I’m super confused on 1/3 times 60

Mathematics

1 answer:

inna [77]2 years ago

3 0

Answer:

Step-by-step explanation:

Basically, Yout take 60 and divide it by the number 3 and you get 20. Doing the inverse operation (opposite).

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The population of a certain animal species you are studying decreases at a rate of 3.5% per year. Only 80 of the animals in the

balu736 [363]

The function:
f ( x ) = x * 0.965^t
where x is the initial amount and t is the number of the years
80 = x * 0.965^3
80 = x * 0.89632
x = 80 : 0.89632 ≈ 89
Answer:
The initial amount of animals was 89.

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If $8700 is invested at 3% annual simple interest, how much should be invested at 6% annual simple interest so that the total ye

True [87]

First off, see how much 8700 as principal, yields at 3% APR
that is $\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\qquad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$8700\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=years\to &1
\end{cases}$

it will yield some amount

subtract that amount from 393
the difference is how much the yield will be on the 6% investment
so

$\bf \qquad \textit{Simple Interest Earned}\\\\
I = Prt\quad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$8700\\
r=rate\to 3\%\to \frac{3}{100}\to &0.03\\
t=years\to &1
\end{cases}
\\\\\\
\implies \boxed{?}\\\\
-----------------------------\\\\
\textit{how much to invest at 6\%?}
\\\\\\
$

$\bf \qquad \textit{Simple Interest Earned}\\\\
(393-\boxed{?}) = Prt\quad 
\begin{cases}
I=\textit{interest earned}\\
P=\textit{original amount deposited}\to& \$\\
r=rate\to 6\%\to \frac{6}{100}\to &0.06\\
t=years\to &1
\end{cases}
\\\\\\
\textit{solve for "P", to see how much should the Principal be}\\\\
\textit{keep in mind that }P+\boxed{?}=393\leftarrow \textit{both yields added}$

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Three consecutive integers have a sum of 48. find the integers

Sholpan [36]

Your answers are 15 16 and 17 they should be right

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3 years ago

What is the smallest positive integer for x so that f(x)=200(2)* is greater than the value of g(x)=500x+400?

Goshia [24]

We have to functions, namely:

$f(x)=200(2)^{x} \ and \ g(x)=500x+400$

So the problem is asking for the smallest positive integer for $x$ so that $f(x)$ is greater than the value of $g(x)$ , that is:

$f(x)\ \textgreater \ g(x) \\ \therefore 200(2)^{x}\ \textgreater \ 500x+400$

Let's solve this problem by using the trial and error method:

$for \ x=1 \\f(1)=400 \\ g(1)=900 \\ Then \ f(1) \ \textless \ g(1) \\ \\ \\ for \ x=2 \\f(2)=800 \\ g(2)=1400\\ Then \ f(2)\ \textless \ g(2) \\ \\ \\ for \ x=3 \\f(3)=1600 \\ g(3)=1900 \\ Then \ f(3)\ \textless \ g(3) \\ \\ \\ for \ x=4 \\f(4)=3200 \\ g(4)=2400 \\ \boxed{Then \ f(4)\ \textgreater \ g(4)}$

So starting $x$ from 1 and increasing it in steps of one we find that:

$f(x)>g(x)$

when $x=4$

That is, the smallest positive integer for $x$ so that the function $f(x)$ is greater than $g(x)$ is 4.

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garik1379 [7]

Answer:

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Step-by-step explanation:

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