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3 years ago

Solve the inequality for y.

Mathematics

2 answers:

Pepsi [2]3 years ago

4 0

Answer:

Isolate the variable by dividing each side by factors that don't contain the variable.

Inequality Form:

y≥9

Interval Notation:

(9,∞)

hope this helps

dsp733 years ago

3 0

63 <= 7y

To solve for y just divide both sides by 7:

Y <= 63/7

Y <= 9

Because 63 is less than , you need to reverse the inequality sign.

The answer becomes y >=9

Now when you replace y with a number from 9 or higher the inequality will be true.

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Assuming all model requirements for conducting the appropriate procedure have beenâ€‹ satisfied, is the mean IQ of the students

Triss [41]

Answer:

Step-by-step explanation:

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3 years ago

Aisha bought 15 apples and oranges for $9.00. Each orange costs $0.50, and each apple costs $0.65. Let x represent the number of

Alecsey [184]

I might be wrong but i think its
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3 years ago

Read 2 more answers

Megan took a cab ride and was charged a fare of $1.50 per mile and an additional charge of $5 for her luggage. If Megan traveled

____ [38]

For example let's say Megan has to pay the cab driver $50. Here's the equation 50=1.50x + 5 First you want to get x by itself on one side. So subtract 5 on both sides 50=1.50x + 5 -5 -5 --------------------------- 45=1.50x Now that you have x on one side; however, you want to divide to get the answer. 45=1.50x ----------------------- 1.50 x = 30 So if Megan pay $50 that mean she rode 30 miles ~by the way I'm sorry if you are looking at the answers and it isn't in the right place, I'm answering this on a tablet~

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2 years ago

Find the value of x.<br> 108°<br> 40°<br> x = [ ? 1°

tensa zangetsu [6.8K]

Answer:

x = 28°

Step-by-step explanation:

Take angle on a straight line which is equal to 180°

180° =(108 + 44 + x)°

180° = 152° + x°

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3 years ago

Find the value of $ 15,000 at the end of one year if it is invested in an account that has an interest rate of 4.95 % and is com

lora16 [44]

A)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{twelve months, thus}
\end{array}\to &12\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{12}\right)^{12\cdot 1}$

b)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{365 days, thus}
\end{array}\to &365\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{365}\right)^{365\cdot 1}$

c)

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$15000\\
r=rate\to 4.95\%\to \frac{4.95}{100}\to &0.0495\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{four quarters, thus}
\end{array}\to &4\\
t=years\to &1
\end{cases}
\\\\\\
A=15000\left(1+\frac{0.0495}{4}\right)^{4\cdot 1}$

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3 years ago