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2 years ago

Plss helpp its due tmr and i dont really get ittt

Mathematics

1 answer:

OlgaM077 [116]2 years ago

4 0

The orthocenter of the triangles ABC are (2, 0) and (3, 3)

<h3>How to determine the orthocenter?</h3>

<u>Question 8</u>

The coordinates are given as:

A(2,0); B(2, 4); C(6,0)

Calculate the slope of BC using:

m = (y2 - y1)/(x2 - x1)

So, we have:

m =(0 - 4)/(6 - 2)

m =-1

The perpendicular slope is:

n = -1/m

n = -1/-1

n = 1

The equation of the perpendicular line is

y = n(x - x1) + y1

Where:

(x1, y1) = (2,0)

So, we have:

y = 1(x - 2) + 0

y = x - 2

Calculate the slope of AC using:

m = (y2 - y1)/(x2 - x1)

So, we have:

m =(0 - 0)/(6 - 2)

m =0

The perpendicular slope is:

n = -1/m

n = undefined

This means that the perpendicular line is a vertical line

The equation of the perpendicular line is

x = x1

Where:

(x1, y1) = (2,4)

So, we have:

x = 2

Substitute x = 2 in y = x - 2

y = 2 - 2

y = 0

So, we have:

(x, y) = (2, 0)

Hence, the orthocenter of ABC is (2, 0)

<u>Question 9</u>

The coordinates are given as:

A(1,1); B(3, 4); C(6,1)

Calculate the slope of BC using:

m = (y2 - y1)/(x2 - x1)

So, we have:

m =(1 - 4)/(6 - 3)

m =-1

The perpendicular slope is:

n = -1/m

n = -1/-1

n = 1

The equation of the perpendicular line is

y = n(x - x1) + y1

Where:

(x1, y1) = (1,1)

So, we have:

y = 1(x - 1) + 1

y = x

Calculate the slope of AC using:

m = (y2 - y1)/(x2 - x1)

So, we have:

m =(1 - 1)/(6 - 1)

m =0

The perpendicular slope is:

n = -1/m

n = undefined

This means that the perpendicular line is a vertical line

The equation of the perpendicular line is

x = x1

Where:

(x1, y1) = (3,4)

So, we have:

x = 3

Substitute x = 3 in y = x

y = 3

So, we have:

(x, y) = (3, 3)

Hence, the orthocenter of ABC is (3, 3)

<h3>How to solve for x?</h3>

<u>Question 18 - 20</u>

Line 2x and x+ 3 are congruent lines

So, we have:

2x = x +3

Solve for x

x= 3

Line 3x and 2x + 5 are congruent lines

So, we have:

3x = 2x +5

Solve for x

x = 5

Line 2x - 3 and x + 2 are congruent lines

So, we have:

2x - 3 = x + 2

Solve for x

x = 5

<u>Question 21 - 23</u>

Line x + 5 and 3x + 7 are congruent lines

So, we have:

x + 5 = 3x + 7

Solve for x

x = -1

Line x + 6 and 5x - 4 are congruent lines

So, we have:

x + 6 = 5x - 4

Solve for x

x = 2.5

Line x + 8 and 5x + 7 are congruent lines

So, we have:

x + 8 = 5x + 7

Solve for x

x = 1/4

You might be interested in

A farm has 8 cows, 28 chickens, 4 pigs, and 12 goats. which ratio is the

QveST [7]

Answer:

Step-by-step explanation:

ratio of pigs to chickens = 4/28, which simplifies to 1/12

ratio of goats to chickens = 12/28, which simplifies to 3/7

ratio of cows to goats = 8/12, which simplifies to 2/3

ratio of pigs to cows = 4/8, which simplifies to 1/2

since 2/3 is the largest, the answer is C

3 0

2 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.