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3 years ago

Amanda received a $80 gift card for a coffee store. She used it in buying some coffee that cost $7.27 per pound. After

Mathematics

1 answer:

Rom4ik [11]3 years ago

5 0

Answer:

She bought 6 pounds of coffee.

Step-by-step explanation:

80-36.38=43.62

so she spent $43.62 on coffee

43.62 divided by 7.27 = 6

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The least common multiple of two numbers is 24. Which could be the two numbers?

jonny [76]

Answer:

Step-by-step explanation:

Because 4x6 is 24 that's it I think

7 0

2 years ago

If a(x) = 3x + 1 and b(x) = x2, what is the value of a(b(5))?

Airida [17]

Answer:

a(b(5)) = 76

Step-by-step explanation:

Evaluate b(5) then substitute the value obtained into a(x)

b(5) = 5² = 25, then

a(25) = 3(25) + 1 = 75 + 1 = 76

4 0

3 years ago

Write a mathematical sentence that expresses the information given below . Use t as your variable name.Necessary type <= to m

miskamm [114]

Answer:

t + 8° > 82°

Step-by-step explanation:

Given that :

Let t = temperature when Susanne checks at 11:00

If t rises by 8 more degrees, it will break the record high temperature for that day

Record high temperature for the day = 82 degree

Hence,

Temperature at 11:00 + 8° > record high temperature for the day

t + 8° > 82°

6 0

3 years ago

Radioactive Decay

SpyIntel [72]

Answer:

Percentage of (226Ra) after 900 years is 68%

Step-by-step explanation:

Let P(t) be the amount of (226Ra) present at any time t

Half life of (226 Ra) = 1599 years

If P₀ is initial amount of (226 Ra) then after 1599 years

P(1599)=P₀/2

Decay i amount of radioactive substance is related to time t as

$\frac{dP}{dt}=kP(t)\\\\\frac{1}{P}\,dP=kdt\\\\Integrating\,\, both\,\,sides\\\\ln|P|=kt+c\\\\P(t)=Ce^{kt}\\\\at \,\, t=0\,\, P(0)=P_{o}\\\\P(0)=Ce^{k0}\\\\P_{o}=C\\\\then\\\\P(t)=P_{o}e^{kt}$

To find value of k

$at\,\, t=1599\,years\\\\P(1599)=\frac{P_{o}}{2}\\\\then\\\\\frac{P_{o}}{2} =P_{o}e^{k(1599)}\\\\\frac{1}{2} =e^{k(1599)}\\\\ln|\frac{1}{2}|=k(1599)\\\\k=\frac{ln|\frac{1}{2}|}{1599}=-4.3\times 10^{-3}\\\\\implies P(t)=P_{o}e^{-4.3\times 10^{-3}t}\\\\at\,\, t=900 \\\\P(900)=P_{o}e^{-4.3\times 10^{-3}(900)}\\\\P(900)=0.68P_{o}$