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Marysya12 [62]
3 years ago
9

Round 5.951 to the place value of the underlined digit ( underlined digit is 9)

Mathematics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

A.) 6.0

Step-by-step explanation:

I dont explain very well, but..

if the next number is above or it is 5, then u round up

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5(5t-2) = -35<br><br>show how to solve the given equation ​
Aloiza [94]
Distribute the 5
25t-10=-35
+10 +10
25t =-25
divide by 25 on both sides
t=-1
8 0
3 years ago
What is 309,099,990 in word form
labwork [276]
Three hundred and nine million, ninetynine thousand, and nine hundred ninety 
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3 years ago
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According to one cosmological theory, there were equal amounts of the two uranium isotopes 235U and 238U at the creation of the
FromTheMoon [43]

Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let P(t) be the amount of ^{235}U and Q(t) be the amount of ^{238}U after t years.

Then, we obtain two differential equations

                               \frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q

where k_1 and k_2 are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

                             \frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt

Now, the variables are separated, P and Q appear only on the left, and t appears only on the right, so that we can integrate both sides.

                         \int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt

which yields

                      \ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2,

where c_1 and c_2 are constants of integration.

By taking exponents, we obtain

                     e^{\ln |P|} = e^{-k_1t + c_1}  \quad e^{\ln |Q|} = e^{-k_12t + c_2}

Hence,

                            P  = C_1e^{-k_1t} \quad Q  = C_2e^{-k_2t},

where C_1 := \pm e^{c_1} and C_2 := \pm e^{c_2}.

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

                                 P(0) = Q(0) = C

Substituting 0 for P in the general solution gives

                         C = P(0) = C_1 e^0 \implies C= C_1

Similarly, we obtain C = C_2 and

                                P  = Ce^{-k_1t} \quad Q  = Ce^{-k_2t}

The relation between the decay constant k and the half-life is given by

                                            \tau = \frac{\ln 2}{k}

We can use this fact to determine the numeric values of the decay constants k_1 and k_2. Thus,

                     4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}

and

                     7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}

Therefore,

                              P  = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q  = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}

We have that

                                          \frac{P(t)}{Q(t)} = 137.7

Hence,

                                   \frac{Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} }{Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}} = 137.7

Solving for t yields t \approx 6 \times 10^9, which means that the age of the  universe is about 6 billion years.

5 0
3 years ago
1∕2x – 2y + 2z –3z + 3∕4x – y
kogti [31]

Answer:

x-3y-z

Step-by-step explanation:

separate common terms and add them

1/2x+3/4x=1x

-2y-y=-3y

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3 years ago
what was the original principal for an 8% simple interest bank account that holds $4340 after 3 years?
strojnjashka [21]

The principal amount is \$3500

<u>SOLUTION: </u>

Given that, simple interest 8\%; \$4340 after 3 years

We have to find the principal.

Now, let the principal amount be p.  

Then, \text { Simple Interest }=\frac{p \times 8 \times 3}{100}=\frac{p \times 24}{100}

Now, we know that, Balance = Principal amount + Simple Interest  

\begin{array}{l}{\rightarrow \$ 4340=p+\frac{24 p}{100}} \\\\ {\rightarrow 4340=p\left(1+\frac{24}{100}\right)} \\\\ {\rightarrow 4340=p \times \frac{100+124}{100}} \\\\ {\rightarrow p=4340 \times \frac{100}{124}} \\\\ {\rightarrow p=35 \times 100} \\\\ {\rightarrow p=3500}\end{array}

5 0
3 years ago
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