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jeyben [28]
3 years ago
5

F(x)=8x+10 g(x)=6x-4 what is f(4)-g(3)

Mathematics
1 answer:
In-s [12.5K]3 years ago
8 0

Answer:

f(4) - g(3) = 28

Step-by-step explanation:

f(x) = 8x + 10

f(4) = 8(4) + 10

f(4) = 32 + 10

f(4) = 42

g(x) = 6x - 4

g(3) = 6(3) - 4

g(3) = 18 - 4

g(3) = 14

f(4) - g(3)

= 42 - 14

= 28

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If you're using the app, try seeing this answer through your browser:  brainly.com/question/2862279

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Simplify the trigonometric expression.

Call the expression E:

\mathsf{E=\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}\qquad\qquad(but~~1-sin^2\,\theta=cos^2\,\theta)}\\\\\\
\mathsf{E=\dfrac{sin\,\theta}{\sqrt{cos^2\,\theta}}}\\\\\\
\mathsf{E=\dfrac{sin\,\theta}{\left|cos\,\theta\right|}}}


\mathsf{E}=\left\{\!\begin{array}{rl}\mathsf{\dfrac{sin\,\theta}{cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta>0}\\\\\mathsf{\dfrac{sin\,\theta}{-\,cos\,\theta}\,,}&\mathsf{\quad if~~cos\,\theta


So basically E is equivalent to \pm\,\mathsf{tan\,\theta}, where the sign depends on which quadrant \theta lies. Shortly,

\mathsf{\dfrac{sin\,\theta}{\sqrt{1-sin^2\,\theta}}}=\left\{\!
\begin{array}{rl}
\mathsf{tan\,\theta,}&\mathsf{\quad if~\theta~lies~either~in~the~1^{st}~or~the~4^{th}~quadrant}\\\\
\mathsf{-\,tan\,\theta,}&\mathsf{\quad if~\theta~lies~either~in~the~2^{nd}~or~the~3^{rd}~quadrant}
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I hope this helps. =)


Tags:  <em>simplify trigonometric trig expression sine cosine tangent sin cos tan absolute value modulus trigonometry</em>

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Four cards are dealt from a standard fifty-two-card poker deck. What is the probability that all four are aces given that at lea
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Answer:

The probability is 0.0052

Step-by-step explanation:

Let's call A the event that the four cards are aces, B the event that at least three are aces. So, the probability P(A/B) that all four are aces given that at least three are aces is calculated as:

P(A/B) =  P(A∩B)/P(B)

The probability P(B) that at least three are aces is the sum of the following probabilities:

  • The four card are aces: This is one hand from the 270,725 differents sets of four cards, so the probability is 1/270,725
  • There are exactly 3 aces: we need to calculated how many hands have exactly 3 aces, so we are going to calculate de number of combinations or ways in which we can select k elements from a group of n elements. This can be calculated as:

nCk=\frac{n!}{k!(n-k)!}

So, the number of ways to select exactly 3 aces is:

4C3*48C1=\frac{4!}{3!(4-3)!}*\frac{48!}{1!(48-1)!}=192

Because we are going to select 3 aces from the 4 in the poker deck and we are going to select 1 card from the 48 that aren't aces. So the probability in this case is 192/270,725

Then, the probability P(B) that at least three are aces is:

P(B)=\frac{1}{270,725} +\frac{192}{270,725} =\frac{193}{270,725}

On the other hand the probability P(A∩B) that the four cards are aces and at least three are aces is equal to the probability that the four card are aces, so:

P(A∩B) = 1/270,725

Finally, the probability P(A/B) that all four are aces given that at least three are aces is:

P=\frac{1/270,725}{193/270,725} =\frac{1}{193}=0.0052

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The correct answer would be A) X^6 Y^5
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