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3 years ago

Evaluate the expression. 42+5⋅32−42÷23 CLEAR CHECK 12.625 222 45 59

Mathematics

1 answer:

lidiya [134]3 years ago

5 0

Answer:

$\frac{4604}{23}$ or 200.173913

Step-by-step explanation:

$42+(5)(32)-\frac{42}{23} \\\\=42+160-\\\frac{42}{23}\\\=202-\frac{42}{23} =\frac{4604}{23}$

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Select the correct answer.

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3 years ago

A carrot has 55 grams of sugar per serving. A banana has 15 grams of sugar per serving. Which statement is true?

GREYUIT [131]

For an equal number of servings, the bananas will have three times as much sugar.

To put it in perspective:
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3 years ago

A museum sold 95 tickets for $1,025. Each adult ticket was $15, and each child ticket was $7. Let a represent the number of adul

Genrish500 [490]

Answer:

a + c = 95........eq1

15a + 7c = 1,025....eq2

Step-by-step explanation:

Given:

Total number of ticket sold = 95

Total amount collected = $1,025

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Number of adult tickets sold = a

Number of child tickets sold = c

So,

Number of adult tickets sold + Number of child tickets sold = Total number of ticket sold

a + c = 95........eq1

and

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3 years ago

Ben goes to the grocery store at a rate of 4 times a week. How many times would he be expected to

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3 years ago

Read 2 more answers

Atmospheric pressure decreases by about 11.8% for every 1000 meters you climb. The pressure at sea level is 1013 millibars (a un

erica [24]

At sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see

$\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I$

so, the inital amount is 1013, when t = 0,

$\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}$

now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.

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3 years ago