All categories

3 years ago

Write an equation of the line that is perpendicular to the line y =

Mathematics

1 answer:

pogonyaev3 years ago

6 0

Answer:

c) y= -3x + 3

Step-by-step explanation:

Given equation

y= 1/3x+6

Here slope m1 = 1/3

We know that the slope of perpendicular lines

m1*m2 = -1

slope of perpendicular line m2 = -1/m1

= -1/ (1/3)

m2 = -3

Also the line passes through (2, -3)

equation of line is given by y = mx+b

y= -3

x= 2

m2 = -3

Applying above values to the equation of line,

-3 = (-3*2)+b

b=3

Hence equation of perpendicular line is y = -3x + 3

You might be interested in

Solve x to y i am very confused

ohaa [14]

Answer:

Step-by-step explanation:

8 0

3 years ago

Solve step by step solution then only i can do it plxx

Anuta_ua [19.1K]

Answer:

<h3>Let's understand the concept:-</h3>

Here angle B is 90°

So $\triangle ABC$ and $\triangle ABD$ Are right angled triangle

So we use Pythagoras thereon for solution

<h3>Required Answer:-</h3>

First in triangle ABC

perpendicular=p=8cm

Hypontenuse =h =10cm

We need to find base=b

According to Pythagoras thereon

${\boxed{\sf b^2=h^2-p^2}}$

Substitutethe values

$\longrightarrow$ $\sf b^2=10^2-p^2$

$\longrightarrow$ $\sf b={\sqrt {10^2-8^2}}$

$\longrightarrow$ $\sf b={\sqrt{100-64}}$

$\longrightarrow$ $\bf b={\sqrt {36}}$

$\longrightarrow$ $\sf b=6$

$\therefore$ $\overline{BC}=6cm$

BD=BC+CD

$\longrightarrow$ $BD=9+6$

$\longrightarrow$ $BD=15cm$

Now in $\triangle ABD$

Perpendicular=p=8cm

Base =b=15cm

We need to find Hypontenuse =AD(x)

According to Pythagoras thereon

${\boxed {\sf h^2=p^2+b^2}}$

Substitute the values

$\longrightarrow$ $\sf h^2=8^2+15^2$

$\longrightarrow$ $\sf h={\sqrt {8^2+15^2}}$

$\longrightarrow$ $\sf h={\sqrt {64+225}}$

$\longrightarrow$ $\sf h={\sqrt {289}}$

$\longrightarrow$ $\sf h=17cm$

$\therefore$ ${\underline{\boxed{\bf x=17cm}}}$

3 0

3 years ago

Monica’s retirement party cost two dollars, plus an additional three dollars for every guess she invites. If there are 16 guests

lesya692 [45]

Answer:

TWO BUCKS?? ballin on a budget

anyways, if there are 16 guests, the party would cost $50. sounds a lot more correct lol

8 0

3 years ago

| x-3 | < x-3 can someone give a step by step process on how to do this?

statuscvo [17]

Answer:

There are no solutions to the inequality.

Step-by-step explanation:

|x - 3| < x – 3

1. Separate the inequality into two separate ones.

(1) x – 3 < x – 3

(2) x – 3 < -(x – 3)

2. Solve each equation separately

(a) Equation (1)

$\begin{array}{rcl}x - 3 & < & x - 3\\x & < & x\\\end{array}\\\text{This is impossible. No solutions exist.}$

(b) Equation (2)

$\begin{array}{rcl}x - 3 & < & -(x - 3)\\x - 3 & < & -x + 3\\x &$

For example, if x = 0, we get

|0 - 3| < 0 - 3 or

3 < -3

4 0

3 years ago

The half-life of the isotope Osmium-183 is 12 hours. Choose the equation below that gives the remaining mass of Osmium-183 in gr

raketka [301]

The given equations are incomprehensible, I'm afraid...

You're given that osmium-183 has a half-life of 12 hours, so for some initial mass M₀, the mass after 12 hours is half that:

1/2 M₀ = M₀ exp(12k)

for some decay constant k. Solve for this k :

1/2 = exp(12k)

ln(1/2) = 12k

k = 1/12 ln(1/2) = - ln(2)/12

Now for some starting mass M₀, the mass M remaining after time t is given by

M = M₀ exp(kt )

So if M₀ = 590 g and t = 36 h, plugging these into the equation with the previously determined value of k gives

M = 590 exp(36k) = 73.75

so 73.75 ≈ 74 g of Os-183 are left.

Alternatively, notice that the given time period of 36 hours is simply 3 times the half-life of 12 hours, so 1/2³ = 1/8 of the starting amount of Os-183 is left:

590/8 = 73.75 ≈ 74

6 0

3 years ago