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3 years ago

The vertices A(-2,-1), B(-3, 2), C(-1, 3), and D(0, 0) form a parallelogram. The vertices A'(-1, -2), B'(2, -3), C'(3,-1),

Mathematics

1 answer:

attashe74 [19]3 years ago

5 0

Answer:

A is correct.

Step-by-step explanation:

The rule for this transformation is that (x,y) will become (y,x) if it flips over y=x OR the y-axis and the x-axis.

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The parent function f(x) = x2 is translated such that the function g(x) = –x2 + 6x – 5 represents the new function.

zhenek [66]

Answer:

g(x) has an axis of symmetry at x = 3.

g(x) is shifted right 3 units from the graph of f(x)

g(x) is shifted up 4 units from the graph of f(x).

Step-by-step explanation:

The parent function is:

$f(x)=x^2$

The transformed function is $g(x)=-x^2+6x-5$ .

This new function can be rewritten in the vertex form as:

$g(x)=-(x-3)^2+4$

This function is obtained by shifting the parent function 3 units right and 4 units up.

The axis of symmetry is x=3;(x=h) and h=3.

There is no horizontal stretch or compression.

The new function is however reflected in the x-axis

8 0

4 years ago

Read 2 more answers

How many different ways can the letters of “football” be arranged

makvit [3.9K]

<h3>Answer: 10,080</h3>

Explanation:

There are 8 letters so there are 8! = 8*7*6*5*4*3*2*1 = 40,320 permutations of those letters. However, the letters "O" and "L" show up twice each, so we must divide by 2! = 2*1 = 2 for each instance this happens.

So,

(8!)/(2!*2!) = (40,320)/(2*2) = (40,320)/4 = 10,080

is the number of ways to arrange the letters of "football".

The reason we divide by 2 for each instance of a duplicate letter is because we can't tell the difference between the two "O"s or the two "L"s. If there was a way to distinguish between them, then we wouldnt have to divide by 2.

3 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

Please help me asap!!

Amiraneli [1.4K]

Answer:

Step-by-step explanation:

6 0

3 years ago

In ΔABC shown below, line segment AB is congruent to line segment BC:

yan [13]

Answer:

∠ABD ≅ ∠CBD

Step-by-step explanation:

<u>Given: </u>line segment AB ≅ line segment BC

<u>Prove:</u> The base angles of an isosceles triangle are congruent.

Statement Reason

1. Segment BD is an angle bisector of ∠ABC - By construction

2. ∠ABD ≅ ∠CBD - Definition of an Angle Bisector

3. Segment BD ≅ segment BD - Reflexive Property

4. ΔABD ≅ ΔCBD - Side-Angle-Side (SAS) Postulate

5. ∠BAC ≅ ∠BCA - CPCTC

6 0

3 years ago

Read 2 more answers