All categories

2 years ago

PLEASE HELP ME Given the equation fine f(6) F(x)=-4x-1

Mathematics

2 answers:

Tcecarenko [31]2 years ago

7 0

Answer:

F(6)=-25

Step-by-step explanation:

F(x) = -4x-1

F(6) means x=6

F(6)=-4(6)-1=-24-1=-25

F(6)=-25

velikii [3]2 years ago

6 0

Answer:

-25 plug in 6 for x and minus one

Step-by-step explanation:

You might be interested in

How to solve 10(x+4)(x-4)

IceJOKER [234]

Answer:

1 Use Difference of Squares:

a^2-b^2=(a+b)(a−b).

10(x^2-4^2)

2 Simplify 4^2 to 16

10(x^2 −16)

3 Expand by distributing terms.

10x^2 −160

Step-by-step explanation:

Hope dis helps and pls mark me as brainlist!!!

§ALEX§

4 0

3 years ago

I WILL MARK 5.0 AND ANYTHING YOU WANT!! NEED THIS ASAP

Dominik [7]

Answer:

119°

Step-by-step explanation:

$m\angle BAD = 62\degree + 57\degree$

(Vertical angles)

$m\angle BAD = 119\degree$

4 0

2 years ago

What is the completely factored form of 12xy-9x-8y+6?

PilotLPTM [1.2K]

Rearrange slightly...

12xy-8y-9x+6 now factor 1st and 2nd pair of terms

4y(3x-2)-3(3x-2)

(4y-3)(3x-2)

8 0

2 years ago

Read 2 more answers

What is the minimum value of 2x + 2y in the feasible region?

Liono4ka [1.6K]

Find and graph the feasible region for the following constraints: x + y < 5. 2x<span> + y > 4 ... y = 10/3. x = 30/3 - 10/3 = 20/3. Intersects at (20/3, 10/3). -x + </span>2y<span> = 0. x - </span>2y = 0.

6 0

3 years ago

Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x

Radda [10]

Answer:

${\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}$

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is $\dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1$

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes are $y = k \pm \dfrac{b}{a}(x - h)$

1. Calculate h and k. The hyperbola is symmetric about the origin, so

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3

3. For 'b': The equation for the asymptote with the positive slope is

$y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x$

Thus, asymptote has the slope of

$\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}$

4. The equation of the hyperbola is

$\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}$

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0

2 years ago