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2 years ago

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Mathematics

2 answers:

Lubov Fominskaja [6]2 years ago

4 0

I’m so sorry I can’t figure it out 0512

Dvinal [7]2 years ago

3 0

Should of payed attention

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Find the new amount of the original amount is 45 grams and the percent of decrease is 36%.

Semenov [28]

Answer:

28.8

Step-by-step explanation:

the original is 45=100%

it reduced by 36% meaning;100%-36%=64%

if45=100%

? =64%

45*64=2880/100=28.8

6 0

3 years ago

ASAP I WILL GIVE BRAINLIEST

hammer [34]

Answer:

8 inches

Step-by-step explanation:

V= Volume

A= Base

H= Height

Formula: H= V÷A

336in÷ 42in²= 8in

3 0

2 years ago

On a coordinate plane, a point is at (negative 4, negative 4) and (negative 4, negative 10).

Valentin [98]

Answer:

Option C.

Step-by-step explanation:

Hey there!

The points are; (-4,-4) and (-4,-10).

Use distance formula.

$d = \sqrt{ {(x2 - x1)}^{2} + {(y2 - y1)}^{2} }$

~ Put all values.

$d = \sqrt{ {( - 4 + 4)}^{2} + ( { - 10 + 4)}^{2} }$

~ Simplify it.

$d = \sqrt{ {0}^{2} + ( { - 6)}^{2} }$

$d = \sqrt{36}$

Therefore, distance between the points is 6 units.

Hope it helps...

6 0

3 years ago

NEED HELP ASAP BRAINLIEST PLEASE

sashaice [31]

Answer:

View Image.

Step-by-step explanation:

The solution is where both graph cross each other.

Is the solution, the point (x, y) that they gave you, the same point as where the two graph cross each other?

If yes then the solution is correct.

If no then the solution is wrong.

6 0

3 years ago

Dinosaur fossils are often dated by using an element other than carbon, like potassium-40, that has a longer half life (in this

Firlakuza [10]

The amount of substance left of a radioactive element of half life, $t_{\frac{1}{2}}$ after a time, t, is given by:

$N(t)=N_0\left( \frac{1}{2} \right)^ \frac{t}{t_{ \frac{1}{2} }}$

Given that potassium-40 has a half life of approximately 1.25 billion years.

The number of years it will take for 0.1% of potassium-40 to remain is obtained as follows:

$0.1=100\left( \frac{1}{2} \right)^ \frac{t}{1.25}} \\ \\ \Rightarrow\left( \frac{1}{2} \right)^ \frac{t}{1.25}}=0.001 \\ \\ \Rightarrow\frac{t}{1.25}\ln\left( \frac{1}{2} \right)=\ln(0.001) \\ \\ \Rightarrow \frac{t}{1.25}= \frac{\ln(0.001)}{\ln\left( \frac{1}{2} \right)} =9.966 \\ \\ t=9.966(1.25)=12.5$

Therefore, the maximum age of a fossil that we could date using 40k is 12.5 billion years.

3 0

3 years ago