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kipiarov [429]
4 years ago
10

Value of the expression down below

Mathematics
1 answer:
zimovet [89]4 years ago
8 0
Hi there. Let's figure this out. 

1.5 (-2.4 + (-5.3))=

1.5(-7.7)=

-11.55
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4 * 19 to the 9th power in scientific notation and result
xxMikexx [17]
The answer would be 1,290,750, 791, 116

in scientific notation its 1.290750791116 * 10 to the 12th power

number name its 1 trillion, 290 billion, 750 million, 791 thousand, 116
6 0
3 years ago
Read 2 more answers
I need help on number 1
ollegr [7]
2.5. is the answer. ..
7 0
4 years ago
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Mark and Tony share a sum of money in the ratio 3:5. What fraction of the money did Tony ratio receive?
gladu [14]
Answer:

Part A is 3/8 of the whole
Part B is 5/8 of the whole

Solution:
Assuming there are no other parts,
the Whole = A + B is the denominator:
Whole = 3 + 5 = 8

Part A = 3 and Part B = 5 are numerators for each fraction.

The fractions are then:
3/8 and 5/8

Meaning:
Part A is 3/8 of the whole
Part B is 5/8 of the whole
4 0
4 years ago
Find the common ratio of the geometric sequence 4, -12,36,...
yulyashka [42]

Step-by-step explanation:

the common ratio of the geometric sequence 4, -12,36,...

= -12/4 = -3

8 0
3 years ago
A breeder reactor converts uranium-238 into an isotope of plutonium-239 at a rate proportional to the amount of uranium-238 pres
topjm [15]

Let U(t) denote the amount of uranium-238 in the reactor at time t. As conversion to plutonium-239 occurs, the amount of uranium will decrease, so the conversion rate is negative. Because the rate is proportional to the current amount of uranium, we have

\dfrac{\mathrm dU}{\mathrm dt}=-kU

where k>0 is constant. Separating variables and integrating both sides gives

\dfrac{\mathrm dU}U=-k\,\mathrm dt\implies\ln|U|=-kt+C\implies U=Ce^{-kt}

Suppose we start some amount u. This means that at time t=0 we have U(0)=u, so that

u=Ce^{-0k}\implies C=u\implies U=ue^{-kt}

We're given that after 10 years, 99.97% of the original amount of uranium remains. This means (if t is taken to be in years) for some starting amount u,

0.9997u=ue^{-10k}\implies k=-\dfrac{\ln(0.9997)}{10}

The half-life is the time t_{1/2} it takes for the starting amount u to decay to half, 0.5u:

0.5u=ue^{-kt_{1/2}}\implies t_{1/2}=-\dfrac{\ln(0.5)}k=\dfrac{10\ln2}{\ln(0.9997)}

or about 23,101 years. Notice that it doesn't matter what the actual starting amount is, the half-life is independent of that.

7 0
3 years ago
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