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diamong [38]
4 years ago
6

Can y’all help me what do I do here please I’ll give you brainliest

Mathematics
1 answer:
exis [7]4 years ago
3 0

Answer:

1, 2, -3, 51

Step-by-step explanation:

0 ÷ 2 = 0

0 + 1 = 1

2 ÷ 2 = 1

1 + 1 = 2

-8 ÷ 2 = -4

-4 + 1 = -3

100 ÷ 2 = 50

50 + 1 = 51

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Perform the operation. (10x2 – 10x – 6) + (x+6)
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Answer:

10x² - 9x

Step-by-step explanation:

(10x² - 10x - 6) + (x + 6)

10x² - 10x - 6 + x + 6

10x² - 9x - 6 + 6

10x² - 9x

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3 years ago
Which statements about the function are true? Choose<br> three options.
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Answer:

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3 years ago
Find the slope and standard form of the line that passes through the points (3, 6) and (4, 5).
qwelly [4]

Answer:

B) m = -1 , x + y = 9

Step-by-step explanation:

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3 years ago
What is the area of a parallelogram with a base of 2 cm and a height of 8 cm?
Vsevolod [243]

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Read 2 more answers
A CAT scan produces equally spaced cross-sectional views of a human organ that provide information about the organ otherwise obt
True [87]

Answer:

1,107 cc

Step-by-step explanation:

The 10 intervals of the scanning are

[0,1.5), [1.5,3), [3,4.5), [4.5,6), [6,7.5), [7.5,9), [9,10.5), [10.5,12), [12,13.5), [13.5,15)

The ideal thing would be to estimate the volume with the Midpoint Rule should be take n=10.

As we must use n=5, we divide the interval [0,15] in five intervals of length 15/5 = 3

[0,3], [3,6], [6,9], [9,12], [12,15] and take their midpoints

1.5, 4.5, 7.5, 10.5, and 13.5.

Now we estimate the volume V of 5 cylinders of height h=3 and area of the base A equals to the area given for the scanning where that midpoints fall

Cylinder 1

Midpoint=1.5, interval of scanning 2nd

A = 18, V= height*area of the base = 18*3 =54  cc

Cylinder 2

Midpoint=4.5, interval of scanning 4th

A = 78, V= height*area of the base = 78*3 =234  cc

Cylinder 3

Midpoint=7.5, interval of scanning 6th

A = 106, V= height*area of the base = 106*3 =318  cc

Cylinder 4

Midpoint=10.5, interval of scanning 8th

A = 129, V= height*area of the base = 129*3 =387  cc

Cylinder 5

Midpoint=13.5, interval of scanning 10th

A = 38, V= height*area of the base = 38*3 =114  cc

And the <em>estimate volume </em>is

54+234+318+387+114=1,107 cm^3

6 0
3 years ago
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