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3 years ago

RS bisects TU at Q. If TQ=5k-9 and QU=15-x

Mathematics

1 answer:

snow_lady [41]3 years ago

4 0

Answer:

x = 4, TU = 22

Step-by-step explanation:

Given that TS bisects TU at Q , then

TQ = QU , substitute values

5x - 9 = 15 - x ( add x to both sides )

6x - 9 = 15 ( add 9 to both sides )

6x = 24 ( divide both sides by 6 )

x = 4

Thus

TU = 5x - 9 + 15 - x = 4x + 6 = 4(4) + 6 = 16 + 6 = 22

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The temperature outside changed from 64°F to 43°F over a period of seven days. If the temperature changed by the same amount eac

sertanlavr [38]

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Step-by-step explanation:

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3 years ago

A sample of a radioactive isotope had an initial mass of 360 mg in the year 1998 and

RUDIKE [14]

Answer:

193 mg

Step-by-step explanation:

Exponential decay formula:

$A_t = A_0e^r^t$
where Aₜ = mass at time t, A₀ = mass at time 0, r = decay constant (rate), t = time

Our known variables are:

1998 to the year 2004 is a total of t = 6 years.
The sample of radioactive isotope has an initial mass of A₀ = 360 mg at time 0 and a mass of Aₜ = 270 mg at time t.

Let's solve for the decay constant of this sample.

$270=360e^-^r^(^6^)$
$270=360e^-^6^r$
$\frac{3}{4} =e^-^6^r$
$\text{ln} (\frac{3}{4} )= \text{ln}(e^-^6^r)$
$\text{ln} (\frac{3}{4} )=-6r$
$r=-\frac{\text{ln}\frac{3}{4} }{6}$
$r=0.04794701$

Using our new variables, we can now solve for Aₜ at t = 7 years, since we go from 2004 to 2011.

Our new initial mass is A₀ = 270 mg. We solved for the decay constant, r = 0.04794701.

$A_t=270e^-^(^0^.^0^4^7^9^4^8^0^1^)^(^7^)$
$A_t=270e^-^0^.^3^3^5^6^2^9^0^7$
$A_t=193.01982213$

The expected mass of the sample in the year 2011 would be 193 mg.

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2 years ago

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Answer:

Step-by-step explanation:

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3 years ago