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inna [77]
2 years ago
11

Multiply the binomials: (w-9) • (w-4)

Mathematics
1 answer:
Novay_Z [31]2 years ago
3 0

Answer:

w² - 13w + 36

Step-by-step explanation:

Given

(w - 9)(w - 4)

Each term in the second factor is multiplied by each term in the first factor, that is

w(w - 4) - 9(w - 4) ← distribute both parenthesis

= w² - 4w - 9w + 36 ← collect like terms

= w² - 13w + 36

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Answer: y=16
explanation: y=128 when x=512, so to get the value of y, you need to divide 512/64=8. dividing 512(x) by 8 times gets 64, so you need to divide 8 on the other side(y). 128/8=16. hope this helped!
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1 year ago
Trigonometry<br> Find AC in the diagram attached.
sladkih [1.3K]

Answer:

2√2

Step-by-step explanation:

< C = 180-15-45 = 120°

AC/sin 45° = AB/sin 120°

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3 years ago
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HELP BRAINLIEST FOR. CORRECT AND ALL ANSWERED ANSWER
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Answer:

1. 6/7

2. 1/2

3. 7/10

4. 39/40

5. 15/28

6. 3/10

7. 7/12

8. 1/6

9. 5/12

10. 1/8

 Step-by-step explanation:

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3 years ago
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Which of the following appear in the diagram below?
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Step-by-step explanation:

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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
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