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artcher [175]
3 years ago
6

Which line has a slope of 1/2 and goes through the point (2,4)?

Mathematics
1 answer:
Pani-rosa [81]3 years ago
7 0
The line equation for its slope and one point is:
y - y1 = m(x - x1)
m is the slope, x1, y1 are the point coordinates, so lets substitute:
y - 4 = (1/2)(x - 2)
y = <span>(1/2)x + 3</span>
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How many solutions does the equation x^2-9x=-8 have?
Elza [17]
X^2-9x=-8
x^2-9x+8=-8+8
x^2-9x+8=0
(x-8)(x-1)=0

x-8=0
x-8+8=0+8
x=8

x-1=0
x-1+1=0+1
x=1

There are 2 solutions, which are 1 and 8 
8 0
3 years ago
Is there a multiple of 3? And why not or why yes?
rodikova [14]
Yes there is because u can multiply it by different numbers except for 1 and 0
6 0
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Read 2 more answers
Please help on questions 8&amp;9!!!!(higher maths vectors)
alekssr [168]
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4 0
3 years ago
For a camping trip, three scoutmasters are each driving a portion of the 150 miles to campground mr. Jimenez drives 3/10 of the
babymother [125]

Answer:

45 miles

Step-by-step explanation:

Jimenez drives 3/10 of the distance:  (3/10)(150 miles) = 45 miles.

Mr. Robertson drives 2/5 of the distance:  (2/5)(150) = 60 miles.

These two people drive 105 miles total.

Mr. Weatherly drives the rest:  (150 miles - 105 miles) = 45 miles.

4 0
3 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
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