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3 years ago

Can someone help i have been asking for 2 hours

Mathematics

1 answer:

yawa3891 [41]3 years ago

6 0

Flipping a point across the y axis, the y coordinate should flip signs (negative if it was positive before, and vice versa). So

reflection across y axis = (-1 5/7, 1).

Similarly, reflecting across the x axis, the sign will also flip. So

reflection across x axis = (1 5/7, -1).

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Read 2 more answers

How much of a radioactive kind of thorium will be left after 14,680 years if you start with

babymother [125]

Answer:

8978 grams

Step-by-step explanation:

The equation to find the half-life is:

$N(t)= N_{0}e^{-kt}$

N(t) = amount after the time <em>t</em>

$N_{0}$ = initial amount of substance

t = time

It is known that after a half-life there will be twice less of a substance than what it intially was. So, we can get a simplified equation that looks like this, in terms of half-lives.

$N(t)= N_{0}e^{-\frac{ln(\frac{1}{2}) }{t_{h} } t}$ or more simply $N(t)= N_{0}(\frac{1}{2})^{\frac{1}{t_{h} } }$

$t_{h}$ = time of the half-life

We know that $N_{0}$ = 35,912, t = 14,680, and $t_{h}$ =7,340

Plug these into the equation:

$N(t) = 35912(\frac{1}{2})^{\frac{14680}{7340} }$

Using a calculator we get:

N(t) = 8978

Therefore, after 14,680 years 8,978 grams of thorium will be left.

Hope this helps!! Ask questions if you need!!

8 0

2 years ago

Use the function f(x) = 2x3 - 3x2 + 7 to complete the exercises.

GenaCL600 [577]

Easy
sub 2 for x

f(2)=2(2)^3-3(2)^2+7
f(2)=2(8)-3(4)+7
f(2)=16-12+7
f(2)=11

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3 years ago