. The outside temperature is 20 F.

Newton's Law of Cooling:

$E(t) - 20 = (E_0 - 20) e^{- k t}$

We're given $E(1)=70 \textrm{ and } E(5)=45$

$70-20 = (E_0 - 20) e^{-k}$

$45-20 =(E_0 - 20) e^{-5k}$

Dividing,

$\dfrac{50}{25}=e^{4k}$

$k = \dfrac 1 4 \ln 2$

Now,

$E(t) - 20 = (E_0 - 20) e^{- k t}$

$E_0 = 20 + (E(t) - 20)e^{k t}$

$E_0 = 20 + (E(1) - 20)e^{k}$

$E_0 = 20 + (70 - 20)e^{\ln 2/4}$

$E_0 = 20 + 50 \cdot 2^{1/4} \approx 79.46 \textrm{ degrees F}$

3 0

3 years ago