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3 years ago

Find the point-slope equation for

Mathematics

1 answer:

Igoryamba3 years ago

4 0

Answer:

y+(21)=-4(x-7)

Step-by-step explanation:

the point slope equation , two point

m=y2-y1/x2-x1

m=23+21/-4-7=44/-11= -4

y+(21)=-4(x-7)

You might be interested in

What is 3x9x-5x=7#hh&

Paha777 [63]

move the 3 to the 7 side. it will become negative.

9x-5x=7-3

4x=4

x=1

7 0

3 years ago

A quadrilateral is formed by the points a(1, -1), b(0, 3), c(5, 5), and d(6, 1). plot the points and use the distance formula to

CaHeK987 [17]

The type of quadrilateral is rectangle.

What is distance formula?

The distance between coordinate P(x1 , y1) and coordinate Q(x2 , y2) is calculated using the distance formula: d = √[(x2 - x1)² + (y2 - y1)²]

Given coordinates: a(1, -1), b(0, 3), c(5, 5), and d(6, 1).

ab= √(0-1)² + (3+1)²

ab= √17

bc= √(5-0)² + (5-3)²

bc= √29

cd= √(6-5)² + (1-5)²

cd= √17

da=√(6-1)² + (1+1)²

da= √29

ac=√(5-1)² + (5+1)²

ac= √16+36

ac= √52

bd =√(6-0)² + (1-3)²

bd= √36+4

bd= √40

Learn more about this concept here:

brainly.com/question/12864470

#SPJ1

7 0

2 years ago

What is the volume of this triangular prism?

kati45 [8]

Answer:

Given:

perpendicular [p]=9cm

base[b]=15cm

length[l]=4cm

we have

the volume of this triangular prism=area of traingle. ×length=1/2 p×b×l

=½×9×15×4=270cm³

270cm³ is a required volume.

4 0

3 years ago

Read 2 more answers

Plz help I’m getting timed

Pani-rosa [81]

Answer:how long u got to answer the question

Step-by-step explanation:

7 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

eduard

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)$

with partial derivatives (set equal to 0)

$\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0$

$\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0$

for each $1\le i\le n$ .

Let $\Lambda$ be the sum of all the multipliers $\lambda_i$ ,

$\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n$

We notice that

$x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0$

so that

$\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0$

We know that $\sum\limits_{i=1}^n{x_i}^2=36$ , so

$\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda$

Solving the first $n$ equations for $x_i$ gives

$1+2\Lambda x_i=0\implies x_i=-\dfrac1{2\Lambda}$

and in particular

$\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}$

It follows that

$-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}$

which gives us

$x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}$

That is, we've found two critical points,

$\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)$

At the critical point with positive signs, $f(x_1,\ldots,x_n)$ attains a maximum value of

$\displaystyle\sum_{i=1}^nx_i=\dfrac{6n}{\sqrt n}=6\sqrt n$

and at the other, a minimum value of

$\displaystyle\sum_{i=1}^nx_i=-\dfrac{6n}{\sqrt n}=-6\sqrt n$

4 0

4 years ago