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3 years ago

5x + 12 + 3x = 6(x + 4) Please Show Work.

Mathematics

2 answers:

Len [333]3 years ago

8 0

5x + 12 + 3x = 6(x + 4)

5x+3x+12=6(x+4)

Multiply the bracket by 6

(6)(x)(6)(+4)=6x+24

5x+3x+12=6x+24

8x+12=6x+24

Move 6x to the other side. Sign changes from +6x to -6x.

8x-6x+12=6x-6x+24

8x-6x+12=24

2x+12=24

Move 12 to the other side. Sign changes from +12 to -12.

2x+12-12=24-12

2x=12

Divide by 2 for both sides.

2x/2=12/2

x=6

Answer: x=6

castortr0y [4]3 years ago

5 0

The answer is x= 6
This is how I got the answer

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How to solve this problem

Pavlova-9 [17]

For this one you can use 3/8 + 3/8 since 3+3 is equals to 6 and the 8 stays as it is.
I hope this helps.
YOU'RE WELCOME :D

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3 years ago

81(p+q)^2-9p-9q Factor this

polet [3.4K]

Answer:

9(p + q)(9p + 9q - 1)

Step-by-step explanation:

Given

81(p + q)² - 9p - 9q ← factor out - 9 from these 2 terms

= 81(p + q)² - 9(p + q) ← factor out 9(p + q) from each term

= 9(p + q)(9(p + q) - 1)

= 9(p + q)(9p + 9q - 1)

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3 years ago

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Given the figure below, find the values of x and z.

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Answer:

I think z = 89 and I don’t know x

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3 years ago

Find the equation of a line that is perpendicular to y = 3x – 5 and passes through the point (1, -3).

Svetradugi [14.3K]

keeping in mind that perpendicular lines have negative reciprocal slopes, let's check for the slope of the equation above

$\begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}\qquad \qquad y = \stackrel{\stackrel{m}{\downarrow }}{3}x-5$

well then therefore

$\stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{3\implies \cfrac{3}{1}} ~\hfill \stackrel{reciprocal}{\cfrac{1}{3}} ~\hfill \stackrel{negative~reciprocal}{-\cfrac{1}{3}}}$

so we're really looking for the equation of a line with slope of -1/3 and that passes through (1, -3 )

$(\stackrel{x_1}{1}~,~\stackrel{y_1}{-3})\qquad \qquad \stackrel{slope}{m}\implies -\cfrac{1}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{-\cfrac{1}{3}}(x-\stackrel{x_1}{1})\implies y+3=-\cfrac{1}{3}x+\cfrac{1}{3} \\\\\\ y=-\cfrac{1}{3}x+\cfrac{1}{3}-3\implies y=-\cfrac{1}{3}x-\cfrac{8}{3}$

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2 years ago

A new medical test provides a false positive result for hepatitis 2% of the time that is a perfectly healthy subject being teste

Oxana [17]

Given that X be the number of subjects who test positive for the disease out of the 30 healthy subjects used for the test.

The probability of success, i.e. the probability that a healthy subject tests positive is given as 2% = 0.02

Part A:

The probability that all 30 subjects will appropriately test as not being infected, that is the probability that none of the healthy subjects will test positive is given by:

 $P(X)={ ^nC_xp^x(1-p)^{n-x}} \\ \\ P(0)={ ^{30}C_0(0.02)^0(1-0.02)^{30-0}} \\ \\ =1(1)(0.98)^{30}=0.5455$


Part B:

The mean of a binomial distribution is given by

 $\mu=np \\ \\ =30(0.02) \\ \\ =0.6$

The standard deviation is given by:

 $\sigma=\sqrt{np(1-p)} \\ \\ =\sqrt{30(0.02)(1-0.02)} \\ \\ =\sqrt{30(0.02)(0.98)}=\sqrt{0.588} \\ \\ =0.7668$


Part C:

This test will not be a trusted test in the field of medicine as it has a standard deviation higher than the mean. The testing method will not be consistent in determining the infection of hepatitis.

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3 years ago