HELP I NEED HELP ASAP! Given f(x)=3(2-x), what is the value of f(-6)?

Is the number of miles to the next service station and the number of kilometers linear

evablogger [386]

I need a photo to tell you the answer

7 0

3 years ago

test the symmetric of a function f(x)=x5+x3+x plz answer my question and you will be mark as brainly

babunello [35]

Answer:

Step-by-step explanation:

2000264(X) b x a =177777082

5 0

3 years ago

Find the exact values of a) sec of theta b)tan of theta if cos of theta= -4/5 and sin<0

Gre4nikov [31]

Answer:

Using trigonometric ratio:

$\sec \theta = \frac{1}{\cos \theta}$

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

From the given statement:

$\cos \theta = -\frac{4}{5}$ and sin < 0

⇒ $\theta$ lies in the 3rd quadrant.

then;

$\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

Using trigonometry identities:

$\sin \theta = \pm \sqrt{1-\cos^2 \theta}$

Substitute the given values we have;

$\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}$

Since, sin < 0

⇒ $\sin \theta = -\frac{3}{5}$

now, find $\tan \theta$ :

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Substitute the given values we have;

$\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}$

Therefore, the exact value of:

(a)

$\sec \theta =-\frac{5}{4}$

(b)

$\tan \theta= \frac{3}{4}$

7 0

3 years ago

What is the slope of a line parallel to the line whose equation is y - x = 5 ? please explain answer

Trava [24]

The slope of a line is c. 1

6 0

3 years ago

Read 2 more answers

PLEASE HELP GUYS i am struggling so much, two questions

ankoles [38]

Answer: the equation of the standard parabola

1) $(y-6)^2 = 4 (x-1)$

The equation of the standard parabola

2) (x+5)^2 = 16(y-2)

Step-by-step explanation:

Explanation

Parabola:-

The set of points in a plane whose distance from a fixed point and a constant ratio to their corresponding perpendicular distance from a fixed straight line is a conic.

Let S be a fixed point and l be a fixed straight line from any point P,the perpendicular PM is drawn to the line 'l'

The locus of P such that $\frac{SP}{PM} = constant$
The fixed point 'S' is called the Focus.
The fixed line'l 'is called the directrix of the conic
The constant ratio is known as the eccentricity, denoted by 'e'
If e=1 , the conic is called a parabola

1) Step 1 :-

Given the focus S = (2,6) and directrix is x=0

we know that $\frac{SP}{PM}=1$

now cross multiplication , we get

$SP = PM$

squaring on both sides,we get

$SP^{2} = PM^2$

step 2:-

now using distance formula is

$\sqrt(((x_{2}-x_{1})^2+(y_{2} -y_{1} )^2)$

Given S =(2,6) and P(x,y) be any point on parabola

$SP^2 = (x-2)^2+(y-6)^2$ ........(1)

Now using perpendicular distance formula

let P(x , y ) be any point on the parabola

$\frac{ax_{1}+by_{1}+c }{\sqrt{a^2+b^2} }$

Given the directrix is x =0 and P(x,y) be any point on parabola

$PM^2 = \frac{x^2}{\sqrt{1}^2 }$ ......(2)

equating equation(1) and equation (2), on simplification

we get $(x-2)^2+(y-6)^2 = x^2$ .....(3)

apply $(a-b)^2 = a^2+ b^2+2 ab$

now the equation (3) is

$(y-6)^2 = 4 x-4$

now the standard form of parabola is

$(y-k)^2 = 4 a(x-h)$

Final answer:-

$(y-6)^2 = 4 (x-1)$

2) Explanation:-

step 1:

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

here the given points of 'x'co- ordinates are equal

Therefore the axis AS is parallel to y- axis

now the standard equation of parabola

$(x-h)^2 = 4 a (y-k)$

now you have to find' a' value

Given vertex of a parabola is A(-5,2) and its focus is S(-5,6)

The distance of $AS = \sqrt{(-5-(-5)^2+(2-6)^2}$

on simplification we get a =4

Final answer :-

the vertex (h,k) = (-5,2) and a=4

$(x-h)^2 = 4 a (y-k)$

The standard parabola is $(x+5)^2 = 16 (y-2)$

5 0

3 years ago