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LiRa [457]
3 years ago
12

Absolute Value 6|n|+3=-9

Mathematics
1 answer:
Lena [83]3 years ago
7 0
Remember you can do anything to an equaiton as long as we do it to both sides

and
absolute value makes whatever is inside positeive
aka
|smileyface|>0
remember that


to solve
|x|=y
assume
x=y and
x=-y
(reason is because |-x|=|x|)

6 times abasolute value of n +3=-9 is it translated to words

6|n|+3=-9
move over, minus 3 both sides
6|n|=-12
divide both sides by 6
|n|=-2
false, absolute value makes it positive
-2 is not positive


no solution
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Graph the linear equation by plotting three points. 2y=-3+2​
Ksivusya [100]

Answer:

Step-by-step explanation:

Convert 2y=-3x+2 into y=mx+b form

\frac{2y}{2} =\frac{-3x+2}{2}

y=\frac{-3x}{2} +1

Y intercept is 1, slope is -3/2

Graph looks like this:

3 0
2 years ago
Identify the class​ width, class​ midpoints, and class boundaries for the given frequency distribution.
Crank

Answer:

a. Class width=4

b.

Class midpoints

46.5

50.5

54.5

58.5

62.5

66.5

70.5

c.

Class boundaries

44.5-48.5

48.5-52.5

52.5-56.5

57.5-60.5

60.5-64.5

64.5-68.5

68.5-72.5

Step-by-step explanation:

There are total 7 classes in the given frequency distribution. By arranging the frequency distribution into the refine form we get,

Class

Interval frequency

45-48 1

49-52 3

53-56 5

57-60 11

61-64 7

65-68 7

69-72 1

a)

Class width is calculated by taking difference of consecutive two upper class limits or two lower class limits.

Class width=49-45=4

b)

The midpoints of each class is calculated by taking average of upper class limit and lower class limit for each class.

M=\frac{lower class limit+upper class limit}{2}

Class

Interval Midpoints

45-48 \frac{45+48}{2}=46.5

49-52 \frac{49+52}{2}=50.5

53-56 \frac{53+56}{2}=54.5

57-60 \frac{57+60}{2}=58.5

61-64 \frac{61+64}{2}=62.5

65-68 \frac{65+68}{2}=66.5

69-72 \frac{69+72}{2}=70.5

c)

Class boundaries are calculated by subtracting 0.5 from the lower class limit and adding 0.5 to the upper class interval.

Class

Interval Class boundary

45-48 44.5-48.5

49-52 48.5-52.5

53-56 52.5-56.5

57-60 56.5-60.5

61-64 60.5-64.5

65-68 64.5-68.5

69-72 68.5-72.5

7 0
3 years ago
K-2k=15+1+6k-2 help me
sveticcg [70]
K - 2k = 15 + 1 + 6k - 2

-k = (15 + 1 - 2) + 6k

-k = 14 + 6k      |subtract 6k from both sides

-7k = 14    |divide both sides by (-7)

k = -2
6 0
3 years ago
What is the probability that a randomly selected passenger hss bronze status and i checked bag
monitta
Uhh not sure that this question is put the right way
4 0
3 years ago
How do you do this question?
Sergeeva-Olga [200]

Answer:

V = 2000r³/3

Step-by-step explanation:

We know that the base is a circular disk, so it creates a circle on the xy plane. It would be in the form x² + y² = r². In other words x² + y² = (5r)². Let's isolate y in this equation now:

x² + y² = (5r)²,

x² + y² = 25r²,

y² = 25r² - x²,

y = √25r² - x² ---- (1)

Now remember that parallel cross sections perpendicular to the base are squares. Therefore Area = length^2. The length will then be = 2√25r² - x² --- (2). Now we can evaluate the integral from -5r to 5r, of [ 2√25r² - x² ]² dx.

\int _{-5r}^{5r}\:\left[\:2\sqrt{\left(25r^2\:-\:x^2\right)}\:\right]\:^2\:dx\\=\int _{-5r}^{5r}4\left(25r^2-x^2\right)dx\\\\= 4\cdot \int _{-5r}^{5r}25r^2-x^2dx\\\\= 4\left(\int _{-5r}^{5r}25r^2dx-\int _{-5r}^{5r}x^2dx\right)\\\\= 4\left(250r^3-\frac{250r^3}{3}\right)\\\\= 4\cdot \frac{500r^3}{3}\\\\= \frac{2000r^3}{3}

As you can see, your exact solution would be, V = 2000r³/3. Hope that helps!

3 0
3 years ago
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