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3 years ago

The only solution to x^2+bx+c=0 is x=5 find a and b

Mathematics

1 answer:

miss Akunina [59]3 years ago

5 0

Answer:

b=-10

c=25

Step-by-step explanation:

if the only solution is 5 so

x^2+bx+c=(x-5)²=x²-10x+25

b=-10

c=25

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Step-by-step explanation:

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13 + 14 + 15 = 42

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3 years ago

(5t^3)^-4 1. 625/t^12 2. 20/t^7 3. 1/625t^12 4. 1/20t^7

rusak2 [61]

Answer:

$\large\boxed{3.\ \dfrac{1}{625t^{12}}}$

Step-by-step explanation:

$\text{Use}\\\\(ab)^n=a^nb^n\\\\(a^n)^m=a^{nm}\\\\a^{-n}=\dfrac{1}{a^n}\\---------------\\\\(5t^3)^{-4}=\dfrac{1}{(5t^3)^4}=\dfrac{1}{5^4(t^3)^4}=\dfrac{1}{625t^{(3)(4)}}=\dfrac{1}{625t^{12}}$

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3 years ago

Element X decays radioactively with a half life of 12 minutes. If there are 160

Salsk061 [2.6K]

Answer:

It would take 75.8 minutes for the element to decay to 2 grams.

Step-by-step explanation:

The number of grams of element x, after t minutes, is given by the following equation:

$X(t) = X(0)(1-r)^{t}$

In which X(0) is the initial amount and r is the decay rate.

There are 160 grams of Element X

This means that X(0) = 160.

$X(t) = X(0)(1-r)^{t}$

$X(t) = 160(1-r)^{t}$

Half life of 12 minutes.

This means that X(12) = 0.5*X(0) = 0.5*160 = 80. So

$X(t) = 160(1-r)^{t}$

$80 = 160(1-r)^{12}$

$(1 - r)^{12} = 0.5$

$\sqrt[12]{(1 - r)^{12}} = \sqrt[12]{0.5}$

$1 - r = 0.9438$

$X(t) = 160(1-r)^{t}$

$X(t) = 160(0.9438)^{t}$

How long would it take for the element to decay to 2 grams?

This is t for which X(t) = 2. So

$X(t) = 160(0.9438)^{t}$

$2 = 160(0.9438)^{t}$

$(0.9438)^{t} = \frac{2}{160}$

$\log{(0.9438)^{t}} = \log{\frac{2}{160}}$

$t\log{0.9438} = \log{\frac{2}{160}}$

$t = \frac{\log{\frac{2}{160}}}{\log{0.9438}}$

$t = 75.8$

It would take 75.8 minutes for the element to decay to 2 grams.

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3 years ago