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Inessa05 [86]
3 years ago
9

If f(1) = 10, what is f(3)?

Mathematics
1 answer:
Margarita [4]3 years ago
5 0

Answer: f(3)=30 my good sir

Step-by-step explanation:

I honestly just guessed

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Find the equation of the parabola with its focus at (5,0) and it’s directrix y=2
Verizon [17]

Answer:

Option A

Step-by-step explanation:

From the question given in the picture attached,

Directrix of the parabola → y = 2

Focus of the parabola → (-5, 0)

Since, focus is below the directrix, parabola will open open downwards.

Equation of the parabola will be in the form of,

y - k = -4p(x - h)²

Here, (h, k) is the vertex of the parabola

p = Distance between vertex and focus or distance between vertex and directrix

Since, distance between vertex and focus = Distance between vertex and directrix

Therefore, vertex of the parabola → (-5, 1)

Now distance (p) between vertex (-5, 0) and focus (-5, 1) = 1 unit

By substituting these values in the equation,

y - 1 = -4(1)[x - (-5)]²

y - 1 = -4(x + 5)²

y = -4(x + 5)² + 1

Option A will be the answer.

3 0
3 years ago
8 2/10- 5 7/10 I really need to know the answer to this
Darina [25.2K]

Answer:

\frac{5}{2}

Step-by-step explanation:

8\frac{2}{10}  -  5\frac{7}{10} = \frac{5}{2}

6 0
3 years ago
What does negative six times three equal
Sergio [31]
It equals -18
Hope this helps
8 0
4 years ago
Read 2 more answers
1, The perimeter of a square is 60 cm. Find the length of each side and its area.
Nitella [24]

Answer:

length of each side = 15cm

area = 225cm²

Step-by-step explanation:

4 0
3 years ago
Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.
leonid [27]

<u>Answer-</u>

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

12x-5y+14=0 \\ 12x-5y-14=0

<u>Solution-</u>

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1

\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1

\Rightarrow 12h-5k-1=\pm 13

\Rightarrow 12h-5k=\pm 14

\Rightarrow 12h-5k=14,\ 12h-5k=-14

\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0

\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0

Two equations are formed because one will be upper from the the given line and other will be below it.

4 0
3 years ago
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