All categories

4 years ago

30 min before test please help !!numbers 13 and 14

Mathematics

1 answer:

anzhelika [568]4 years ago

3 0

For 13. just add all the sides.

4 + 21 + 23 + 7 + 9 = 64

For 14. multiply width and height of both shapes then add them together.

1.7 x 5.2 = 3.74 ( rectangle area )
3.0 x 3.0 divided by 2 = 4.5 ( triangle area )

Add the two answers

3.74 + 4.5 = 8.24

You might be interested in

A 40 of foxes in a wildlife preserve quadruples in size every 13 years. The function y=40•4^4, where x is the number of 13-year

SashulF [63]

Answer:

640 foxes

Step-by-step explanation:

The actual formula is $y = 40 * 4^{x}$

Now that we have the real formula we would need to find out how many 13 year periods exist in the 26 years that are being used in the question. We calculate this by simply dividing 26 by 13.

26 / 13 = 2 periods

Therefore, now that we know there are 2 periods of 13 years in the 26-year span, we plug this value into the formula and solve for y...

$y = 40 * 4^{x}$

$y = 40 * 4^{2}$

$y = 40 * 16$

$y = 640$

Finally, we can see that there should be 640 foxes after 26 years.

7 0

3 years ago

I need help on this

sdas [7]

Answer:

Step-by-step explanation:

yes

8 0

3 years ago

The surface area of this cone is 734.76 square meters. What is the slant height of this cone?

o-na [289]

Answer:

l= 234/r

Step-by-step explanation:

Given data

Surface area of cone=734.76 square meters.

The formula for the lateral surface area is

S.A=πrl

substitute the surface area

734.76 = 3.14*r*l

734.76/3.14=rl

234=rl

l= 234/r

Hence, with a value for r, the expression for the slant height l is given as

l= 234/r

5 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

8 divided 4,000 <br> 8 divided 4,000

aliina [53]

Answer:

500

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers