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3 years ago

You and your family left the house for a vacation at 8 A.M. You promptly fell asleep in the

Mathematics

2 answers:

poizon [28]3 years ago

8 0

Answer:

201 miles.

Step-by-step explanation:

If you've traveled 124 miles in 2 hours, than in one hour must've been 62 miles. 62 multiplied by 3 is 186, which is how much miles have been traveled in 3 hours. Plus the 15 minutes would probably be 201 miles. Hope this helps!

Misha Larkins [42]3 years ago

4 0

Answer is the miles times speed I think

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Element X is a radioactive isotope such that its mass decreases by 26% every day. If an experiment starts out with 810 grams of

ioda

Answer:

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

The function to represent the mass of the sample after t days is $A(t) = 810(0.74)^t$

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance is given by:

$A(t) = A(0)(1-r)^t$

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Hourly rate of change:

Decreases 26% by day. A day has 24 hours. This means that $A(24) = (1-0.26)A(0) = 0.74A(0)$ ; We use this to find r.

$A(t) = A(0)(1-r)^t$

$0.74A(0) = A(0)(1-r)^{24}$

$(1-r)^{24} = 0.74$

$\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}$

$1 - r = (0.74)^{\frac{1}{24}}$

$1 - r = 0.9875$

$r = 1 - 0.9875 = 0.0125$

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

Starts out with 810 grams of Element X

This means that $A(0) = 810$

Element X is a radioactive isotope such that its mass decreases by 26% every day.

This means that we use, for this equation, r = 0.26.

The equation is:

$A(t) = A(0)(1-r)^t$

$A(t) = 810(1 - 0.26)^t$

$A(t) = 810(0.74)^t$

The function to represent the mass of the sample after t days is $A(t) = 810(0.74)^t$

5 0

3 years ago

Find the numerical value of the log expression. \log a=7\hspace{15px} \log b=10\hspace{15px} \log c=3 loga=7logb=10logc=3 \log \

bija089 [108]

Answer:

Step-by-step explanation:

$\log a=7\hspace{15px} \log b=10\hspace{15px} \log c=3\\\\ \log \dfrac{a^8c^2}{b^3}=8\log{a}+2\log{c}-3\log{b}=(8)(7)+(2)(3)-(3)(10)\\\\=56+6-30=\boxed{32}$