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Greeley [361]
3 years ago
6

In the figure, BD¯¯¯¯¯ bisects ∠ABC .

Mathematics
1 answer:
pashok25 [27]3 years ago
5 0
The bisector BD is perpendicular to the line segment AC then the angle formed in between is 90°. We can solve for DC using SOH CAH TOA theorem or Pythagorean theorem.
Since ∠BDA is 90°, we can examine that AD is equal to 2units. Proving this, we have BD^2=BD^2+AD^2.
3^2=DB^2+2^2
DB=5units
Therefore, DC=AC-AD and the answer is 2 units.

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A) chaim used an invalid reason to justify the congruence of a pair of sides or angles.
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Answer:

C

Step-by-step explanation:

Side Side Angle does not exist.

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3 years ago
Larry scored 10 points less than 3 times the number of points that Ross scored. Larry scored 10 points. How many points did Ross
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The answer is 20. you'll have to multiply 10 times 3 and get 30. Then you'll subtract the 10 from the 30 and get 20
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A circle has the order pairs (-1, 2) (0, 1) (-2, -1) what is the equation . Show your work.
olga55 [171]
We know that:

(x-a)^2+(y-b)^2=r^2

is an equation of a circle.

When we substitute x and y (from the pairs we have), we'll get a system of equations:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}

and all we have to do is solve it for a, b and r.

There will be:

\begin{cases}(-1-a)^2+(2-b)^2=r^2\\(0-a)^2+(1-b)^2=r^2\\(-2-a)^2+(-1-b)^2=r^2\end{cases}\\\\\\
\begin{cases}1+2a+a^2+4-4b+b^2=r^2\\a^2+1-2b+b^2=r^2\\4+4a+a^2+1+2b+b^2=r^2\end{cases}\\\\\\
\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\\\\\


From equations (II) and (III) we have:

\begin{cases}a^2+b^2-2b+1=r^2\\a^2+b^2+4a+2b+5=r^2\end{cases}\\--------------(-)\\\\a^2+b^2-2b+1-a^2-b^2-4a-2b-5=r^2-r^2\\\\-4a-4b-4=0\qquad|:(-4)\\\\\boxed{-a-b-1=0}

and from (I) and (II):

\begin{cases}a^2+b^2+2a-4b+5=r^2\\a^2+b^2-2b+1=r^2\end{cases}\\--------------(-)\\\\a^2+b^2+2a-4b+5-a^2-b^2+2b-1=r^2-r^2\\\\2a-2b+4=0\qquad|:2\\\\\boxed{a-b+2=0}

Now we can easly calculate a and b:

\begin{cases}-a-b-1=0\\a-b+2=0\end{cases}\\--------(+)\\\\-a-b-1+a-b+2=0+0\\\\-2b+1=0\\\\-2b=-1\qquad|:(-2)\\\\\boxed{b=\frac{1}{2}}\\\\\\\\a-b+2=0\\\\\\a-\dfrac{1}{2}+2=0\\\\\\a+\dfrac{3}{2}=0\\\\\\\boxed{a=-\frac{3}{2}}

Finally we calculate r^2:

a^2+b^2-2b+1=r^2\\\\\\\left(-\dfrac{3}{2}\right)^2+\left(\dfrac{1}{2}\right)^2-2\cdot\dfrac{1}{2}+1=r^2\\\\\\\dfrac{9}{4}+\dfrac{1}{4}-1+1=r^2\\\\\\\dfrac{10}{4}=r^2\\\\\\\boxed{r^2=\frac{5}{2}}

And the equation of the circle is:

(x-a)^2+(y-b)^2=r^2\\\\\\\left(x-\left(-\dfrac{3}{2}\right)\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}\\\\\\\boxed{\left(x+\dfrac{3}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{5}{2}}
7 0
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What is a formula for the nth term of the given sequence?<br> 36, 24, 16...
SpyIntel [72]

Answer:

The formula to find the nth term of the given sequence is 54 · \frac{2}{3} ^{n}

Step-by-step explanation:

The formula for nth term of an geometric progression is :

a_{n} = \frac{a_{1}(r^{n})}{r}

In this example, we have a_{1} = 36 (the first term in the sequence) and

r = \frac{2}{3} (the rate in which the sequence is changing).

Knowing what the values for r and a_{1} are, now we can solve.

a_{n} = \frac{a_{1}(r^{n})}{r} = \frac{36 (\frac{2}{3} ^{n}) }{\frac{2}{3} } = 54 · \frac{2}{3} ^{n}

Therefore, the formula to find the nth term of the given sequence is

54 · \frac{2}{3} ^{n}

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gregori [183]
The answer is 243 for the question above
5 0
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