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alekssr [168]
3 years ago
5

Manuel is making cookies for the holidays.each batch of cookies he will make requires 3/4 cup of sugar. He has a total of 2 5/8

cups of sugar
Mathematics
1 answer:
Alchen [17]3 years ago
7 0

Answer:

3 1/2 batch of cookies

Step-by-step explanation:

(2 5/8) / (3/4) = (21/8) / (3/4) = (21/8) * (4/3) = (7/2) * (1/1) = 7/2 = 3 1/2 batch of cookies

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3 years ago
How do I type in a equation
Gnesinka [82]

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6 0
3 years ago
The half-life of a radioactive substance is the time it takes for a quantity of the substance to decay to half of the initial am
posledela

Step-by-step walkthrough:

a.

Well a standard half-life equation looks like this.

N = N_0 * (\frac{1}{2})^{t/p

N_0 is the starting amount of parent element.

N is the end amount of parent element

t is the time elapsed

p is a half-life decay period

We know that the starting amount is 74g, and the period for a half-life is 2.8 days.

Therefore you can create a function based off of the original equation, just sub in the values you already know.

N(t) = 74g * (\frac{1}{2})^{t/2.8days

b.

This is easy now that we have already made the function. Here we just reuse it, but plug in 2.8 days.

N(t) = 74g * (\frac{1}{2})^{t/2.8days} = N(2.8days) = 74g * (\frac{1}{2})^{2.8days/2.8days}\\= 74g * \frac{1}{2}  =  37g

c.

Now we just gotta do some algebra. Use the original function but this time, replace N(t) with 10g and solve algebraically.

10g = 74g * (\frac{1}{2})^{t/2.8days}\\\\\frac{10g}{74g} = (\frac{1}{2})^{t/2.8days}

Take the log of both sides.

log(\frac{5}{37}) = log((\frac{1}{2})^{t/2.8days})

Use the exponent rule for log laws that, log(b^x) = x*log(b)

log(\frac{5}{37}) = \frac{t}{2.8days} * log(\frac{1}{2})

\frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = \frac{t}{2.8days}

2.8 * \frac{log(\frac{5}{37})}{log(\frac{1}{2})}  = t

slap that in your calculator and you get

t = 8.1 days

7 0
2 years ago
What is the solution to the system of equations below
nekit [7.7K]

Answer:

There is no solution.

6 0
2 years ago
Find sec theta if theta is in quadrant 4 and sin theta= -1/5
lara31 [8.8K]

Answer:

\frac{5}{2\sqrt{6} }

Step-by-step explanation:

Since Θ is in fourth quadrant then cosΘ > 0, as is secΘ

Given

sinΘ = - \frac{1}{5}, then

cosΘ = \sqrt{1-(-1/5)^2}

         = \sqrt{1-\frac{1}{25} } = \sqrt{\frac{24}{25} } = \frac{2\sqrt{6} }{5}

Hence

secΘ = \frac{1}{\frac{2\sqrt{6} }{5} } = \frac{5}{2\sqrt{6} }

5 0
3 years ago
Read 2 more answers
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