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MA_775_DIABLO [31]
3 years ago
15

What are the cordinates for point A?

Mathematics
1 answer:
TEA [102]3 years ago
3 0

Answer:

b im pretty sure

Step-by-step explanation:


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Help asap! giving brainliest! Check all that apply
Oksana_A [137]

Answer: F

Step-by-step explanation:

becuse we do not have the thing so the answer would be F

4 0
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How many times does 23 go into 194?
USPshnik [31]
The answer is approximately 0.11857
7 0
2 years ago
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Calculate what is 365 {365- (365: (365+365)}}?
Delicious77 [7]

Step-by-step explanation:

365{365-(365:(365+365)}}

= 365{365-(365:730)}

= 365{365-0.5}

= 365{364.5}

= 133,042.5

3 0
3 years ago
Ye has his own business. He checks his sales receipts three times a day, his afternoon sales were 50$ more than his mourning sal
IRINA_888 [86]

Answer:

The answer is 150

Step-by-step explanation:

It's 150 because ye evening sales are three times of his afternoon sales.

SO that means you have to times 50x3=150 or you can add 50+50+50=150. there is your answer.

3 0
3 years ago
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Hi! Please help me, will mark brainliest!
PIT_PIT [208]

Answer:

p(t) = 0 for t = 1

p(t) = 1 for t = 1/8 = 8^-1

Step-by-step explanation:

the graph you will have to do yourself.

just go there and type in

-  log_{8}(x)

well, don't type "log" in letters.

you start by typing the "-" sign, and then you need to look up the functions by clicking on the "funcs" button and look for the log functions .

pick the

log_{a}

option. and then simply enter 8 as the first parameter in the {} brackets and x as the second in the () brackets.

and then you see.

any logarithm is 0 for x (or t) = 1.

because any a⁰ = 1.

and the logarithm gives you that exponent of the base number that leads to the given x value.

in other words : a logarithm is the inverse function of an exponential function.

the exponential function is

y = a^x

and the logarithm then determines

x =  log_{a}(y)

that is all.

and

-  log_{8}(x)  = 1

means that the logarithm itself delivered -1.

and 8^-1 = 1/8

so, p(t) = 1 for t = 1/8

5 0
2 years ago
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