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3 years ago

Question 3 of 10 Which of the following is(are) the solution(s) to |15x+2| = 8?

Mathematics

2 answers:

joja [24]3 years ago

8 0

Answer:

x = 6/5 x = -2

Step-by-step explanation:

|5x+2| = 8

There are 2 solutions

5x+2 = 8 and 5x+2 = -8

Subtract 2 from each side

5x+2 -2 = 8-2 and 5x+2-2 = -8-2

5x= 6 5x = -10

Divide by 5

5x/5 = 6/5 5x/5 = -10/5

x = 6/5 x = -2

kondor19780726 [428]3 years ago

7 0

Answer:

Step-by-step explanation:

|5x+2|=8

1) |5×(-2)+2|=8

|-10+2|=8

|-8|=8

8=8

2) |5×6/5+2|=8

|6+2|=8

|8|=8

8=8

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Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
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16sin^5(x)+2sin(x)-12sin^3(x)=0
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\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
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$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
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~~~~~~~~~~~~\textit{quadratic formula}
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\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
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sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
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\frac{1}{2}\\\\
\frac{1}{4}
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sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
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\frac{\pi }{6}~,~\frac{5\pi }{6}\\
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\approx~0.252680~radians\\
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\approx~14.47751~de grees\\
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180-14.47751\\
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3 0

4 years ago

Which expression has a value of 1/36?

Nataly [62]

Answer:

1/36) = 0.0277777777778

(1/108)^3 = 7.9383224102 x 10^-7

(1/9)^4 = 0.000152415790276

(1/6)^2 = 0.0277777777778

(1/2)^5 = 0.03125

The only one that matches with the value of 1/36 is (1/6)^2. Therefore, your answer is . (1/6)^2

8 0

2 years ago

In a running competition, a bronze, silver and gold medal must be given to the top three girls and top three boys. If 15 boys an

denis-greek [22]

Answer:

There are a total of possible 1820 results.

Step-by-step explanation:

Since there are two competitions, one for boys and one for girls, and we want all the possible results we will calculate the possible combinations for the boys and multiply them by the possible combinations for the girls. This is shown below:

$Boys = \frac{15!}{3!(15 - 3)!} = \frac{15!}{3!12!} = \frac{15*14*13*12!}{3!12!}\\Boys = \frac{15*14*13}{3*2*1}\\Boys = \frac{15*14*13}{3*2*1}\\Boys = 455$