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2 years ago

F(x) = 3x^2 +12x+7 in standard form

Mathematics

1 answer:

ivann1987 [24]2 years ago

7 0

Answer:

y = ax^2 + bx + c => F(x) = 3x^2 +12x+7; F(x) = 3x^2 +12x+7 is already in standard form for the equation of a parabola.

Step-by-step explanation:

Actually, F(x) = 3x^2 +12x+7 is already in standard form, y = ax^2 + bx + c. Double check the instructions for this problem. Did you perhaps want "vertex form?"

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Which of the following is the domainof the given relation?{(-6, 3), (-4 ,5), (0, 0)}A {0, 4, -6}B {0, 3, 5)C {3, 4, 0}D {-6, -4,

Bas_tet [7]

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11 months ago

If (14,25) and (9,-10) are two anchor points on a trend line, then find the equation of the line.

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Step-by-step explanation:

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3 years ago

arlik [135]

Given:

In a right angle triangle θ is an acute angle and $\tan\theta =\dfrac{3}{5}$ .

To find:

The value of $\cos \theta$ .

Solution:

In a right angle triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

We have,

$\tan\theta =\dfrac{3}{5}$

It means the ratio of perpendicular to base is 3:5. Let 3x be the perpendicular and 5x be the base.

By using Pythagoras theorem,

$Hypotenuse=\sqrt{Perpendicular^2+base^2}$

$Hypotenuse=\sqrt{(3x)^2+(5x)^2}$

$Hypotenuse=\sqrt{9x^2+25x^2}$

$Hypotenuse=\sqrt{34x^2}$

$Hypotenuse=x\sqrt{34}$

In a right angle triangle,

$\cos \theta=\dfrac{Base}{Hypotenuse}$

$\cos \theta=\dfrac{5x}{x\sqrt{34}}$

$\cos \theta=\dfrac{5}{\sqrt{34}}$

Therefore, the value of $\cos \theta$ is $\dfrac{5}{\sqrt{34}}$ .

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2 years ago