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coldgirl [10]
3 years ago
7

Least common multiple of z^2+6z+9 and z^2+z-6

Mathematics
1 answer:
Firdavs [7]3 years ago
5 0
Hint:    \bf \begin{array}{ccllll}
z^2+6z+9&z^2+z-6\\
\downarrow &\downarrow \\
(z+3)(z+3)&\underline{(z+3)(z-2)}
\end{array}
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Write an equation or inequality to represent the scenario:
aniked [119]
<h3><u>The value of x is equal to 1.</u></h3><h3><u>6(x + 2) = 20x - 2</u></h3>

<em><u>Distributive property.</u></em>

6x + 12 = 20x - 2

<em><u>Add 2 to both sides.</u></em>

6x + 14 = 20x

<em><u>Subtract 16x from both sides.</u></em>

14 = 14x

<em><u>Divide both sides by x.</u></em>

x = 1

6 0
3 years ago
Last week’s and this week’s low temperatures are shown in the table below. Low Temperatures for 5 Days This Week and Last Week L
STatiana [176]

Question:

Temperatures for 5 Days This Week and Last Week

Low Temperatures This Week (Degrees Fahrenheit) 4 10 6 9 6 Low Temperatures Last Week (Degrees Fahrenheit) 13 9 5 8 5 Which measures of center or variability are greater than 5 degrees? Select three choices.

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

d)the mean absolute deviation of this week’s temperatures

e) the mean absolute deviation of last week’s temperatures

Answer:

a) the mean of this week’s temperatures

b) the mean of last week’s temperatures

c) the range of this week’s temperatures

Step-by-step explanation:

I would be verifying the options a, b, and c in my answer above through calculations which are shown below.

We were given the following data:

Low Temperatures This Week (Degrees Fahrenheit)

4, 10, 6, 9, 6

Low Temperatures Last Week (Degrees Fahrenheit)

13, 9, 5, 8, 5

We are to find which measures of center or variability are greater than 5 degrees.

Option a

The mean of this week’s temperatures

(4+ 10+ 6+9+ 6) °F ÷ 5 = 35 °F ÷5 = 7°F

Option a is correct because it measures of center or variability which is 7 °F is higher than 5°F

Option b

The mean of last week’s temperatures

(13+ 9 + 5 + 8 + 5) °F = 40°F ÷ 5 = 8°F

Option b is correct , because its measure of variability which is 8°F is greater than 5°F.

Option c

the range of this week’s temperatures

This week's temperature is given as

(4, 10, 6, 9, 6) °F

Range is defined as the difference between the highest number and the lowest number

Range of this week's temperature = (10 - 4) °F = 6°F

Hence Option c is correct because it measures of center or variability which is 6°F is greater than 5°F

From the above calculations we can accurately confirm that options a, b, and c are correct because the measures of their center or variability is greater than 5°F

3 0
3 years ago
Read 2 more answers
Neo mixes 600ml of black paint with white paint in order to get grey paint. He thinks the colour came out too dark, so he decrea
butalik [34]

Answer:

the black paint is of 200 ml

Step-by-step explanation:

first he wanted grey paint by mixing 600 ml of both colour so he reduced black paint by 7:12 ratio an we get difference of 3 with that ratio so dividing 600 by 3 and we get amount of blacl paint removed

4 0
2 years ago
Darryl earns $10 per hour tutoring students after school. If Darryl tutored for 25 hours this month, how much money did he earn
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He made $250 bc ten times twenty five is 250
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3 years ago
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Convert the equation:
Licemer1 [7]

Answer:

(x+1)²/6 - (y+3)²/12 = 1

Step-by-step explanation:

The standard form of writing the equation of an hyperbola is expressed as;

(x-h)²/a² - (y-b)²/b² = 1 where (h,k) is the centre of the hyperbola.

Given the equation:

6x²-3y²+12x-18y-3 = 0

We are to convert it to the standard form of writing the equation of a hyperbola.

Collecting the like terms will give;

(6x²+12x)-(3y²+18y)-3 = 0

Divide through by 3

(2x²+4x)-(y²+6y)-1 = 0

Completing the square of the equation in parenthesis and adding the constants to the other side of the equation:

(2x²+4x)-(y²+6y+(6/2)²)-1 = 0+(6/2)²

(2x²+4x)-(y²+6y+9)-1 = 9

2x²+4x -{(y+3)²} = 9+1

2x²+4x - (y+3)² = 10

Divide through by 2

x²+2x -(y+3)²/2 = 5

(x²+2x+(2/2)²)-(y+3)²/2 = 5+(2/2)²

(x²+2x+1) - (y+3)²/2 = 5+1

(x+1)²-(y+3)²/2 = 6

Divide through by 6

(x+1)²/6 - (y+3)²/12 = 1

The resulting equation is the required standard form of a hyperbola with centre at (-1, -3)

8 0
3 years ago
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