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nydimaria [60]
3 years ago
7

the approximate Half-Life of titanium-44 is 63 years. how long will it take a sample to loose 75% of its titanium-44? explain Wh

y.
Mathematics
1 answer:
lesya692 [45]3 years ago
6 0
126 years....it takes half of the sample to decay in 63 yrs. then half of the sample remaining, ( half of 50% is 25%) takes another 63 years which would total at 126
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Help please<br>2x-3y=12<br><br>find y if x is 0
mafiozo [28]
For this, you simply plug in 0 for "x" and solve:

2(0) - 3y = 12  multiply 2x0 to get 0
     0 - 3y = 12  0 - 3y is -3y
         -3y = 12  Isolate your variable by diving both sides by -3
            y = -4   you're left with y = -4

Answer: if x is 0, y is -4


7 0
3 years ago
8w^2+8w+9 <br> what's the value if w=2
artcher [175]

Answer:

Substitute the w‘s in the equation for 2.

8(2)^2+8(2)+9

32+16+9

57

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7 0
2 years ago
2x + 3y = 10
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4 0
3 years ago
Read 2 more answers
Describe the steps you would use to factor 2x3 + 5x2 – 8x – 20 completely. Then state the factored form.
WARRIOR [948]
The polynomial 2x^3 + 5x^2-8x-20 may have solutions which are the divisors of -20, therefore -20 has the following divisors: \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20.
If x=1, then 2\cdot1^3 + 5\cdot1^2-8\cdot1-20=-20\neq 0,
if x=-1, then 2\cdot(-1)^3 + 5\cdot(-1)^2-8\cdot(-1)-20=-9\neq 0,
if x=2, then 2\cdot2^3 + 5\cdot2^2-8\cdot2-20=0, then x=2 is a solution and you have the first factor (x-2). 
If x=-2, then 2\cdot(-2)^3 + 5\cdot(-2)^2-8\cdot(-2)-20=0, then x=-2 is a solution, so you have the second factor (x+2).
Since x-2 and x+2 are two factors of 2x^3 + 5x^2-8x-20 , then the polynomial x^2-4 is a divisor of 2x^3 + 5x^2-8x-20 and dividing the polynomial 2x^3 + 5x^2-8x-20 by x^2-4 you obtain 
 2x^3 + 5x^2-8x-20=(x-2)(x+2)(2x+5).











5 0
3 years ago
Read 2 more answers
Compare and Contrast how you would factor x^2−9 and x^2+9. For full credit, make sure you explain why they are factored differen
marishachu [46]

The factored form of the expression x^2 - 9 is (x+3)(x-3)

The expression x^2 + 9 has no real roots hence cannot be factored

Quadratic equations are equations with a degree of 2.

Given the quadratic function x^2-9 \ and \ x^2+9

For the function x^2 - 9

=x^2-9\\=x^2-3^2

Using the difference of two squares:

a^2-b^2=(a-b)(a+b)\\x^2-3^2=(x-3)(x+3)

For the expression x^2 + 9

The expression x^2 + 9 has no real roots hence cannot be factored

Learn more on factorization here: brainly.com/question/20293447

7 0
2 years ago
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