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Leno4ka [110]
3 years ago
13

If (6,-4) is dilated by a scale factor of -2 the new point will be located at?

Mathematics
1 answer:
Sphinxa [80]3 years ago
3 0
(4,-2) Y/x With the s/f of 2

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Y=3x-8 in function notation
julia-pushkina [17]

To write it in function notation you'll convert it into a function, substituting y by f(x):


y = 3x - 8 = f(x) = 3x - 8



Hope it helped,



BioTeacher101

3 0
3 years ago
Find the product.
IrinaK [193]
(2n+1)(2n-1)(n+5)
=(4n^2+2n-2n-1)(n+5)
=(4n^2-1)(n+5)
=4n^3-n+20n^2-5
4n^3+20n^2-n-5
4 0
2 years ago
What is x^2 times 4x
Flura [38]
Say x is 1 1^2=1  4x=4 4*1=4  I believe that is what you were asking

6 0
3 years ago
Find the z-scores that bound the middle 74% of the area under the standard normal curve.
aniked [119]

Since 74% of the middle area is bounded, this means that there is 13% on the left side, and another 13% on the right side.

P (left) = 0.13

P (right ) = 1 - 0.13 = 0.87

 

At this P values, the z scores are approximately:

z score (left) = -2.22

<span>z score (right) = 1.13</span>

5 0
2 years ago
The fish population in a certain lake rises and falls according to the formula F = 3000(23 + 11t − t2). Here F is the number of
Gemiola [76]

Answer:

a) On January 1, 2017 the fish population will be the same as the initial population.

b) On September 18th, 2018 the fish population will be zero.

Step-by-step explanation:

Hi there!

a) First, let´s write the function:

F(t) = 3000(23 + 11t − t²)

The population on January 1, 2006 is the population at t = 0. Then:

F(0) = 3000(23 + 11· 0 - 0²)

F(0) = 3000 · 23 = 69000

This will be the population every time at which t² - 11t = 0. Then let´s find the other value of t (besides t = 0) that makes that expression to be zero:

t² - 11t = 0

t(t - 11) = 0

t = 0

and

t - 11 = 0

t = 11

On January 1, 2017 (2006 + 11), the fish population will be the same as the initial population.

b) We have to obtain the value of t at which F(t) = 0

F(t) = 3000(23 + 11t − t²)

0 = 3000(23 + 11t − t²)

divide both sides of the equation by 3000

0 = 23 + 11t − t²

Let´s solve this quadratic equation using the quadratic formula:

a = -1

b = 11

c = 23

x = [-b ± √(b² - 4ac)] / 2a

x = 12.8  ( the other value of x is negative and therefore discarded).

After 12.8 years all the fish in the lake will have died.

If 1 year is 12 months, 0.8 years will be:

0.8 years · 12 months/year = 9.6 months

If 1 month is 30 days, 0.6 month will be:

0.6 month · 30 days / month = 18 days

All the fish will have died after 12 years, 9 months and 18 days from January 1, 2006. That is, on September 18th, 2018.

7 0
3 years ago
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