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Over [174]
3 years ago
10

How many hundredths are in 539.26? Is the answer 6? Or is 6 just in the hundredth’s spot

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
3 0

Answer:

its 600 because its in the hundredths place so it would be 6 hundredths

Step-by-step explanation:

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Five numbers has a mean of 12. When one (1) number is removed the mean is 11. What is the value of the number removed?
bonufazy [111]

Answer:

16

Step-by-step explanation:

The mean of a group of values is calculated as

mean = \frac{sum}{count}

Given 5 numbers with a mean of 12, then

\frac{sum}{5} = 12 ( multiply both sides by 5 )

sum = 60

let the number removed be x , then

\frac{60-x}{4} = 11 ( multiply both sides by 4 )

60 - x = 44 ( subtract 60 from both sides )

- x = - 16 ( multiply both sides by - 1 )

x = 16

The number removed was 16

6 0
3 years ago
6(y+5) distributive property
galben [10]

Answer:

6y + 30

Step-by-step explanation:

<u>Step 1:  Distribute</u>

6(y + 5)

6y + 30

Answer:  6y + 30

4 0
3 years ago
Read 2 more answers
Am I correct? ( top question)
EastWind [94]

Answer:

72

Step-by-step explanation:

<1 and <2 are not equal to each other

Let the angle  directly above angle 2 (on the right side of the line) be angle 3

<1 and <3 are corresponding angle, which means they are equal

<2 and <3 are supplementary angles since the form a line

<2 + <3 = 180

We know <1 = <3

<1 + <3 = 180

We are given <1 = 2x+12  and <2 = 3x+18

2x+12 + 3x+18 = 180

Combine like terms

5x+30 = 180

Subtract 30 from each side

5x+30-30 = 180-30

5x= 150

Divide each side by 5

5x/5 = 150/5

x=30

<1 = 2x+12

Substitute 30 in for x to find angle 1

   = 2*30 +12

   =60+12

  = 72

6 0
3 years ago
Find an equation in standard form for the hyperbola with vertices at (0, ±6) and foci at (0, ±9).
Dvinal [7]
<span>For given hyperbola:
center: (0,0)
a=7 (distance from center to vertices)
a^2=49
c=9 (distance from center to vertices)
c^2=81
c^2=a^2+b^2
b^2=c^2-a^2=81-49=32
Equation of given hyperbola:


..
2: vertices (0,+/-3) foci (0,+/-6)
hyperbola has a vertical transverse axis
Its standard form of equation:  , (h,k)=(x,y) coordinates of center
For given hyperbola:
center: (0,0)
a=3 (distance from center to vertices)
a^2=9
c=6 (distance from center to vertices)
c^2=36 a^2+b^2
b^2=c^2-a^2=36-9=25
Equation of given hyperbola:
</span>
4 0
3 years ago
Find the missing side . round the the nearest tenth
lys-0071 [83]

S\frac{O}{H} C\frac{A}{H} T\frac{O}{A}

Sin=\frac{Opposite}{Hypotenuse} Cos=\frac{Adjacent}{Hypotenuse} Tan=\frac{Opposite}{Adjacent}

7) You have the opposite. You need the adjacent.

Tan=\frac{Opposite}{Adjacent}

Tan57=\frac{12}{x}

x=\frac{12}{Tan57}

x = 7.79289... =

<h2>7.8</h2><h2 />

8) You have the hypotenuse. You need the Opposite.

Sin=\frac{x}{Hypotenuse}

Sin37=\frac{x}{13}

x = sin37 × 13

x = 7.8235... =

<h2>7.8</h2><h2 />

9) You have the hypotenuse. You need the opposite.

Sin=\frac{x}{Hypotenuse}

Sin59=\frac{x}{11}

x = sin59 × 11

x = 9.4288.... =

<h2>9.4</h2><h2 />

10) You have the hypotenuse. You need the opposite.

Sin=\frac{x}{Hypotenuse}

Sin53=\frac{x}{11}

x = sin53 × 11

x = 8.7459.... =

<h2>8.7</h2>
7 0
2 years ago
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