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2 years ago

What is the length of the major axis of the ellipse (x-7)^2/4+(y+3)^2/16=1

1 answer:

6 0

$\bf \begin{array}{llll}
\cfrac{(x-{{ h}})^2}{{{ a}}^2}+\cfrac{(y-{{ k}})^2}{{{ b}}^2}=1\\\\ \cfrac{(x-{{ h}})^2}{{{ b}}^2}+\cfrac{(y-{{ k}})^2}{{{ a}}^2}=1
\end{array}\quad 
\begin{cases}
\textit{major axis}=a+a\\
a=\textit{the larger denominator}\\
b=\textit{smaller denominator}
\end{cases}\\\\
-----------------------------\\\\
\cfrac{(x-7)^2}{4}+\cfrac{(y+3)^2}{16}=1\implies \cfrac{(x-7)^2}{2^2}+\cfrac{(y+3)^2}{4^2}=1\quad 
\begin{cases}
a=4\\
b=2
\end{cases}$

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ABC DEF AB corresponds to DF. True False

Alex Ar [27]

it's false

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3 years ago

Read 2 more answers

In the figure, triangle CAE is an enlargement of triangle CBD with scale factor of 4/3. The area of the smaller triangle is 9cm2

balandron [24]

Answer:

16 cm^2

Step-by-step explanation:

Given

$\triangle CAE$ -- Bigger Triangle

$\triangle CBD$ -- Smaller Triangle

$k = \frac{4}{3}$ --- Scale factor

Area of CBD = 9

Required

Determine the area of CAE

The area of triangle CBD is:

$A_1 = \frac{1}{2}bh$

$\frac{1}{2}bh = 9$

The area of CAE is:

$A_2 = \frac{1}{2}BH$

Where:

$B = \frac{4}{3}b$ and

$H = \frac{4}{3}h$

The above values is the dimension of the larger triangle (after dilation).

So, we have:

$A_2 = \frac{1}{2}*\frac{4}{3}b * \frac{4}{3} * h$

$A_2 = \frac{1}{2}*\frac{4}{3} * \frac{4}{3} *b* h$

$A_2 = \frac{1}{2}*\frac{16}{9} *b* h$

Re-order

$A_2 = \frac{16}{9}*\frac{1}{2}* b* h$

$A_2 = \frac{16}{9}*\frac{1}{2}bh$

Recall that:

$\frac{1}{2}bh = 9$

$A_2 = \frac{16}{9}*9$

$A_2 = 16$

Hence, the area is 16 cm^2

3 0

3 years ago

What’s the answer plzzzzzzzz?

maks197457 [2]

Step-by-step explanation:

please forgive me if I have done something wrong there I am in a hurry I have to go ccooking. if there's something wrong there you can tell me I check it out when I come back good luck.