Determine the number of lines of symmetry for the figure

Write an expression of the calculation of the products of five and two and five and one

lana [24]

(5)(2)(5)(1) could be an answer

4 0

3 years ago

Solve this problem y=(x-3)^2 when x=9

atroni [7]

Y=36

it basically says y=(9-3)^2
which is y=6^2
which is then y=36

3 0

3 years ago

3(x+2)-2(×+1) i need help with this problem

zhuklara [117]

Question:

What is 3(x + 2) - 2(x + 1)?

Steps:

When you solve this type of equation always use the distributive property formula to solve this equation!

Formula: a(b + c) = ab + ac

So, what is 3(x + 2) ?

3x + 6

Also, what is -2(x + 1)?

-2x - 2

Then, put it back into the equation for you to solve this problem.

So, this would be the new equation:

3x + 6 - 2x - 2

Then, what you want to do is put variables next to each other and the numbers next to each other so it will be easier to simplify:

3x - 2x + 6 - 2

Then, simplify:

x + 4 is the answer!

Since 3x - 2x = x and 6 - 2 = 4

Hope that helped!!

~Serina

7 0

3 years ago

Identify the value of x that makes each pair of ratios equivalent

Monica [59]

I would be B. because from 4 to 20 is times 5 so you just divide 45 by 5 and you get 9.
HOPE THIS HELPS!!
FEEL FREE TO FOLLOW ME IF YOU NEED ANYMORE HELP IN THE FUTURE

8 0

3 years ago

The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

$\frac{10000(0.2)^1}{10000(0.2)^0}$

$\frac{0.2^1}{0.2^0}=\frac{0.2}{1}=0.2$

0.2=20%

2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

8 0

3 years ago