All categories

2 years ago

Help!!!!!!!!!!!!!!!!!!!!!!!

Mathematics

1 answer:

Nataly [62]2 years ago

4 0

Hello.

Firstly, let's isolate x on the equation:

$\mathsf{f(x) = 3x + 3} \\ \\ \mathsf{y = 3x + 3} \\ \\ \mathsf{-3x = -y + 3} \\ \\ \mathsf{x = \dfrac{-y + 3}{3}}$

Now, we can replace 'x' for 'y' and vice-versa:

$\mathsf{f^{-1}(x) = \frac{-x + 3}{3}`}$

Hope I helped.

You might be interested in

3. [25 points]. Answer each of the following questions in the context of algorithms.

Nimfa-mama [501]

Answer:

also its 25

Step-by-step explanation:

hope i helped

6 0

2 years ago

AB and BC form a right angle at point B if A = (-3,-1) and B = (4,4) what is the equation if BC

sergejj [24]

Answer:

$7x+5y-48=0$

Step-by-step explanation:

Refer the attached figure .

We are given a line AB

Point A $=(x_1,y_1) = (-3,-1)$

Point B $=(x_2,y_2) = (4,4)$

Formula for slope 'm' when two points are given :

$m=\frac{y_2-y_1}{x_2-x_1}$

Substituting thew values

$m=\frac{4-(-1)}{4-(-3)}$

$m=\frac{5}{7}$

Since we are given that AB is perpendicular to BC

So, By property - perpendicular slopes are negative reciprocals of each other.

$\Rightarrow m_1\times m_2 =-1$

$\Rightarrow \frac{5}{7}\times m_2 =-1$

$\Rightarrow m_2 = \frac{-7}{5}$

So, Slope for line BC : $m_2 = \frac{-7}{5}$

Point B = (4,4)

Formula of equation of line : $y-y_1=m(x-x_1)$

$y-4=\frac{-7}{5}(x-4)$

$5(y-4)=-7(x-4)$

$5y-20=-7x+28$

$7x+5y-20-28=0$

$7x+5y-48=0$

Hence the equation of line BC : $7x+5y-48=0$

5 0

2 years ago

Can you please answer this question

tamaranim1 [39]

$3tan^{2} \theta +7sec\theta=3$
First I converted the equation terms into sine and cosine.
$tan^{2}\theta = \frac{sin^{2}\theta}{cos^{2}\theta}$ and $sec\theta= \frac{1}{cos\theta}$
Substitution:
$\frac{3sin^2\theta}{cos^2\theta} + \frac{7}{cos\theta} =3$
Common Denominator Created:
$\frac{3sin^2\theta}{cos^2\theta} + \frac{7cos\theta}{cos^2\theta} =3$
Multiply each term by the LCD:
$3sin^2\theta+7cos\theta=3cos^2\theta$
Substitution: Recall ⇒ $sin^2\theta =1-cos^2\theta$
$3(1-cos^2\theta)+7cos\theta=3cos^2\theta$
Distribute and collect all terms on one side:
$6cos^2\theta-7cos\theta-3=0$
Factor and set each factor equal to 0:
$(2cos\theta-3)(3cos\theta+1)=0$
$2cos\theta-3=0$ ⇒ $theta=cos^{-1} \frac{3}{2}$
$3cos\theta+1=0$ ⇒ $theta=cos^{-1} \frac{-1}{3}$
The 2nd factor provides only possible answer 109.5 degrees

4 0

3 years ago

Two sides of a triangle are 8 m and 9 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Find the rat

Lelechka [254]

<h2>Rate at which area is increasing is 1.08 m²/s.</h2>

Step-by-step explanation:

Area of triangle is with side a and b and angle C between them is given by

A = 0.5 ab SinC

Here we need to find how area changes with a and b fixed and C is changing,

$\frac{dA}{dt}=\frac{d}{dt}\left (0.5 absinC\right )\\\\\frac{dA}{dt}=0.5ab\frac{d}{dt}\left (sinC\right )\\\\\frac{dA}{dt}=0.5abcosC\frac{dC}{dt}$

We have

a = 8 m

b = 9 m

$C=\frac{\pi}{3}rad\\\\\frac{dC}{dt}=0.06rad/s$

Substituting

$\frac{dA}{dt}=0.5\times 8\times 9\times cos\left ( \frac{\pi}{3}\right )\times 0.06\\\\\frac{dA}{dt}=1.08m^2/s$

Rate at which area is increasing is 1.08 m²/s.

4 0

3 years ago

3. Two quantities, x and y, are in a

frez [133]

Answer:

y = x + 1.8

Step-by-step explanation:

6 0

2 years ago