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Tju [1.3M]
3 years ago
13

In the late nineteenth century, the scientist Marie Curie performed experiments that led to the discovery of radioactive substan

ces. A radioactive substance is a substance that gives off radiation as it decays. Scientists describe the rate at which a radioactive substance decays as its half-life. The half-life of a substance is the amount of time it takes for one-half of the substance to decay. Radium has a half-life of 1600 years. How much radium will be left from a 1000-gram sample after 1600 years?
Mathematics
1 answer:
gtnhenbr [62]3 years ago
5 0

Answer:

Step-by-step explanation:

Half life is the time taken by a radioactive substance to reduce of half of its original size

If the half life of radium is 1600, this means that the original radioactive substance N0 have reduced to N0/2 after 1600 years

if we are given N0 = 1000grams

The amount of the substance that will remain after 1600 years will be N0/2 i.e 1000/2 = 500g

<em>Hence 500 grams of radium will be left after 600years</em>

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What is 1*727/3+6*2-5=? pIz help!!!
fiasKO [112]

Answer:

its 249 1/3

Step-by-step explanation:

=727/3 + 6 x 2 - 5

=727/3 + 12 - 5

= 748/3

= 249 1/3

4 0
3 years ago
K is between J and M. L is
xeze [42]

Step-by-step explanation:

given,

K is between J and M, L is between K and M, M is between K and N. JN = 14, KM=4 and JK= KL= LM

NL=?

according to the question,

J-------K-------L------M--------N

We know,

KM= 4

KL+LM=4

KL+KL=4 (LM=KM)

2KL=4

KL=2

SO,

JK=KL=2 and LM=KL=2

we have ,

JN=14

JK+KM+MN= 14

2+4+MN=14

MN=8

Now,

NL=LM+MN= 2+8= 10

7 0
3 years ago
Grade 10 Math. Solve for y. Will mark right answer brainliest :)​
Dima020 [189]

Answer:

y=5, y=\frac{38}{11}

Step-by-step explanation:

Hi there!

We are given the equation

\frac{y+2}{y-3}+\frac{y-1}{y-4}=\frac{15}{2} and we need to solve for y

first, we need to find the domain, which is which is the set of values that y CANNOT be, as the denominator of the fractions cannot be 0

which means that y-3≠0, or y≠3, and y-4≠0, or y≠4

\frac{y+2}{y-3} and \frac{y-1}{y-4} are algebraic fractions, meaning that they are fractions (notice the fraction bar), but BOTH the numerator and denominator have algebraic expressions

Nonetheless, they are still fractions, and we need to add them.

To add fractions, we need to find a common denominator

One of the easiest ways to find a common denominator is to multiply the denominators of the fractions together

Let's do that here;

on \frac{y+2}{y-3}, multiply the numerator and denominator by y-4

\frac{(y+2)(y-4)}{(y-3)(y-4)}; simplify by multiplying the binomials together using FOIL to get:

\frac{y^{2}-2y-8}{y^{2}-7y+12}

Now on \frac{y-1}{y-4}, multiply the numerator and denominator by y-3

\frac{(y-1)(y-3)}{(y-4)(y-3)}; simplify by multiplying the binomials together using FOIL to get:

\frac{y^{2}-4y+3}{y^{2}-7y+12}

now add \frac{y^{2}-2y-8}{y^{2}-7y+12} and \frac{y^{2}-4y+3}{y^{2}-7y+12} together

Remember: since they have the same denominator, we add the numerators together

\frac{y^{2}-2y-8+y^{2}-4y+3}{y^{2}-7y+12}

simplify by combining like terms

the result is:

\frac{2y^{2}-6y-5}{y^{2}-7y+12}

remember, that's set equal to \frac{15}{2}

here is our equation now:

\frac{2y^{2}-6y-5}{y^{2}-7y+12}=\frac{15}{2}

it is a proportion, so you may cross multiply

2(2y²-6y-5)=15(y²-7y+12)

do the distributive property

4y²-12y-10=15y²-105y+180

subtract 4y² from both sides

-12y-10=11y²-105y+180

add 12 y to both sides

-10=11y²-93y+180

add 10 to both sides

11y²-93y+190=0

now we have a quadratic equation

Let's solve this using the quadratic formula

Recall that the quadratic formula is y=(-b±√(b²-4ac))/2a, where a, b, and c are the coefficients of the numbers in a quadratic equation

in this case,

a=11

b=-93

c=190

substitute into the formula

y=(93±√(8649-4(11*190))/2*11

simplify the part under the radical

y=(93±√289)/22

take the square root of 289

y=(93±17)/22

split into 2 separate equations:

y=\frac{93+17}{22}

y=\frac{110}{22}

y=5

and:

y=\frac{93-17}{22}

y=\frac{76}{22}

y=\frac{38}{11}

Both numbers work in this case (remember: the domain is y≠3, y≠4)

So the answer is:

y=5, y=\frac{38}{11}

Hope this helps! :)

5 0
3 years ago
PLEASE ANSWER ASAP!!!!
matrenka [14]

Answer:

68

Step-by-step explanation:

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2 years ago
In the 3D map shown below, what is the depth of the area marked by the red "x"?
Sedbober [7]

Answer:

2m

Step-by-step explanation:

got it right on edg

7 0
3 years ago
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