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2 years ago

Please answer this with full explanation.

Mathematics

2 answers:

fenix001 [56]2 years ago

4 0

Answer:

heeeeeeereeeeeeee

Step-by-step explanation:

Gnoma [55]2 years ago

3 0

Mitressofdark name is like mistress's. and your rank is in genius but your brain is like a child. what a poooooooooooor.

l o. o. s. e. r.

ha. ha. ha. ha.

mistress mean a genius one. but you are not

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Simplify the expression completely.

ivolga24 [154]

You can’t simplify it any further. 288 1/4 is already simplified.

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3 years ago

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Put 5x + 5y = -3,500 into a slope-intercept form

GREYUIT [131]

Slope intercept form is y=mx+b
y=-x+700

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2 years ago

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Dennis wants to buy a card for his wife. Dennis calculates the amount of the card as $4.50. The actual price of a card is $4. Wh

Alex17521 [72]

Answer:

12.5%

Step-by-step explanation:

$4.50-$4/$4 x 100%

0.5/4x100%

50/4= 25/2 = 12.5%

8 0

3 years ago

What is the solution to (5m)² - (2n-3m)³ if m=-3 and n=5

Nimfa-mama [501]

Answer:

-6634

Step-by-step explanation:

(5m)² - (2n-3m)³ if m=-3 and n=5

= (5 * (-3))^2 - (2 * 5 - 3 * (-3))^3

= (-15)^2 - (10 - (-9))^3

= 225 - (19)^3

= 225 - 6859

= -6634

7 0

3 years ago

Dubnium-262 has a half-life of 34 s. How long will it take for 500.0 grams to

dmitriy555 [2]

Answer:

the time taken for the radioactive element to decay to 1 g is 304.8 s.

Step-by-step explanation:

Given;

half-life of the given Dubnium = 34 s

initial mass of the given Dubnium, m₀ = 500 grams

final mass of the element, mf = 1 g

The time taken for the radioactive element to decay to its final mass is calculated as follows;

$1 = 500 (0.5)^{\frac{t}{34}} \\\\\frac{1}{500} = (0.5)^{\frac{t}{34}}\\\\log(\frac{1}{500}) = log [(0.5)^{\frac{t}{34}}]\\\\log(\frac{1}{500}) = \frac{t}{34} log(0.5)\\\\-2.699 = \frac{t}{34} (-0.301)\\\\t = \frac{2.699 \times 34}{0.301} \\\\t = 304.8 \ s$

Therefore, the time taken for the radioactive element to decay to 1 g is 304.8 s.

4 0

2 years ago

Please answer this with full explanation.​

Please answer this with full explanation.