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Radda [10]
3 years ago
7

What is the result of multiplying 2.5x10^10 by 3.5x10^-2?

Mathematics
2 answers:
Arturiano [62]3 years ago
8 0

Answer:

The result is 8.75 x10^8

Step-by-step explanation:

When you multiply numbers in scientific notation, you do the following:  

Multiply the base numbers normally:  

2.5* 3.5 = 8.75  

The exponents with base 10 are added and the base stays the same in the result:  

10^10x10^-2= 10^(10+(-2))=10^8.  

The answer is the multiplicaton of both 8.75*10^8.

mojhsa [17]3 years ago
5 0
8.75^8 when you multiply you add your exponents and when you divide you subtract your exponents

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Solve 4.62 + (−12.3).<br><br> 56.83<br> 5.85<br> −7.68<br> −16.92
ikadub [295]

The sum of the given decimal numbers is -7.68

<h3>Sum of decimal numbers</h3>

Decimal numbers are numbers that contains decimal points. Given the expression below;

4.62 + (−12.3).

Convert to fraction to have:

4.62 + (−12.3) = 462/100 - 123/10

Find the LCM

4.62 + (−12.3) = 462-1230/100

4.62 + (−12.3) = -768/100

4.62 + (−12.3) =-7.68

Hence the sum of the given decimal numbers is -7.68

Learn more on sum of numbers here; brainly.com/question/25734188

#SPJ1

4 0
1 year ago
Matches the description in the table below. 
tatiyna
I think it is A because i think it is the only one that makes sense
5 0
3 years ago
(hg^(-3))/(w^(-4) k^0 )
Svetlanka [38]
One may note you never quite asked anything, now, assuming simplification,

\bf ~~~~~~~~~~~~\textit{negative exponents}\\\\&#10;a^{-n} \implies \cfrac{1}{a^n}&#10;\qquad \qquad&#10;\cfrac{1}{a^n}\implies a^{-n}&#10;\qquad \qquad &#10;a^n\implies \cfrac{1}{a^{-n}}&#10;\\\\&#10;-------------------------------\\\\&#10;\cfrac{hg^{-3}}{w^{-4}k^0}\implies \cfrac{hw^4}{g^3(1)}\implies \cfrac{hw^4}{g^3}
4 0
2 years ago
Since at t=0, n(t)=n0, and at t=∞, n(t)=0, there must be some time between zero and infinity at which exactly half of the origin
Airida [17]
Answer: t-half = ln(2) / λ ≈ 0.693 / λ

Explanation:

The question is incomplete, so I did some research and found the complete question in internet.

The complete question is:

Suppose a radioactive sample initially contains N0unstable nuclei. These nuclei will decay into stable nuclei, and as they do, the number of unstable nuclei that remain, N(t), will decrease with time. Although there is no way for us to predict exactly when any one nucleus will decay, we can write down an expression for the total number of unstable nuclei that remain after a time t:

N(t)=No e−λt,

where λ is known as the decay constant. Note that at t=0, N(t)=No, the original number of unstable nuclei. N(t) decreases exponentially with time, and as t approaches infinity, the number of unstable nuclei that remain approaches zero.

Part (A) Since at t=0, N(t)=No, and at t=∞, N(t)=0, there must be some time between zero and infinity at which exactly half of the original number of nuclei remain. Find an expression for this time, t half.

Express your answer in terms of N0 and/or λ.

Answer:

1) Equation given:

N(t)=N _{0} e^{-  \alpha  t} ← I used α instead of λ just for editing facility..

Where No is the initial number of nuclei.

2) Half of the initial number of nuclei: N (t-half) =  No / 2

So, replace in the given equation:

N_{t-half} =  N_{0} /2 =  N_{0}  e^{- \alpha t}

3) Solving for α (remember α is λ)

\frac{1}{2} =  e^{- \alpha t} &#10;&#10;2 =   e^{ \alpha t} &#10;&#10; \alpha t = ln(2)

αt ≈ 0.693

⇒ t = ln (2) / α ≈ 0.693 / α ← final answer when you change α for λ




4 0
3 years ago
Find the x-intercept and y-intercept for this equation 6x –y = -12. Plz help!!
12345 [234]
Y intercept = (0,12)
x intercept = (-2,0)
3 0
3 years ago
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