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Alecsey [184]
3 years ago
14

Evaluate log1∕6 36. please help

Mathematics
1 answer:
Sholpan [36]3 years ago
4 0

\log_{\tfrac 16}   36 \\\\\\=\dfrac{\ln 36}{ \ln \left(\tfrac 16\right)}\\\\\\=\dfrac{ \ln 6^2}{\ln (6^{-1})}\\\\\\=\dfrac{2 \ln 6}{-1 \ln 6}\\\\= - 2

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I need help ASAP!!
Dmitrij [34]
Suggestion:

The width is the x value and the hight is the y value.

Plot the points on the graph as decimal if that helps.

The point at the bottom left is the origin (0,0)

I hope this helps.
5 0
2 years ago
Gina visits her local farm stand to buy apples and oranges to make a fruit salad. She has​ $10.00 to spend. If she buys only ora
icang [17]

Answer:

29 oranges rounding up but the exact answer is 29.41176

Step-by-step explanation:

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3 0
3 years ago
Which zero pair could be added to the function f(x)=x^2+12x+6 so that the function can be written in vertex form?
andrew11 [14]
<span>f(x) = x</span>² <span>+ 12x + 6  </span>→ y = x² + 12x + 6<span>

Let us convert the standard form into vertex form.

1) Complete the squares. Isolate x</span>² and x terms.
<span>y - 6 = x</span>² + 12x
<span>
2) Create the perfect square trinomial. Whatever number is added on one side must also be added on the other side. 

y - 6 + 36 = x</span>² + 12x + 36<span>
y + 30 = (x + 6)</span>²
<span>y = (x + 6)</span>² - 30  ← Vertex form
<span>
To check:

y = (x + 6) (x + 6) - 30
y = x</span>² + 6x + 6x + 36 - 30
<span>y = x</span>² + 12x + 6<span>

The zero that could be added to the given function is 36, -36</span>
4 0
3 years ago
I need help like really bad in the next 10 mins or I fail math please help me-
shutvik [7]
The answer should be 16!!
5 0
3 years ago
Read 2 more answers
Guys please help i’m blanking!!!<br> Select all the functions whose graphs include the point (16,4)
Gnom [1K]
D and E —> substitute the x or y of the point given into the functions to test if it is correct
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3 years ago
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