0.6 reoccurring decimal as a fraction in the simplest term​

Answer:

10.2

Step-by-step explanation:

Length of arc=(2*pi*r)*(theta/360)

Length of arc=(2*pi*3)*(195/360)=10.2

It will take Fred 42 minutes to type and spell check 60 pages.

Given the following data;

To find how long it takes him to type and spell check 60 pages;

We would use direct proportion to express the work described and solve for the unknown variable (time).

[tex]9 \;minutes : 13 \;pages\\\\X \;minutes : 60 \;pages[/tex]

Cross-multiplying, we have;

[tex]9[/tex] × [tex]60[/tex] [tex]= 13[/tex] × [tex]X[/tex]

[tex]540 = 13X\\\\X = \frac{540}{13}[/tex]

X = 41.54 ≈ 42 minutes

Therefore, it will take Fred 42 minutes to type and spell check 60 pages.

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The answer is prime! I hope this helps you out!

Answer:

23 is a prime number. Reason: Prime number are those numbers which are divisible by 1 and itself. Example: 5 is divisible by 1 and 5 only.

[tex]\\ \sf\longmapsto 2(3x-7)+4(3x+2)=6(5x+9)+3[/tex]

[tex]\\ \sf\longmapsto 6x-14+12x+8=30x+54+3[/tex]

[tex]\\ \sf\longmapsto 6x+12x-14+8=30x+57[/tex]

[tex]\\ \sf\longmapsto 18x+6=30x+57[/tex]

[tex]\\ \sf\longmapsto 30x-18x=6-57[/tex]

[tex]\\ \sf\longmapsto 12x=51[/tex]

[tex]\\ \sf\longmapsto x=\dfrac{51}{12}[/tex]

Answer:

The choose C. 2

[tex]( \frac{4 ^{ \frac{5}{4} } \times 4^{ \frac{1}{4} } }{4 ^{ \frac{1}{2} } } )^{ \frac{1}{2} } = 2[/tex]

Answer:

2

Step-by-step explanation:

inside the parenthesis is equivalent to 4^(5/4 + 1/4 - 1/2) = 4¹ = 4

√4 = 2

.・。.・゜✭・Srry but which language is this!?

9514 1404 393

Answer:

  π = 2A/r²

Step-by-step explanation:

Multiply by the inverse of the coefficient of π.

  A = π(r²/2)

  π = 2A/r²

Answer:

sum of two numbers is 30

Step-by-step explanation:

let three numbers are x,y,z

average =x+y+z/3

x+y+z/3=16

x+y+z=48......(1)

The sum of two numbers is 18.

according to condition:

let x=18

subtitute x=18 in (1)

18+y+z=48

y+z=48-18

y+z=30

Answer:

1080

Step-by-step explanation:

PEMDAS says multiply and divide from left to right

9*5*6*3÷3*4*6÷6

45*6*3÷3*4*6÷6

270*3÷3*4*6÷6

810÷3*4*6÷6

270*4*6÷6

1080*6÷6

6480÷6

1080

Answer:

Hello,

Answer C

Step-by-step explanation:

Since ln(1)=0

if x=1 then y=4 ==> y=ln(x)+4

y=ln(x) is translated up for 4 units.

9514 1404 393

Answer:

  (a) 5.82 cm (correctly shown)

  (b) 10.53 cm

Step-by-step explanation:

a) The length BC can be found from the law of sines:

  BD/sin(C) = BC/sin(D)

  BC = BC·sin(C)/sin(D) = (6 cm)sin(48°)/sin(50°) ≈ 5.82 cm

__

b) The angle ABD is the sum of the angles shown:

  angle ABD = 50° +48° = 98°

We know the lengths BA and BD and the included angle ABD, so we can use the law of cosines to find AD.

