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3 years ago

Solve the inequality and graph its solution: 5n – 10 > 25

Mathematics

2 answers:

solniwko [45]3 years ago

8 0

N is 3. you subtract the 10 from the 25, then divide it by 5.

kozerog [31]3 years ago

5 0

5n > 35 (Addition property of equality)
n > 7 (division property of equality)

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Simplify 176.4-23.72

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Answer is 152.68 simple subtraction problem

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3 years ago

What is .0252525252525252525... as a simplified fraction? Will give Brainliest.

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Step-by-step explanation:

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3 years ago

Raymond had a balance of −$467.25 in his bank account, and Roger had a balance of −$356.75 in his bank account. Which statement

GaryK [48]

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3 years ago

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

Determine whether the equation represents direct, inverse, joint, or combined variation.

Dominik [7]

Answer: The retort to the peculiar proposed interrogate is definitively identified as ‘Joint Variation’, interchangeably acknowledged as ‘Combined Variation’.

Step-by-Step Explanation:

Given the mathematical statement,
Y = 4x^3z^3

Such coherently demonstrates the principle of a foreign variable, ‘y’, which whom varies directly as the product of at least two distinct or distinguishable quantities occur.

The mentioned is the precise definition of Joint Variation, where ‘y’ is jointly proportional to the product of the minimum agglomerate of two quantities, whilst a presence of a nonzero constant, k.

*I hope this helps.

5 0

3 years ago