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3 years ago

Find the radius of circle if AB = 24 and OC = 5

Mathematics

1 answer:

Likurg_2 [28]3 years ago

8 0

Answer:

Step-by-step explanation:

Since OCB forms a right triangle, you can find the length of side OB and therefore the radius by simply using the pythagorean theorem. The first step is to note that, since AB=24 and OC bisects that, that both CB and AC have length 12. Therefore, the radius is $BO=\sqrt{12^2+5^2}=13$ . Hope this helps!

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Pranav bought four candy bars for a total<br> of $12. How much did each candy bar<br> cost?

anastassius [24]

You have to divide the total cost $12 by the 4 bars, so $12/4 bars = $3 per bar

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2 years ago

sam's music teacher told him to practice his trombone for X amount of hours. he spent X amount of hours practicing did he practi

Crank

Yes if he practices X amount of hours and is told to practice X amount of hours than he practiced long enough

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3 years ago

Given the following table, determine the missing value.

sergeinik [125]

We know that
Based on the table
Percent%=[1-Decay factor]*100%
so

for decay factor=0.98
Percent%=[1-0.98]*100%----> 2%

for decay factor=0.50
Percent%=[1-0.50]*100%---->50 %

for decay factor=0.64
Percent%=[1-0.64]*100%----> 36%

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Percent%=[1-0.23]*100%----> 77%

therefore

the answer is
36%

6 0

3 years ago

Given the functionf ( x ) = x^2 + 7 x + 10/ x^2 + 9 x + 20

vladimir1956 [14]

<em>x = -4 is a vertical asymptote for the function.</em>

<h2>Explanation:</h2>

The graph of $y=f(x)$ is a vertical has an asymptote at $x=a$ if at least one of the following statements is true:

$1) \ \underset{x\rightarrow a^{-}}{lim}f(x)=\infty\\ \\ 2) \ \underset{x\rightarrow a^{-}}{lim}f(x)=-\infty \\ \\ 3) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty \\ \\ 4) \ \underset{x\rightarrow a^{+}}{lim}f(x)=\infty$

The function is:

$f(x)=\frac{x^2+7x+10}{x^2+9x+20}$

First of all, let't factor out:

$f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20} \\ \\ f(x)=\frac{x(x+5)+2(x+5)}{x(x+5)+4(x+5)} \\ \\ f(x)=\frac{(x+5)(x+2)}{(x+5)(x+4)} \\ \\ f(x)=\frac{(x+2)}{(x+4)}, \ x\neq 5$

From here:

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ right: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(-4^{+}+2)}{(-4^{+}+4)} \\ \\ \\ The \ numerator \ is \ negative \ and \ the \ denominator \\ is \ a \ small \ positive \ number. \ So: \\ \\ \underset{x\rightarrow -4^{+}}{lim}\frac{(x+2)}{(x+4)}=-\infty$

$\bullet \ When \ x \ approaches \ -4 \ on \ the \ left: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=? \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(-4^{-}+2)}{(-4^{-}+4)} \\ \\ \\ The \ numerator \ is \ a \ negative \ and \ the \ denominator \\ is \ a \ small \ negative \ number \ too. \ So: \\ \\ \underset{x\rightarrow -4^{-}}{lim}\frac{(x+2)}{(x+4)}=+\infty$

Accordingly:

$x=-4 \ is \ a \ vertical \ asymptote \ for \\ \\ f(x)=\frac{x^2+5x+2x+10}{x^2+5x+4x+20}$

<h2>Learn more:</h2>

Vertical and horizontal asymptotes: brainly.com/question/10254973

#LearnWithBrainly

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3 years ago

Triangle ABC has vertices at A(3, 8) , B(11, 8) , and C(9, 12) . Triangle FGH has vertices at F(1, 3) , G(9, 3) , and H(7, 7) .

Alex17521 [72]

Hmm, (x,y)
x is horizontal or left right
y is vertical or up down

so

we see that A to F is move left 2 and down 5
B to G is left 2 and down 5
C to H is left 2 and down 5

so translatete ABC left 2 and down 5 and
translate FGH right 2 and up 5

not 1st option
2nd works
3rd works
4th is false

2nd and 3rd

6 0

3 years ago

Read 2 more answers