PlSSS HELP IF YOU TURLY KNOW THISS

I need help plzz. <br><br> Dilate this by 1/3

leva [86]

For this shape there are four points. The plotted ordered pairs are:
M = (3, 9)
N = (6, 0)
O = (3, 6)
P = (-3, 0)

All you have to do is multiply the BOTH x- and y-values by 1/3 on EACH ordered pair. And if it’s easier, you can divide by 3 as well. It’s the same thing.

M = (3, 9) = (3*1/3, 9*1/3) = (1, 3)
N = (6, 0) = (6*1/3, 0*1/3) = (2, 0)
O = (3, 6) = (3*1/3, 6*1/3) = (1, 2)
P = (-3, 0) = (-3*1/3, 0*1/3) = (-1, 0)

(1, 3) , (2, 0) , (1, 2) , and (-1, 0) are your points.

8 0

3 years ago

Read 2 more answers

Factor 225x2 - 1. A) (15x +1)(15x + 1) B) (15x - 1)(15x + 1) C) (15x - 1)(15x - 1) D) (125x - 1)(125x + 1)

MrRa [10]

Answer:

B)

Step-by-step explanation:

225x^2 - 1=

(15x)^2 - 1^2=(*)

(15x-1)(15x+1)

A^2 - B^2 =(A-B)(A +B)............. (*)

4 0

3 years ago

Read 2 more answers

Janie bought a shirt at American Eagle and spent a total of $36.04, paying 6% in sales tax. The shirt was marked down 15% and wa

ss7ja [257]

Answer:

40·(1 - 0.15)·0.06 = 2.04

or

36.04·0.06/1.06 = 2.04

3 0

3 years ago

Read 2 more answers

A consulting firm has received 2 Super Bowl playoff tickets from one of its clients. To be fair, the firm is randomly selecting

Soloha48 [4]

Answer:

Therefore the only statement that is not true is b.)

Step-by-step explanation:

There employees are 6 secretaries, 5 consultants and 4 partners in the firm.

a.) The probability that a secretary wins in the first draw

= $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secreataries}{total \hspace{0.1cm} number \hspace{0.1cm} of \hspace{0.1cm} employees} = \frac{6}{15}$

b.) The probability that a secretary wins a ticket on second draw. It has been given that a ticket was won on the first draw by a consultant.

p(secretary wins on second draw | consultant wins on first draw)

= $\frac{p((consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)\cap( secretary\hspace{0.1cm} wins\hspace{0.1cm} on second \hspace{0.1cm}draw))}{p(consultant \hspace{0.1cm} wins \hspace{0.1cm} on \hspace{0.1cm} first \hspace{0.1cm}draw)}$

= $\frac{\frac{5}{15} \times \frac{6}{14}}{\frac{5}{15} } = \frac{6}{14}$ .

The probability that a ticket was won on the first draw by a consultant a secretary wins a ticket on second draw = $\frac{6}{15}$ is not true.

The probability that a secretary wins on the second draw = $\frac{number \hspace{0.1cm} of \hspace{0.1cm} secretaries \hspace{0.1cm} remaining } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} remaining} = \frac{6 - 1}{15 - 1} = \frac{5}{14}$

c.) The probability that a consultant wins on the first draw =

$\frac{number \hspace{0.1cm} of \hspace{0.1cm} consultants \hspace{0.1cm} } { number \hspace{0.1cm} of \hspace{0.1cm} employees \hspace{0.1cm} } = \frac{5 }{15} = \frac{1}{3}$