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user100 [1]
4 years ago
15

What is the value of f(3) in the function f(x)=5/6x+8

Mathematics
1 answer:
natita [175]4 years ago
3 0

For this case we have a function of the form y = f (x), where f (x) = \frac {5} {6} x + 8

We must find the value of the function when x = 3, that is, we must find f (3). So:

f (3) = \frac {5} {6} (3) + 8\\f (3) = \frac {15} {6} +8\\f (3) = \frac {5} {2} +8\\f (3) = 2.5 + 8\\f (3) = 10.5

Sof (3) = 10.5

Answer:

f (3) = 10.5

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snow_lady [41]

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it is a 45-45-90 triangle and in that the two legs are equal to each other so both are 2 and then the hypotenuse is x times the square root of 2

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3 years ago
) Find AD if BC = 13x + 279.<br>AD = 2x + 135, AC = 11x + 267,<br>and BD = 63.​
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6 0
3 years ago
Hi Can you please help me
dimaraw [331]

Answer:

x = 68°

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  • given ∠CBE = 22°
  • ∠ABD is parallel to ∠CBE, thus has same angle of 22°
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5 0
3 years ago
Read 2 more answers
(A)using geometry vocabulary, describe a sequence of transformations that maps figure P (-1,2)(-1,4) (-4,2) (-4,4) onto figure Q
andrey2020 [161]

Before we proceed on determining the transformation happening on this problem, it's better to see first the location of the figure by drawing it in a cartesian coordinate plane. We have

If we observe the figures and the coordinates of the plot, we can see that there is a difference of 1 on the x coordinates of P and y coordinates of Q. Therefore, the first transformation that we consider here is the movement of figure P by 1 unit to the left. We have

\begin{gathered} P_1=(-1-1,2_{})=(-2,2) \\ P_2=(-1-1,4)=(-2,4) \\ P_3=(-4-1,2)=(-5,2) \\ P_4=(-4-1,4)=(-5,4) \end{gathered}

This transformation changes the location of figure P into

The next transformation will be the rotation of the red dotted figure on the figure above by 90 degrees counterclockwise. With this transformation, the coordinates will transform as

P_{ccw,90}=(-y,x)

Hence, for the rotation, we have the new coordinates.

\begin{gathered} P_1^{\prime}=(-2,-2) \\ P_2^{\prime}=(-4,-2) \\ P_3^{\prime}=(-2,-5) \\ P_4^{\prime}=(-4,-5) \end{gathered}

The transformed image, which is represented as NMPO, will now be at

For the last transformation, we will be reflecting the figure NMPO over the <em>y</em> axis. This changes the coordinates as

P_{\text{rotation,y}-\text{axis}}=(-x,y)

We now have the new coordinates:

\begin{gathered} P^{\doubleprime}_1=(2,-2)=Q_1_{}_{} \\ P_2^{\doubleprime}=(4,-2)=Q_3 \\ P_3^{\doubleprime}=(2,-5)=Q_2 \\ P_4^{\doubleprime}_{}=(4,-5)=Q_4_{} \end{gathered}

As you can see, they have the same coordinates as figure Q.

The mapping rules for the sequence described above are as follows:

First transformation (moving one unit to the left (x-1,y))

\begin{gathered} P_1(-1,2)\rightarrow P_1(-1-1,2)\rightarrow P_1(-2,2) \\ P_2(-1,4)\rightarrow P_1(-1-1,4)\rightarrow P_2(-2,4) \\ P_3(-4,2)\rightarrow P_1(-4-1,2)\rightarrow P_3(-5,2) \\ P_4(-4,4)\rightarrow P_1(-4-1,4)\rightarrow P_4(-5,4) \end{gathered}

Second transformation (rotation counter clockwise (-y,x))

\begin{gathered} P_1(-2,2)\rightarrow P^{\prime}_1(-2,-2)_{} \\ P_2(-2,4)\rightarrow P^{\prime}_2(-4,-2) \\ P_3(-5,2)\rightarrow P^{\prime}_3(-2,-5)_{} \\ P_4(-5,4)\rightarrow P^{\prime}_4(-4,-5)_{} \end{gathered}

Third Transformation (reflection over y-axis (-x,y))

\begin{gathered} P^{\prime}_1(-2,-2)\rightarrow P^{\doubleprime}_1(-(-2),-2)\rightarrow P^{\doubleprime}_1=(2,-2)=Q_1 \\ P^{\prime}_2(-4,-2)\rightarrow P^{\doubleprime}_1(-(-4),-2)\rightarrow P^{\doubleprime}_1=(4,-2)=Q_3 \\ P^{\prime}_3(-2,-5)\rightarrow P^{\doubleprime}_1(-(-2),-5)\rightarrow P^{\doubleprime}_1=(2,-5)=Q_2 \\ P^{\prime}_4(-4,-5)\rightarrow P^{\doubleprime}_1(-(-4),-5)\rightarrow P^{\doubleprime}_1=(4,-5)=Q_4 \end{gathered}

7 0
1 year ago
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