All categories

2 years ago

How can I tell when figures are similar?

2 answers:

8 0

Two figures that have the same shape are said to be similar. When two figures are similar, the ratios of the lengths of their corresponding sides are equal.

Anestetic [448]2 years ago

4 0

Two figures that have the same shape are said to be similar. When two figures are similar, the ratios of the lengths of their corresponding sides are equal. To determine if the triangles below are similar, compare their corresponding sides.

You might be interested in

What measurement is equelvent to 3 meters

Lelu [443]

Answer:

The Chi and Zhang were units of length equivalent to approximately 25 centimeters (9.8 inches) and 3 meters (9.8 feet) respectively.

Step-by-step explanation: 300 centimeters if you need metric measurements.

4 0

2 years ago

Jake cuts out five-pointed stars. They are different sizes, but they

Verdich [7]

Answer:

Step-by-step explanation: 12L + 1/2 = 72.5

72.5 - 24 - 8.5 + 20

72.5 -24 = 48.5 - 8.5 = 40

40 + 20 = 60

3 0

2 years ago

Is 3y^2-2y already simplified

geniusboy [140]

Yes it is already simplified because there is nothing else that you are able to do to it.

4 0

3 years ago

Read 2 more answers

manuel deposits $10000 for 12 yr in an account paying 4% compounded annually.He then puts this total amount on deposit in anothe

Naddik [55]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
\quad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$10000\\
r=rate\to 4\%\to \frac{4}{100}\to &0.04\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{annually, thus once}
\end{array}\to &1\\
t=years\to &12
\end{cases}
\\\\\\
A=10000\left(1+\frac{0.04}{1}\right)^{1\cdot 12}\implies A=1000(1.04)^{12}\\\\\\ A\approx 16010.32$

he then turns around and grabs that money and sticks it for another 9 years,

$\bf ~~~~~~ \textit{Compound Interest Earned Amount}
\\\\
A=P\left(1+\frac{r}{n}\right)^{nt}
~~
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to &\$16010.32\\
r=rate\to 5\%\to \frac{5}{100}\to &0.05\\
n=
\begin{array}{llll}
\textit{times it compounds per year}\\
\textit{semi-annually, thus twice}
\end{array}\to &2\\
t=years\to &9
\end{cases}
\\\\\\
A=16010.32\left(1+\frac{0.05}{2}\right)^{2\cdot 9}\implies A=16010.32(1.025)^{18}
\\\\\\
A\approx 24970.64$

add both amounts, and that's how much is for the whole 21 years.

6 0

3 years ago

CALCULUS: Determine which function is a solution to the differential equation y ' − y = 0.

Montano1993 [528]

C: none of these are solutions to the given equation.

• If y(x) = e², then y is constant and y' = 0. Then y' - y = -e² ≠ 0.

• If y(x) = x, then y' = 1, but y' - y = 1 - x ≠ 0.

The actual solution is easy to find, since this equation is separable.

y' - y = 0

dy/dx = y

dy/y = dx

∫ dy/y = ∫ dx

ln|y| = x + C

y = exp(x + C )

y = C exp(x) = C eˣ

8 0

2 years ago