Ask question

Ask question

All categories

Sergeeva-Olga [200]

3 years ago

14

15 POINTS

1 answer:

Lunna [17]3 years ago

4 0

f(n)= f(n-1) · 0.6+48, n>0

You might be interested in

Help???!!??????!??!!???!????!??

svet-max [94.6K]

$x_{vertex}= \frac{-2}{2 \cdot 3}=- \frac{2}{6}= - \frac{1}{3} \\\\\\ 
y_{vertex}=3 \cdot \left(- \frac{1}{3} \right)^2+2 \cdot \left(- \frac{1}{3} \right)+1=3 \cdot\frac{1}{9} - \frac{2}{3} +1=\frac{1}{3} - \frac{2}{3} +1= \frac{2}{3} \\ \\ \\ \boxed {\text {Answer:} \ \left(- \frac{1}{3}; \frac{2}{3} \right)}$

6 0

3 years ago

Is (6,3) a solution to this system of equations y=3x - 3 3x - y =3

aleksandr82 [10.1K]

Answer:

It is not a solution

Step-by-step explanation:

Plug the point into the equations and check to see if they are true

y=3x - 3

3 = 3(6) -3

3 = 18-3

3 = 15

False

We do not need to check the other equation since this is false

3 0

2 years ago

Shannon is selling boxes cookies and candy for a fundraiser. She's sold each box of cookies for $5.25. She also sold one one box

defon

Answer:

The minimum boxes of cookies is 39

Step-by-step explanation:

Let

x ----> the number of boxes of cookies sold

y ----> the number of boxes of candies sold

we know that

The word "at least" means "greater than or equal to"

so

The inequality that represent this problem is

$5.25x+8.75y\geq 200$

The solution is the shaded area above the solid line $5.25x+8.75y=200$

using a graphing tool

The solution is the shaded area -----> see the attached figure

Find out the minimum boxes of cookies needed to sell to reach the goal

assuming only cookies are sold

For y=0

$5.25x+8.75(0)\geq 200$

$5.25x\geq 200$

solve for x

$x\geq 38.1$

Round up

therefore

The minimum boxes of cookies is 39

5 0

2 years ago

What of these numbers is a prime number 1,11,21,51

Zepler [3.9K]

11. It is not the multiple of any numbers smaller than it

7 0

3 years ago

Find the slope of the line that passes through each pair of point E(4,-4),F(2,2)

elena-s [515]

This has a slope of 3, you can plot it on a graph and count rise over run.

3 0

3 years ago

Read 2 more answers