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Rashid [163]
2 years ago
11

Which ordered pair would form a proportional relationship with the point graphed below? 40 (-20,40) 30 20 10 -40-30-20-190 10 20

30 40 x -20 -30 -40 0 (-40, 20) 0 (-10.-20) O (15, -30) O (5, -15)​

Mathematics
1 answer:
sasho [114]2 years ago
6 0

Answer:

(15,-30)

Step-by-step explanation:

By forming a proportional relationship with the marked point, what we are saying is that the same dilation if applied to each coordinates of the marked point will form the new point

The marked point is given as (-20,40)

If we dilate this by -0.75 units

That will give (15, -30)

So this forms a proportional relationship with the marked point

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Answer:

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3 years ago
If 13cos theta -5=0 find sin theta +cos theta / sin theta -cos theta​
Ivahew [28]

Step-by-step explanation:

<h3>Need to FinD :</h3>

  • We have to find the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0.

\red{\frak{Given}} \begin{cases} & \sf {13\ cos \theta\ -\ 5\ =\ 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \big\lgroup Can\ also\ be\ written\ as \big\rgroup} \\ & \sf {cos \theta\ =\ {\footnotesize{\dfrac{5}{13}}}} \end{cases}

Here, we're asked to find out the value of (sinθ + cosθ)/(sinθ - cosθ), when 13 cosθ - 5 = 0. In order to find the solution we're gonna use trigonometric ratios to find the value of sinθ and cosθ. Let us consider, a right angled triangle, say PQR.

Where,

  • PQ = Opposite side
  • QR = Adjacent side
  • RP = Hypotenuse
  • ∠Q = 90°
  • ∠C = θ

As we know that, 13 cosθ - 5 = 0 which is stated in the question. So, it can also be written as cosθ = 5/13. As per the cosine ratio, we know that,

\rightarrow {\underline{\boxed{\red{\sf{cos \theta\ =\ \dfrac{Adjacent\ side}{Hypotenuse}}}}}}

Since, we know that,

  • cosθ = 5/13
  • QR (Adjacent side) = 5
  • RP (Hypotenuse) = 13

So, we will find the PQ (Opposite side) in order to estimate the value of sinθ. So, by using the Pythagoras Theorem, we will find the PQ.

Therefore,

\red \bigstar {\underline{\underline{\pmb{\sf{According\ to\ Question:-}}}}}

\rule{200}{3}

\sf \dashrightarrow {(PQ)^2\ +\ (QR)^2\ =\ (RP)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ (5)^2\ =\ (13)^2} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ +\ 25\ =\ 169} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 169\ -\ 25} \\ \\ \\ \sf \dashrightarrow {(PQ)^2\ =\ 144} \\ \\ \\ \sf \dashrightarrow {PQ\ =\ \sqrt{144}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{PQ\ (Opposite\ side)\ =\ 12}}}}_{\sf \blue{\tiny{Required\ value}}}}

∴ Hence, the value of PQ (Opposite side) is 12. Now, in order to determine it's value, we will use the sine ratio.

\rightarrow {\underline{\boxed{\red{\sf{sin \theta\ =\ \dfrac{Opposite\ side}{Hypotenuse}}}}}}

Where,

  • Opposite side = 12
  • Hypotenuse = 13

Therefore,

\sf \rightarrow {sin \theta\ =\ \dfrac{12}{13}}

Now, we have the values of sinθ and cosθ, that are 12/13 and 5/13 respectively. Now, finally we will find out the value of the following.

\rightarrow {\underline{\boxed{\red{\sf{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}}}}}}

  • By substituting the values, we get,

\rule{200}{3}

\sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\Big( \dfrac{12}{13}\ +\ \dfrac{5}{13} \Big)}{\Big( \dfrac{12}{13}\ -\ \dfrac{5}{13} \Big)}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ {\footnotesize{\dfrac{\dfrac{17}{13}}{\dfrac{7}{13}}}}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{13} \times \dfrac{13}{7}} \\ \\ \\ \sf \dashrightarrow {\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{\cancel{13}} \times \dfrac{\cancel{13}}{7}} \\ \\ \\ \dashrightarrow {\underbrace{\boxed{\pink{\frak{\dfrac{sin \theta\ +\ cos \theta}{sin \theta\ -\ cos \theta}\ =\ \dfrac{17}{7}}}}}_{\sf \blue{\tiny{Required\ value}}}}

∴ Hence, the required answer is 17/7.

6 0
2 years ago
I need help with my pre-calculus homework, please show me how to solve them step by step if possible. The image of the problem i
Anna [14]

We know that the law of sines states that:

\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}

For simplicity, let:

\beta=m\angle A_1BC

In triangle A1BC this leads to:

\begin{gathered} \frac{\sin31}{5}=\frac{\sin\beta}{6} \\ \sin\beta=\frac{6}{5}\sin31 \\ \beta=\sin^{-1}(\frac{6}{5}\sin31) \\ \beta=38.174 \end{gathered}

Therefore:

m\angle A_1BC=38.174

Now, triangle A2BC is isosceles which means that both the base angles are equal, since angle CA1B and A2A1B are supplementary we have:

m\angle CA_2B=180-38.174=141.826

8 0
1 year ago
A cereal company's cost, in thousands of dollars, is represented by the function C(x)=2x+4500 and its revenue, in thousands of d
jekas [21]

Given , revenue function : R(x)= 5x

Cost function C(x) = 2x+4500

Profit  = Revenue - Cost

Hence, P = R(x) - C(x)

Total revenue for x = 12000 is R(12000) = 5(12000) = 60000

Total cost for x = 12000, C(12000) = 2(12000)+4500 = 28500

Profit = 60000-28500 = 31500

Answer = 31500

6 0
2 years ago
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goldfiish [28.3K]
Factor it out first so:

y=(x+13)(x-2)

Then you know that x= -13 and 2 to make the equation 0

Hope this helps!
4 0
2 years ago
Read 2 more answers
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