All categories

2 years ago

Which number is greater than 555,621,000? A. 5.0 x 10 B. 6.2 x 10 C. 5.55621 x 10 D. 9.99 x 10

Mathematics

2 answers:

ivolga24 [154]2 years ago

6 0

Answer: I believe D is correct

Step-by-step explanation:

xeze [42]2 years ago

5 0

Answer:

None of these

Step-by-step explanation:

None of these are correct because you would need to multiply 5 x 10 to a certain power. 5.55621 x 10 to the 8th power would do it.

You might be interested in

PLEASE HELP. I NEED ALL!!!!

Iteru [2.4K]

1. y= 6
2. y= 12
3. y=18
4. y= 24
5. y=5
6. y= 10
7. y=15
8. y= 20
9. y=0
10. y= 0.2
11. y=0.4
12. y=0.6

Hope this helps!

7 0

3 years ago

Read 2 more answers

Hey !! Can someone please help me with this??

Katyanochek1 [597]

Answer:

$8^2$ and 4 x 4 x 4

Step-by-step explanation:

4 to the third power, is the same as 4 x 4 x 4.

This is then equal to 64.

8 squared is also equal to 64, making it also an answer.

None of the other ones work.

<em>Hope this helps and have a nice day!! :) :] :D</em>

7 0

3 years ago

I need help please !

Orlov [11]

It is A so you have it right

4 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST IF CORRECT!!

alexgriva [62]

Answer:

4 miles

Step-by-step explanation:

5 0

3 years ago

The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)

Lemur [1.5K]

Answer:

Cardiac output: $F=0.055 L\s$

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output: $F=\frac{A}{\int\limits^T_0 {c(t)} \, dt}$ ---1

A = 3 mg

$\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt$

Do, integration by parts

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}$

Substitute the value in 1

Cardiac output: $F=\frac{3}{54.49}$

Cardiac output: $F=0.055 L\s$

Hence Cardiac output: $F=0.055 L\s$

4 0

3 years ago