  AD² = BA² +BD² -2·BA·BD·cos(98°)

  AD² ≈ 8² +5.82² -2(8)(5.82)(-0.139173) ≈ 110.8411

  AD ≈ √110.8411 ≈ 10.53 . . . . cm

Answer:

y - 9 = 2 (x - 2)

Step-by-step explanation:

y2 - y1 / x2 - x1      9 - 3 / 2 - (-1)      6/3    = 2

y - 9 = 2 (x - 2)

Answer:

Swimming pool

Step-by-step explanation:

Let's say that we invite x people. The cost of the bowling alley will be $15 for the room, and we add $5 dollars for each person. Therefore, we can represent the cost of the bowling alley as

5 * x + 15

Similarly, the cost of the swimming pool is

4 * x + 12

To find the maximum of the function, one thing we can do is set the expressions of the cost equal to 40. This will give us the amount of people we can invite for exactly 40 dollars. Because cost increases with amount of people, this will give us the maximum number of people we can invite with $40.

We thus have

5 * x+ 15  = 40

subtract 15 from both sides to isolate the x and its coefficient

25 = 5 * x

divide both sides by 5 to isolate x

25/5 = x = 5

We can therefore invite 5 people to the bowling alley

4 * x+ 12  = 40

subtract both sides by 12 to isolate the x and its coefficient

28 = 4 * x

divide both sides by 4 to isolate x

x = 7

Therefore, we can invite 7 people to the swimming pool. As 7 > 5, we can invite more people to the swimming pool.

A quicker way to solve this could be by looking at the cost of each location. Because the bowling alley costs more both per person and for the room, and there is no way to decrease the cost except by decreasing the amount of people, there is no way that the bowling alley could get as many people as the swimming pool with the same amount of money

Answer:

[tex]{ \sf{ \underline{ {x}^{2} + 5x - 6 }}}[/tex]

Step-by-step explanation:

[tex](x + 6)(x - 1) \\ = {x}^{2} - x + 6x - 6 \\ = {x}^{2} + 5x - 6[/tex]

Answer:

Step-by-step explanation:

Temperature is a function of the time of day. There can be only one temperature at any particular time.

In contrast, a particular temperature can occur more than one time in a day, so time is not a function of temperature.

No piece of information is supplied to choose from ; However, the requirements for the test statistic of the hypothesis in question is given below.

Answer:

Kindly check explanation

Step-by-step explanation:

The hypothesis in the question above which is to be tested is a one sample proportion test ;

The test statistic formula for a one sample proportion test is given as :

Test statistic = (Phat - P) / √[P(1 - P) / n] ;

Where;

Phat = sample proportion

P = population proportion

n = sample size

Sample proportion = x / n ;

Therefore, once the parameters (Sample proportion, population proportion and sample size are given) then obtaining the test statistic is possible.

Answer:

h(x) = 2x^2 +14x -60

Step-by-step explanation:

A quadratic is of the form

h(x) = ax^2 + bx +c

h(3) = h(-10) = 0

This tells us that the zeros are at x=3 and x = -10

We can write the equation in the form

h(x) = a( x-z1)(x-z2) where z1 and z2 are the zeros

h(x) = a(x-3) (x- -10)

h(x) = a(x-3) (x+10)

FOIL

h(x) = a( x^2 -3x+10x-30)

h(x) = a(x^2 +7x -30)

Let a = 2

h(x) = 2x^2 +14x -60

It means

zeros are 3 and -10

Form equation

Multi ply by 2

Option D

Answer:

angleABC=isosceles triangle

angleB=(180-50)÷2=65

angle B=angleX(alternative angle)

angleB=65degree

Answer:

the equation of the circle is x^2 + y^2 < 36

NOT LESS OR EQUAL cause of the dotted lines

and the theory behind this is because the square root of 36 is +-6 so when the equation is less than +-6 the shade cannot go outside these point, if you know what i mean

hope that answers your question :)

Answer:

