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Anarel [89]
3 years ago
7

there are 20 total students in Kim's class. 12 of the students are boys and 8 are girls. which of the following reqesents the pa

rt of the class that is boys​
Mathematics
1 answer:
diamong [38]3 years ago
3 0

Answer:

Step-by-step explanation:

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(x^2 + 5x + 2) – (5x^2 – x – 2) =
zlopas [31]

Answer:

{ \tt{( {x}^{2} + 5x + 2) - (5 {x}^{2} - x - 2)  }} \\  = { \tt{(1 - 5) {x}^{2}  + (5 + 1)x + (2 + 2)}} \\  = { \tt{ - 4 {x}^{2} + 6x + 4 }} \\  = { \tt{ - 2(2 {x}^{2} - 3x - 2) }}

3 0
3 years ago
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How many x -intercepts does the parabola whose function is f(x)=3 x 2 +4x+4
artcher [175]

Answer:

They are none x- intercepts on the parabola. To find the x-intercept, substitute in 0 for y and solve for x.

3 0
3 years ago
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Evaluate the expression for the given value of the variable.<br><br> |4b-8|+|-1-b^2|+2b^3;b=2
Elena-2011 [213]

In this question we already know the value of 'b' so put the value b=2 in the equation

=|4(2)-8| + |-1-(2)^2| + 2(2)^3

=|8-8| + |-1-4| + 2(8)

=0 -5 +16

=11

4 0
4 years ago
If you are allowed to use numbers 1 – 20 and need to choose the passcode of an exact 4 digit code, how many possibilities are th
VMariaS [17]

Since in a pass code, the placement of the digits is important, therefore this means that to solve for the total number of possibilities we have to make use of the principle of Permutation. The formula for calculating the total number of possibilities using Permutation is given as:

P = n! / (n – r)!

where,

n = is the total amount of numbers to choose from = 20

r = is the total number of digits needed in the passcode = 4

 

Therefore solving for the total possibilities P:

P = 20! / (20 – 4)!

P = 20! / 16!

P = 116,280

 

<span>Hence there are a total of 116,280 possibilities of pass codes.</span>

4 0
3 years ago
I can't find the value of Y​
Lubov Fominskaja [6]

114 = 19y

This is true because they are vertical angles. Now, let's divide both sides by 19

6 = y

5 0
3 years ago
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