.hello :
an equation of the circle Center at the w(a,b) and ridus : r is :
(x-a)² +(y-b)² = r²
in this exercice : r = 1
The points (-18,15) and (-20,15) lie on a circle with a radius of 1:
(-18-a)²+(15-b)² = 1 ....(1)
(-20-a)² +(15-b)² = 1 ....(2)
solve this system :
(1) -(2) : (-18-a)² - (-20-a)² =0
(-18-a)² =(-20-a)² =0
( -18-a = -20-a) or (-18-a = - (-20-a))
1 ) ( -18-a = -20-a) no solution confused : -18=-20
2 ) -18-a =20+a
-2a =38
a = -19
subst in (1) :(-18+19)²+(15-b)² =1
(15-b)² = 0.... 15-b = 0 .... b = 15
the center is :w(-19,15)

6 0

3 years ago

If the teacher has 30 students and 10% left,how much students does she have now

garri49 [273]

Answer:

She has 27 students now.

Step-by-step explanation:

Given :

The teacher has 30 students

10% students left.

To find : how much students does she have now?

Solution :

Since we are given that teacher has 30 students .

And 10% of students left

⇒ 10% of 30

⇒ $\frac{10}{100} *30$

⇒ $\frac{300}{100}$

⇒ $3$

Thus, 3 students left .

She had 30 students . three students left.

So, now she have 30 - 3 = 27 students .

Hence she has 27 students now.

8 0

3 years ago

Read 2 more answers

Order -1 2/3, 1- 3/8, -1 4/5, and -1 9/20

Elis [28]

From least to greatest?

5 0

3 years ago

Read 2 more answers

If T: (x, y) → (x + 6, y + 4), then T-1: (x,y) → _____.

alexgriva [62]

T is a linear transformation from R²→R² with basis {(1,0),(0,1)}
T: (x,y)→(x+6,y+4)

A function from one vector space to another that preserves the underlying (linear) structure of each vector space is called a linear transformation.

Then the vector (1,0) goes to (1+6,4)=(7,4)=7(1,0)+4(0,1)

and the vector (0,1) goes to (6,1+4)=(6,5)=6(1,0)+5(0,1)

So, the matrix of the transformation is

$\left[\begin{array}{ccc}7&6\\4&5\end{array}\right]$

The inverse of the matrix is

$\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]$

So, the Inverse Transformation is given by

$T^{-1}(x,y)=\left[\begin{array}{ccc}\frac{5}{11}&\frac{-6}{11}\\ \\\frac{-4}{11}&\frac{7}{11}\end{array}\right]\left[\begin{array}{ccc}x\\y\end{array}\right] =(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

So, no option is correct. And the answer is

$T^{-1}(x,y)=(\frac{5x-6y}{11}, \frac{-4x+7y}{11})$

Learn more about linear transformations here-

brainly.com/question/13005179

#SPJ10

5 0

2 years ago

*20 points*

20 points