Ans:

i) y-int (0 , 2) and x-int (-2/3 , 0)

ii) y = 3x + 2

Step-by-step explanation:

we know:

m = 3 & point (0,2)

i

y-int is (0,2) because where x is 0 it intersects the y-axis

ii

from this we can sub it into the point slope formula which is:

y - y1 = m (x - x1)          (y1 and x1 being the values from a point lying on the line)

y - 2 = 3 (x - 0)

y - 2 = 3x

y = 3x + 2     < this is the equation

if you also need to find the x-int then:

to find x-int make y = 0 and solve for x

y = 3x + 2                     3x = - 2

0 = 3x + 2                       x = -2/3

therefore x-int is (-2/3,0)

Answer:

x = -1/4

Step-by-step explanation:

[tex]81^{x}[/tex] = 1/3

[tex]3^{4}^{x}[/tex] = [tex]3^{-1}[/tex]

4x = -1

x = -1/4

Answer:

x = - [tex]\frac{1}{4}[/tex]

Step-by-step explanation:

Using the rules of exponents

[tex](a^m)^{n}[/tex] = [tex]a^{mn}[/tex] , [tex]a^{-m}[/tex] = [tex]\frac{1}{a^{m} }[/tex]

Given

[tex]81^{x}[/tex] = [tex]\frac{1}{3}[/tex] [ note that 81 = [tex]3^{4}[/tex] ] , then

[tex](3^4)^{x}[/tex] = [tex]3^{-1}[/tex] , that is

[tex]3^{4x}[/tex] = [tex]3^{-1}[/tex]

Since the bases on both sides are equal, both 3, then equate exponents

4x = - 1 ( divide both sides by 4 )

x = - [tex]\frac{1}{4}[/tex]

The equivalent resistance of the network shown is: [tex]R_{eq}=16.463\Ohm[/tex]

So, first of all, we can start by determining what each of the resistances is equal to. In this case we can start by saying that [tex]R_{1}=11R_{11}[/tex] This means that:

[tex]R_{1}=11(3\Ohm)[/tex]

Therefore:

[tex]R_{1}=33\Ohm[/tex]

We can now use this value to find the value of the other resistances. The given condition for the resistances can be represented with the following formula:

[tex]R_{n}=\frac{R_{1}}{n}[/tex]

so:

[tex]R_{2}=\frac{R_{1}}{2}=\frac{33}{2}\Ohm[/tex]

[tex]R_{3}=\frac{R_{1}}{3}=\frac{33}{3}\Ohm=11 \Ohm[/tex]

[tex]R_{4}=\frac{R_{1}}{4}=\frac{33}{4}\Ohm[/tex]

[tex]R_{5}=\frac{R_{1}}{5}=\frac{33}{5}\Ohm[/tex]

[tex]R_{6}=\frac{R_{1}}{6}=\frac{33}{6}\Ohm=\frac{11}{2}\Ohm[/tex]

[tex]R_{7}=\frac{R_{1}}{7}=\frac{33}{7}\Ohm[/tex]

[tex]R_{8}=\frac{R_{1}}{8}=\frac{33}{8}\Ohm[/tex]

[tex]R_{9}=\frac{R_{1}}{9}=\frac{33}{9}\Ohm=\frac{11}{3}\Ohm[/tex]

[tex]R_{10}=\frac{R_{1}}{10}=\frac{33}{10}\Ohm[/tex]

[tex]R_{11}=3\Ohm[/tex]

So now that we know what each resistance is equal to, we can go ahead and analyze the circuit.

In this case we can see that [tex]R_{10}[/tex] and [tex]R_{11}[/tex] are parallel, so we can calculate their equivalent resistance.

[tex]R_{1011}=\frac{R_{10}R_{11}}{R_{10}+R_{11}}[/tex]

Which yields:

[tex]R_{1011}=\frac{(\frac{33}{10}\Ohm)(3\Ohm)}{\frac{33}{10}\Ohm+3\Ohm}[/tex]

[tex]R_{1011}=\frac{11}{7}\Ohm[/tex]

Now, [tex]R_{8}[/tex], [tex]R_{9}[/tex] and [tex]R_{1011}[/tex] are connected in series, so we can calculate their equivalent resistance like this:

[tex]R_{891011}=R_{8}+R_{9}+R_{1011}[/tex]

[tex]R_{891011}=\frac{33}{8}+\frac{11}{3}+\frac{11}{7}[/tex]

[tex]R_{891011}=\frac{1573}{168}\Ohm[/tex]

Now,  we can see that [tex]R_{7}[/tex] and [tex]R_{891011}[/tex] are parallel, so we can calculate their equivalent resistance.

