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UkoKoshka [18]
3 years ago
15

Solve the equation log (x + 20) = 3

Mathematics
1 answer:
AleksandrR [38]3 years ago
7 0

Answer:

Step-by-step explanation:

(x+20)=3\\    -20       -20\\x=-17

Hope this helps please give brainliest!

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Marta_Voda [28]
To fill the squares look for the one filled in column/row/diagonal and figure out its sum.
All other columns/rows/diagonals must be equal to that, so whenever there is a column/row/diagonal with a single missing value you can add the others up and calculate the missing value.
So you have to find those single missing value columns/rows/diagonals, fill them out and continue with the next one until the whole square is filled in.


8 0
3 years ago
What is 1/3÷1/2 math answers?
Natali5045456 [20]

Answer:

Step-by-step explanation:

1/3 * 2/1 = 2/3

5 0
3 years ago
Read 2 more answers
Step-by-step solution for: (√2+√10)^2
azamat
Note:  √a * √a  = a
           √a * √b  = √ab

(√2 + √10)²  =  (√2 + √10)(√2 + √10)
                    =  √2(√2 + √10) + √10(√2 + √10)
                    =   √2*√2 + √2*√10 + √10*√2 + √10*√10
                    =       2      + √20  + √20 + 10
                    =      (2 + 10) + (√20 + √20)
                    =        12 +  2√20

√20 = √(4 *5) = √4 * √5 = 2√5

                     =        12 +  2√20  = 12 + 2(2√5)
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5 0
3 years ago
Find f(-2) for the function f(x) = 3r2 - 2x +7.<br> 0-13<br> O-1<br> O 1<br> O 23
viva [34]

Step-by-step explanation:

f(x) = 3x² - 2x + 7

f(-2) = 3(-2)² -2(-2) + 7

= 3(4) + 4 + 7

= 23

8 0
3 years ago
adiocarbon dating of blackened grains from the site of ancient Jericho provides a date of 1315 BC ± 13 years for the fall of the
Zigmanuir [339]

Answer:

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659 and \left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

Step-by-step explanation:

The equation of the isotope decay is:

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

14-Carbon has a half-life of 5568 years, the time constant of the isotope is:

\tau = \frac{5568\,years}{\ln 2}

\tau \approx 8032.926\,years

The decay time is:

t = 1315\,years + 2007\,years \pm 13\,years (There is no a year 0 in chronology).

t = 3335 \pm 13\,years

Lastly, the relative amount is estimated by direct substitution:

\frac{m(t)}{m_{o}} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\mp\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{-\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{min} \approx 0.659

\left(\frac{m(t)}{m_{o}} \right)_{max} = e^{-\frac{3335\,years}{8032.926\,years} }\cdot e^{\frac{13\,years}{8032.926\,years} }

\left(\frac{m(t)}{m_{o}} \right)_{max} \approx 0.661

4 0
3 years ago
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