Can someone please help me make a story i need it done tonight!!

Mathematics

1 answer:

bezimeni [28]3 years ago

4 0

For the first;
For every four marbles, Izuku placed 2x more in the jar.

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A quadrilateral has two right angles. The measure of the third angle is 99 degrees. What is the measure of the fourth angle.

kogti [31]

since it is a quadrilateral, all the sides will add to 360. If two angles are 90, and one is 99, add them and subtract from 360. The last angle should be 81 degrees.

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3 years ago

Read 2 more answers

Please answer and explain!!!

meriva

When you have f(3), that means X is equal to 3.

The problem states if x <-3 then f(x) would be -2.
If x is between -1 and -3, then f(x) would be 4x.
If x is greater than -1, f(x) would be x^2.

3 is greater then -1, so f(3) would be x^2, now replace x with 3 to get 3^2, which equals 9.

f(3) = 9

6 0

3 years ago

At an interest rate of 8% compounded annually, how long will it take to double the following investments?

Paladinen [302]

Let's see, if the first one has a Principal of $50, when it doubles the accumulated amount will then be $100,

recall your logarithm rules for an exponential,

$\bf \textit{Logarithm of exponentials}\\\\
log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-------------------------------\\\\
\qquad \textit{Compound Interest Earned Amount}
\\\\
$

$\bf A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$100\\
P=\textit{original amount deposited}\to &\$50\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
100=50\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 100=50(1.08)^t
\\\\\\
\cfrac{100}{50}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
$

$\bf log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------\\\\
$

now, for the second amount, if the Principal is 500, the accumulated amount is 1000 when doubled,

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$1000\\
P=\textit{original amount deposited}\to &\$500\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}
\\\\\\
1000=500\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 1000=500(1.08)^t
\\\\\\
$

$\bf \cfrac{1000}{500}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t\\\\
-------------------------------$

now, for the last, Principal is 1700, amount is then 3400,

$\bf \qquad \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\to &\$3400\\
P=\textit{original amount deposited}\to &\$1700\\
r=rate\to 8\%\to \frac{8}{100}\to &0.08\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annnually, thus once}
\end{array}\to &1\\
t=years
\end{cases}$

$\bf 3400=1700\left(1+\frac{0.08}{1}\right)^{1\cdot t}\implies 3400=1700(1.08)^t
\\\\\\
\cfrac{3400}{1700}=1.08^t\implies 2=1.08^t\implies log(2)=log(1.08^t)
\\\\\\
log(2)=t\cdot log(1.08)\implies \cfrac{log(2)}{log(1.08)}=t\implies 9.0065\approx t$

8 0

4 years ago