[tex]R_{7891011}=\frac{R_{7}R_{891011}}{R_{7}+R_{891011}}[/tex]

Which yields:

[tex]R_{7891011}=\frac{(\frac{33}{7}\Ohm)(\frac{1573}{168})}{\frac{33}{7}\Ohm+\frac{1573}{168}\Ohm}[/tex]

[tex]R_{7891011}=\frac{4719}{1505}\Ohm[/tex]

Now, [tex]R_{5}[/tex], [tex]R_{6}[/tex] and [tex]R_{7891011}[/tex] are connected in series, so we can calculate their equivalent resistance like this:

[tex]R_{567891011}=R_{5}+R_{6}+R_{7891011}[/tex]

[tex]R_{567891011}=\frac{33}{5}\Ohm+\frac{11}{2}\Ohm+\frac{4719}{1505}\Ohm[/tex]

[tex]R_{567891011}=15.236\Ohm[/tex]

Next,  we can see that [tex]R_{4}[/tex] and [tex]R_{567891011}[/tex] are parallel, so we can calculate their equivalent resistance.

[tex]R_{4567891011}=\frac{R_{4}R_{567891011}}{R_{4}+R_{567891011}}[/tex]

Which yields:

[tex]R_{4567891011}=\frac{(\frac{33}{4}\Ohm)(15.236\Ohm)}{\frac{33}{4}\Ohm+15.236\Ohm}[/tex]

[tex]R_{4567891011}=5.352\Ohm[/tex]

Now, [tex]R_{2}[/tex], [tex]R_{3}[/tex] and [tex]R_{4567891011}[/tex] are connected in series, so we can calculate their equivalent resistance like this:

[tex]R_{234567891011}=R_{2}+R_{3}+R_{4567891011}[/tex]

[tex]R_{234567891011}=\frac{33}{2}\Ohm+11\Ohm+5.352\Ohm[/tex]

[tex]R_{234567891011}=32.852\Ohm[/tex]

Finally, we can see that [tex]R_{1}[/tex] and [tex]R_{234567891011}[/tex] are parallel, so we can calculate their equivalent resistance.

[tex]R_{eq}=\frac{R_{1}R_{234567891011}}{R_{1}+R_{234567891011}}[/tex]

Which yields:

[tex]R_{eq}=\frac{(33\Ohm)(32.852\Ohm)}{33\Ohm+32.852\Ohm}[/tex]

[tex]R_{eq}=16.463\Ohm[/tex]

See attached picture for images on how to reduce the circuit.

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Answer:

the cost of filling the pool is $386.3

Step-by-step explanation:

Given;

dimension of the rectangular pool, = 25 yd by 30 yd by 5.1 ft

covert the given dimensions to feet;

1 yd = 3 ft

25 yd = 25 x 3 ft = 75 ft

30 yd = 30 x 3 ft = 90 ft

The volume of the rectangular pool in cubic feet (ft³);

Volume = 75 ft  x 90 ft  x 5.1 ft

Volume = 34,425 ft³

Convert the volume to gallons;

1 ft³ = 7.481 gallons

10,125 ft³ = 34,425 x 7.481 gallons

               = 257,533.425 gallons

The cost per a 1000 gallons is $1.50, then cost of the 257,533.425 gallons is calculated as;

[tex]cost = \frac{\$ 1.50}{1000 \ gallons} \times 257,533.425 \ gallons = \$ 386.3[/tex]

Therefore, the cost of filling the pool is $386.3

Answer:

35 full bags

Step-by-step explanation:

$305-$35.75=$269.25

$269.25 divided by $7.50=35.9 bags (round off to lowest for number of full                                            bags)