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ddd [48]
2 years ago
10

How to write 14.05 in expanded form ?

Mathematics
2 answers:
Mama L [17]2 years ago
6 0
You write 10+4+.05.
this is because you add each value separately <span />
jonny [76]2 years ago
6 0

Answer:

Expanded form 10+4+\frac{0}{10}+\frac{5}{100}.

Step-by-step explanation:

Given  : 14.05

To find : How to write 14.05 in expanded form

Solution : We have given 14.05

We can see 1 is at tens place = 10.

4 is at the ones place = 4

0  is at tenth place  = \frac{0}{10}.

5 is at the hundredth place  = \frac{5}{100}.

Expanded form  :

10+4+\frac{0}{10}+\frac{5}{100}.

Therefore, Expanded form 10+4+\frac{0}{10}+\frac{5}{100}.

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Is the expression completely factored: (5 - x)(6 - 5x)
Naya [18.7K]

Answer:

5x² - 31x + 30

Step-by-step explanation:

(5 - x)(6 - 5x)

= 30 - 25x - 6x + 5x²

= 30 - 31x + 5x²

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Can someone help me with this? It’s unit rates with price. Question: 6 feet of ribbon for $19.20​
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Step-by-step explanation:

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8 0
3 years ago
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Two different radioactive isotopes decay to 10% of their respective original amounts. Isotope A does this in 33 days, while isot
Andrews [41]

Answer:

The approximate difference in the half-lives of the isotopes is 66 days.

Step-by-step explanation:

The decay of an isotope is represented by the following differential equation:

\frac{dm}{dt} = -\frac{t}{\tau}

Where:

m - Current mass of the isotope, measured in kilograms.

t - Time, measured in days.

\tau - Time constant, measured in days.

The solution of the differential equation is:

m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }

Where m_{o} is the initial mass of the isotope, measure in kilograms.

Now, the time constant is cleared:

\ln \frac{m(t)}{m_{o}} = -\frac{t}{\tau}

\tau = -\frac{t}{\ln \frac{m(t)}{m_{o}} }

The half-life of a isotope (t_{1/2}) as a function of time constant is:

t_{1/2} = \tau \cdot \ln2

t_{1/2} = -\left(\frac{t}{\ln\frac{m(t)}{m_{o}} }\right) \cdot \ln 2

The half-life difference between isotope B and isotope A is:

\Delta t_{1/2} = \left| -\left(\frac{t_{A}}{\ln \frac{m_{A}(t)}{m_{o,A}} } \right)\cdot \ln 2+\left(\frac{t_{B}}{\ln \frac{m_{B}(t)}{m_{o,B}} } \right)\cdot \ln 2\right|

If \frac{m_{A}(t)}{m_{o,A}} = \frac{m_{B}(t)}{m_{o,B}} = 0.9, t_{A} = 33\,days and t_{B} = 43\,days, the difference in the half-lives of the isotopes is:

\Delta t_{1/2} = \left|-\left(\frac{33\,days}{\ln 0.90} \right)\cdot \ln 2 + \left(\frac{43\,days}{\ln 0.90} \right)\cdot \ln 2\right|

\Delta t_{1/2} \approx 65.788\,days

The approximate difference in the half-lives of the isotopes is 66 days.

4 0
3 years ago
Read 2 more answers
Please help asap it will help very much
qwelly [4]

Answer:

180 cm²

Step-by-step explanation:

From inspection of the diagram, the surface area is made up of the area of 4 congruent triangles and the area of one square.

Area of a square = x² (where x is the length of one side)

⇒ area of the square = 6² = 36 cm²

Area of a triangle = 1/2 x base x height

⇒ area of one triangle = 1/2 x 6 x 12 = 36 cm²

Total surface area = 4 x area of one triangle + area of the square

                               = 4 x 36 + 36

                               = 144 + 36

                               = 180 cm²

5 0
2 years ago
Read 2 more answers
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