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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

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

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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.

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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