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PolarNik [594]
4 years ago
10

Solving polynomial(2x+8)(-3y-8)​

Mathematics
2 answers:
kakasveta [241]4 years ago
7 0

Answer:

-6xy - 16x -24y -64

Step-by-step explanation:

(2x+8)(-3y-8) =

To find the answer,

First multiply, the first number on both sides,

2x * -3y = -6xy

Then the first number on the left side and the second number on the right side,

2x * -8 = -16x

Then the second number on the left side and the first number on the right side,

8 * -3y = -24y

Then the second number on the left side and the second number on the right side,

8 * -8 = -64

Now add all the answers,

-6xy -16x -24y -64

faust18 [17]4 years ago
3 0

Answer:

-6xy-16x-24y-64

Step-by-step explanation:

(2x+8)(-3y-8)​

Foil

First 2x*-3y = -6xy

outer -8*2x = -16x

inner -3y *8 = -24y

last -8*8 = -64

Add them together

-6xy-16x-24y-64

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Express the rate as a unit.<br> 2.75 yards of fabric for $22
pantera1 [17]

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Answer:

  $8 /yd

Step-by-step explanation:

You probably want dollars per yard. To get that, divide dollars by yards:

  $22/(2.75 yd) = $8 /yd

3 0
4 years ago
Plss help me out!!!!!!
abruzzese [7]

Answer:

its D because they all match the sides

6 0
3 years ago
Which equation correctly applies the distributive property?
Step2247 [10]

Answer: Only Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Distributive property states that

a\times (b+c+d)=a\times b+a\times c+a\times d

According to option 'C', we get that

a = 0.5

b = 0.5

c = 0.3

d = 0.2

So, it is written as

0.5(0.5+0.3+0.2)=0.5\times 0.5+0.5\times 0.3+0.5\times 0.2

Hence, only Option 'C' is correct.

3 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
Michael is trying to hang Christmas lights on his house. His house is 17 ft tall and the ladder leaning is 34 degrees above the
Readme [11.4K]

Answer:

34 feet

Step-by-step explanation:

let length of ladder be x

\ \sin(34)   =  \frac{17}{x}

x \sin(34)  = 17

x =  \frac{17}{ \sin(34) }

x = 32.131083564

8 0
3 years ago
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