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Naya [18.7K]
3 years ago
6

How do I simplify this problem?

Mathematics
1 answer:
masya89 [10]3 years ago
6 0
Expanding (x+h)^3, we get (5x^3+15x^2h+15xh^2+5h^3-5x^3)/h=
(15x^2h+15xh^2+5h^3)/h= 15x^2+15xh+h^2
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Find the missing factor 2x^2+15x+7=(2x+1)(?)
Genrish500 [490]

Answer:

? = (x + 7)

Step-by-step explanation:

2x^2 + 15x + 7 = (2x + 1) (?)

1. First you need to figure out what multiplied by 2x will give you 2x^2, which is x.

2. Then you need to find what would give you 15x. In this case because you have the 1 in (2x + 1), you already know that there is going to be an extra x. So, what times 2x gives you 14x? It's 7.

3. Lastly you need to make sure that 1 multiplied with 7 is equal to 7 and combine like terms.

4. Check: (2x + 1) (x + 7) =

2x^2 + 14x + 1x + 7 =

2x^2 + 15x + 7

5 0
3 years ago
Given f(x)=1/4(5-x)2 what is the value of f(11)
iren [92.7K]

Answer:

f(x)=1/4(5-x)²

Step-by-step explanation:

f(11)=1/4(5-11)²

1/4(-6)²

1/4(36)

=9

7 0
3 years ago
Read 2 more answers
a cookie jar contains four oatmeal raisin, one sugar, nine chocolate chip, and six peanut butter cookies. if one is chosen at ra
VikaD [51]

Step-by-step explanation:

There are a total of 4 + 1 + 9 + 6 = 20 cookies.  So the probabilities of each type for a random cookie are:

P(oatmeal raisin) = 4/20 = 1/5

P(sugar) = 1/20

P(chocolate chip) = 9/20

P(peanut butter) = 6/20 = 3/10

8 0
3 years ago
Jack wants to share his candy with friends. Right now he has 4 1/3 pieces of chocolate. How many friends can he give 1/2 a candy
cricket20 [7]

Answer:

8 Friends

Step-by-step explanation:

First convert 4 1/3 to a mixed fraction

4 1/3 --> 13/3

Then divide by 1/2 to find how many friends per 1/2 candy bar with 13/3 candy bars

(13/3)/(1/2)

=(13*2)/(3*1)

=26/3

8.66667

Since you can't have 0.66667 of a person you get 8 friends to give candy to

3 0
3 years ago
40 POINTS PLS HELP IMEDIATLY
MakcuM [25]
Part A)
If f(x) - 3 is the new equation, it means there is a vertical translation of f(x) down 3 units. The y-intercept will decrease by 3 units. Areas of increasing on the function may be lessened as the function is being translated down 3 units. The areas of decrease will increase because the function is being translated down. End behaviour will not change from a translation as long as the function is continuous at each end, (not a finite function with end points). The evenness or oddness of f(x) will not change either.

Part B:
The y-intercept will be flipped horizontally about the x-axis and multiplied by 2. This will mean that if the y-intercept was positive, it will now be negative and vice versa. The increasing and decreasing regions of the graph will be flipped, so anywhere f(x) was positive will now be negative and vice versa. They will also be double what they were before because all values are multiplied by 2. The end behaviour will switch. If f(x) was from Quad1->Quad3 for example, it will now be Quad2->Quad4 because of the flip at the x-axis. The evenness and oddness of the function will not change seeing as the degree of f(x) is not affected.
6 0
3 years ago
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