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snow_tiger [21]
3 years ago
8

Given f(x)=x^3+5x+k, and x+2 is a factor of f(x), then what is the value of k?

Mathematics
2 answers:
8_murik_8 [283]3 years ago
7 0

Answer: Hopes this helps

Step-by-step explanation:

Marat540 [252]3 years ago
6 0
3x^3+0x^2-2x+k:x-2=3x^2+6x+10
3x^2-6x^2
6x^2-2x
6x^2-12x
10x+k
10x-20
k+20
Thus k=-20
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A girl scout troop sold cookies. If the girls sold 5 more boxes the second week than they did the first, and if they doubled the
mihalych1998 [28]
We can let x be the number of boxes sold for the first week. We can as well express the number of boxes for the second and third week through x using the statements provided.

Since the girl scout sold 5 more boxes on the second week, we have (x + 5) number of boxes sold for the second week. Now, for the third week, since it's double that of the second week, we have 2(x + 5). Thus, we have the following:

first week: x
second week: x + 5
third week: 2(x + 5)

Given that the total number of boxes sold for the three weeks is 431. We have

x + (x + 5) + 2(x + 5) = 431
x + x + 5 + 2x + 10 = 431
4x + 15 = 431
4x = 431 - 15
4x = 416
x = 104

We have now the value of x. Using this, we can find the values for the second and third week.
x + 5 = `104 + 5 = 109
2(x + 5) = 2(109) = 218

Answer: 
first week: 104
second week: 109
third week: 218



5 0
3 years ago
HELP ME POR FAVOR PLEASE PLEASEE
Basile [38]

Answer:

                 x=8\frac34\ \,,\quad y=8\frac12

Step-by-step explanation:

             \Delta TRS\sim \Delta LJK \quad\implies\quad \dfrac{|TR|}{|LJ|}=\dfrac{|RS|}{|JK|}=\dfrac{|ST|}{|KL|}\\\\\dfrac{4}{7}=\dfrac{5}{x}=\dfrac{6}{y+2}\\\\\\\dfrac{4}{7}=\dfrac{5}{x}\quad\implies\quad4x=5\cdot7\quad\implies\quad x=\dfrac{35}4=8\frac34\\\\\\\dfrac{5}{x}=\dfrac{6}{y+2}\quad\implies\quad 5(y+2)=6\cdot\dfrac{35}4\\\\5y+10=3\cdot\dfrac{35}2\\\\5y\ =\ \dfrac{3\cdot35}2-10\\ \div5\qquad\quad \div5\\\\y=\dfrac{3\cdot7}2-2=10\frac12-2=8\frac12

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4 years ago
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