Ask question

Ask question

All categories

2 years ago

6

After a dilation centered at the origin, the image of CD is C'D'.

1 answer:

RSB [31]2 years ago

5 0

Answer: 3/2

Step-by-step explanation: just believe me

You might be interested in

What is the purpose of simultaneous equations?

sveta [45]

Answer:

a set of two or more equations, each containing two or more variables whose values can simultaneously satisfy both or all the equations in the set, the number of variables being equal to or less than the number of equations in the set.

Step-by-step explanation:

It works because of two properties of equations: Multiplying (or dividing) the expression on each side by the same number does not alter the equation. Adding two equations produces another valid equation: e.g. 2x = x + 10 (x = 10) and x − 3 = 7 (x also = 10).

5 0

3 years ago

One jar of crushed ginger costs $2. How<br> many jars can you buy for $4?

Lina20 [59]

Answer:

2 jars

Step-by-step explanation:

If one jar costs $2 and you have $4 then $2+$2=$4

8 0

2 years ago

Read 2 more answers

How do u solve for x please and thank you!! Need ASAP

gavmur [86]

Answer:

Step-by-step explanation:

both sides are the same length so 20+6x+3=18+7x+2

all you have to do is get the variables and non-variables on opposite sides of the equation so 3=x

5 0

3 years ago

A circle with an area 9pi has a sector with a central angle of 1/9pi what is the area of the sector

ludmilkaskok [199]

Answer:

½pi or 0.5pi units²

Step-by-step explanation:

Area : angle

9pi : 2pi

x : pi/9

9pi/2pi = x/(pi/9)

4.5 × pi/9 = x

x = ½pi or 0.5pi units²

8 0

3 years ago

Compound interest In Exercise,$3000 is invested in an account at interest rate r,compounded continuously.Find the time required

xenn [34]

Answer:

(a) 8.15

(b) 12.92

Step-by-step explanation:

Given: P = $3000, r = 0.085

$A = Pe^{rt}$

Where

A is the Amount

P is the Principal

r is the rate

t is the time

(a) For the amount to double, A = 2 × P

A = 2 × $3000

A = $6000

$6000 = 3000e^{0.085t}$

$\frac{6000}{3000} = e^{0.085t}$

$2 = e^{0.085t}$

Take $log_{e}$ of both sides

$log_{e}2 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

∴ $ln2 = 0.085t$

$t = \frac{ln2}{0.085}$

$t = \frac{0.693}{0.085}$

t = 8.15

(b) For the amount to double, A = 3 × P

A = 3 × $3000

A = $9000

$9000 = 3000e^{0.085t}$

$\frac{9000}{3000} = e^{0.085t}$

$3 = e^{0.085t}$

Take $log_{e}$ of both sides

$log_{e}3 = log_{e}e^{0.085t}$

But $log_{e}e = 1$

∴ $ln3 = 0.085t$

$t = \frac{ln3}{0.085}$

$t = \frac{1.0986}{0.085}$

t = 12.92

5 0

3 years ago