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3 years ago

I need help with this question thx

Mathematics

2 answers:

RideAnS [48]3 years ago

6 0

Answer: -5x^3 + 5y^3

Step-by-step explanation: The other parts cancelled out

alukav5142 [94]3 years ago

4 0

Answer:

= -5x³ + 5y³

Step-by-step explanation:

4y³ - 5x³ - 1 - 4y³ - 6 + 7 + 5y³

= -5x³ + (4y³ - 4y³ + 5y³) + (- 1 - 6 + 7)

= -5x³ + (0 + 5y³) + (- 7 + 7)

= -5x³ + 5y³

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Estimate 60% of 52 (WITH WORK!)

Maru [420]

Answer:

Step-by-step explanation:

52 is 100%

you divide 52 by 10 to get 10%

5.2*6(because 60%)

=31.2

3 0

3 years ago

Please solve. best answer gets rewarded! show step by step answers.

Rina8888 [55]

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3 years ago

Calcula el ancho de un terreno rectangular si el largo mide 20 metros y la superficie es de 280 metros cuadrados

Anna35 [415]

Answer:14 m

Step-by-step explanation:

Dada

larga $l=20\ m$

superficie $A=280\ m^2$

Suponga que el ancho es w

La superficie está dada por el producto de la longitud y la anchura.

$\therefore A=lw\\\Rightarrow 280=20\times w\\\Rightarrow w=14\ m$

Por lo tanto, el ancho es de 14 m.

4 0

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: