Mia walks her dog twice a day. Her evening walk is two and a half times as far as her morning walk. At the end of the week she says she walked her dog 30 miles. How long is her morning walk?

please mark this answer as brainlist

Answer:

The two odd numbers are 15 and 17

Step-by-step explanation:

Given

Let the odd numbers be represented with x and y

Let x be the greater number

[tex]x + 5y = 92[/tex]

Required

Find x and y

Since x and y are consecutive odd numbers and x is greater, then

[tex]x = y + 2[/tex]

Substitute y + 2 for x in [tex]x + 5y = 92[/tex]

[tex]y + 2 + 5y = 92[/tex]

Collect Like Terms

[tex]y + 5y = 92 - 2[/tex]

[tex]6y = 90[/tex]

Divide both sides by 6

[tex]\frac{6y}{6} = \frac{90}{6}[/tex]

[tex]y = \frac{90}{6}[/tex]

[tex]y = 15[/tex]

Substitute 15 for y in [tex]x = y + 2[/tex]

[tex]x = 15 + 2[/tex]

[tex]x = 17[/tex]

Hence; the two odd numbers are 15 and 17

Answer:

Maths

Step-by-step explanation:

Answer:

The two odd numbers are 15 and 17

Step-by-step explanation:

Given

Let the odd numbers be represented with x and y

Let x be the greater number

Required

Find x and y

Since x and y are consecutive odd numbers and x is greater, then

Substitute y + 2 for x in  

Collect Like Terms

Divide both sides by 6

Substitute 15 for y in  

Hence; the two odd numbers are 15 and 17

Step-by-step explanation:

I think this might be the correct answer

The number of pupils that like neither tasty-time nor ice cream is 6 if in a class of 70 pupils, 36 like tasty time, 34 like ice cream, 6 like both tasty times.

It is defined as the diagram that shows a logical relation between sets.

The Venn diagram consists of circles to show the logical relation.

We have:

In a class of 70 pupils, 36 like tasty time, 34 like ice cream, 6 like both tasty time.

Total = 70 pupils

Number of like tasty time = 36

Number of like ice cream = 34

Number of like both = 6

Let x be the total number of pupils that like neither tasty-time nor ice cream

The number of pupils that like ice cream only =  34 - 6 = 28

The number of pupils that like tasty-time only = 36 - 6 = 30

From the Venn diagram:

28 + 30 + 6 + x = 70

x = 70 - 64

x = 6

Thus, the number of pupils that like neither tasty-time nor ice cream is 6 if in a class of 70 pupils, 36 like tasty time, 34 like ice cream, 6 like both tasty times.

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Answer:

y=8 and x=8 ✔3. the check is meant to be square root symbol lol

Answer: 600 cm³

Step-by-step explanation:

Volume of square pyramid = (1/3) · b² · h

Therefore, the volume can be calculated as

[tex]\frac{1}{3} *10^{2}*18=\frac{1}{3}*100*18=\frac{1}{3}*1800=\frac{1800}{3}=600[/tex]

Answer

900

Step-by-step explanation:

it is pyramid and finding its volume we use the 3 components that is length, with and the height

1/2 base x width x height

5 x 18 x 10 = 900cm³

Answer:

Step-by-step explanation:For what values of does the equation (2 + 1) have two 2

+ 2 = 10 − 6

real and equal roots?

A quadratic equation that has 2 roots that are equal means that the discriminant  of the equation is 0. The value of k that makes the equation have 2 real and equal roots is [tex]k = \frac{5}{6}[/tex]

Given that:

[tex](2k + 1)x^2 + 2x = 10x - 6[/tex]

Rewrite as:

[tex](2k + 1)x^2 + 2x - 10x + 6 = 0[/tex]

[tex](2k + 1)x^2 - 8x + 6 = 0[/tex]

A quadratic equation is represented as:

[tex]ax^2 + bx + c = 0[/tex]

By comparison:

[tex]a=2k+1\\ b =-8\\\ c =6[/tex]

If an equation has 2 real and equal roots, then:

[tex]b^2 = 4ac[/tex]

So, we have:

[tex](-8)^2 = 4 \times (2k + 1) \times 6[/tex]

[tex]64 = 4 \times (2k + 1) \times 6[/tex]

Divide by 4

[tex]16 = (2k + 1) \times 6[/tex]

Divide by 6

[tex]\frac{16}{6} = (2k + 1)[/tex]

[tex]\frac{8}{3} = 2k + 1[/tex]

Subtract 2 from both sides

[tex]2k = \frac{8}{3} -1[/tex]

Take LCM

[tex]2k = \frac{8-3}{3}[/tex]

[tex]2k = \frac{5}{3}[/tex]

Divide by 2

[tex]k = \frac{5}{6}[/tex]

Hence, the value of k that makes the equation have 2 real and equal roots is [tex]k = \frac{5}{6}[/tex]

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Answer:

True

Step-by-step explanation:

We assume that is true for the nth case and prove it for the n+1 case

and show that it is true for the case when n=1

Complete  Question

The complete question is shown on the first uploaded image

Answer:

[tex]X = 1.2857[/tex]

Step-by-step explanation:

Generally from the table the contribution to the chi-square statistic of the number of individuals in the survey aged 20 - 22 who strongly disagreed about whether they observe packing information before purchasing items is mathematically represented as

            [tex]X = \frac{(10 - 7)^2 }{7}[/tex]

            [tex]X = 1.2857[/tex]

Step-by-step explanation:

{ x| x all real numbers}

The domain of f(x) = 5x – 7 will be all real numbers, as the function is a straight line, with no discontinuities, thus undefined at no value of x.

The domain of the function f(x) = 5ˣ - 7 is {x | x is a real number}.

A function is a relation from a set A to a set B where the elements in set A only maps to one and only one image in set B. No elements in set A has more than one image in set B.

Given function is,

f(x) = 5ˣ - 7

The domain of the function is the set of all values of x such that the function is defined.

If we use any number for x, the function will be defined.

So the domain is the set of all real numbers.

So the correct option is last one {x | x is a real number}.

Hence the domain of the given function is {x | x is a real number}.

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Answer:

The values is  

Step-by-step explanation:

From the question we are told that

  The population mean is  [tex]\mu = 2.45[/tex]

    The  standard deviation is  [tex]\sigma = 0.35 \ mi[/tex]

     The random value is  [tex]x = 2.03[/tex]

The standardized score for a binding site position of 2.03 microns is mathematically represented as

       [tex]z-score = \frac{x - \mu}{ \sigma }[/tex]

=>      [tex]z-score = \frac{2.03 - 2.45}{ 0.35}[/tex]

=>    [tex]z-score = -1.2[/tex]

The distance between each of the students' houses are:

Aakash & Bao Ying live 9 blocks away from each other

Aakash & Chris live 3 blocks away from each other

Aakash & Donna live 66 blocks away from each other

Bao Ying & Chris live 6 blocks away from each other

Bao Ying & Donna live 57 blocks away from each other

Chris & Donna live 63 blocks away from each other

To solve this question, we represent each with the first letter of their names.

So, we have:

[tex]A= 5\ west[/tex] --- Aakash

[tex]B= 4\ east[/tex] ---- Bao Ying

[tex]C = 2\ west[/tex] --- Chris

[tex]D = 61\ east[/tex] ---- Donna

I've added the picture and the number line that represent the scenario as an attachment.

To calculate how far each lives from each other's, we start by representing west as negative and east as positive.

So, the representations become:

[tex]A = -5[/tex]

[tex]B = 4[/tex]

[tex]C = -2[/tex]

[tex]D = 61[/tex]

The absolute difference of each represents how far they live from one another.

Aakash & Bao Ying         Aakash & Chris                 Aakash & Donna

[tex]AB = |-5 -4| =9[/tex]          [tex]AC = |-5 --2| =3[/tex]         [tex]AD = |-5 -61| =66[/tex]

Bao Ying & Chris                 Bao Ying & Donna

[tex]BC = |4 --2| =6[/tex]                [tex]BD = |4 -61| =57[/tex]

Chris & Donna

[tex]CD = |-2 -61| =63[/tex]

This means that

Aakash & Bao Ying live 9 blocks away from each other

Aakash & Chris live 3 blocks away from each other

Aakash & Donna live 66 blocks away from each other

Bao Ying & Chris live 6 blocks away from each other

Bao Ying & Donna live 57 blocks away from each other

Chris & Donna live 63 blocks away from each other

To carry out the above computations from the number line, we simply count the numbers from one to the other.

For instance:

There are 3 numbers from Akash to Chris.

This means that: Aakash & Chris live 3 blocks away from each other

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Answer: 48

Step-by-step explanation: 60-12=48

Answer:

x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I

Step-by-step explanation:

1−2sin2  x≤−sin x    ⇒    (2sin x+1)(sin x−1)≥0

sin x≤−1/2    or    sin x≥1

−5π/6+2nπ≤x≤−π/6+2nπ    or    , n ϵ I x=(4n+1)π/2, n ϵ I⇒    -5π6+2nπ≤x≤-π6+2nπ    or    , n ϵ I x=4n+1π2, n ϵ I     (as sin x = 1 is valid only)

In general⇒    In general    x∈[2nπ−5π/6, 2nπ−π/6]∪{(4n+1)π/2}, n ϵ I

Answer:

the other person is right

you should try putting the WHOLE question

Step-by-step explanation:

Answer:

2^(1/6) (cos(-pi/12)+i sin(-pi/12))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Step-by-step explanation:

Let's convert to polar form.

-2i=2(cos(A)+i sin(A) )

There is no real part so cos(A) has to be zero and since we want -2 and we already have 2 then we need sin(A)=-1 so let's choose A=-pi/2.

So z=2(cos(-pi/2)+i sin(-pi/2)).

There are actually infinitely many ways we can write this polar form which we will need.

z=2(cos(-pi/2+2pi k)+i sin(-pi/2+2pi k))

where k is an integer

Now let's find the 6 6th roots or z.

2^(1/6) (cos(-pi/12+2pi k/6)+i sin(-pi/12+2pi k/6))

Reducing

2^(1/6) (cos(-pi/12+pi k/3)+i sin(-pi/12+pi k/3))

Plug in k=0,1,2,3,4,5 to find the 6 6th roots.

k=0:

2^(1/6) (cos(-pi/12+pi (0)/3)+i sin(-pi/12+pi (0)/3))

=2^(1/6) (cos(-pi/12)+i sin(-pi/12))

k=1:

2^(1/6) (cos(-pi/12+pi/3)+i sin(-pi/12+pi/3))

2^(1/6) (cos(3pi/12)+i sin(3pi/12))

k=2:

2^(1/6) (cos(-pi/12+2pi/3)+i sin(-pi/12+2pi/3))

2^(1/6) (cos(7pi/12)+i sin(7pi/12))

k=3:

2^(1/6) (cos(-pi/12+3pi/3)+i sin(-pi/12+3pi/3))

2^(1/6) (cos(11pi/12)+i sin(11pi/12))

k=4:

2^(1/6) (cos(-pi/12+4pi/3)+i sin(-pi/12+4pi/3))

2^(1/6) (cos(15pi/12)+i sin(15pi/12))

2^(1/6) (cos(5pi/4)+i sin(5pi/4))

k=5:

2^(1/6) (cos(-pi/12+5pi/3)+i sin(-pi/12+5pi/3))

2^(1/6) (cos(19pi/12)+i sin(19pi/12))

Answer:

4a+6b

Step-by-step explanation:

4a -2a +6b +2a

Combine like terms

4a-2a+2a+6b

4a+6b

Answer:

4a + 6b

Step-by-step explanation:

4a - 2a + 6b + 2a

= 4a + 6b

(because -2a and +2a get cancelled by each other)

Answer:

One half, or 1/2.

There are an equal amount of odd numbers as there are even numbers on the spinner.

Answer:

C. 1/2

One-half

Step-by-step explanation:

Volume of unit cube= (1)³ unit³= 1 unit³

Answer:

E(w) = 1600000

v(w) = 240000

Step-by-step explanation:

given data

sequence = 1 million iid  (+1 and +2)

probability of transmitting a +1 =  0.4

solution

sequence will be here as

P{Xi = k } = 0.4              for k = +1

                  0.6              for k = +2

and define is

x1  + x2 + ................ + X1000000

so for expected value for W

E(w) = E( x1  + x2 + ................ +  X1000000 )   ......................1

as per the linear probability of expectation

E(w) = 1000000 ( 0.4 × 1 + 0.6 × 2)

E(w) = 1600000

and

for variance of W

v(w) = V ( x1  + x2 + ................ + X1000000 )    ..........................2

v(w) = V x1  + V x2 + ................  + V  X1000000

here also same as that xi are i.e d so cov(xi, xj ) = 0 and i ≠ j

so

v(w) = 1000000 ( v(x) )

v(w) = 1000000 ( 0.24)

v(w) = 240000

Answer:

840 ( D )

Step-by-step explanation:

GIVEN DIGITS : 1,2,3,4,5,6,7,8  

Number of odd numbers = 4

Number of even numbers = 4

therefore the number of odd numbers with 4 different digits can be formed by the same way the number of even numbers ( without repetition )

Hence the number of ways odd numbers with 4 different digits = Total number of ways of forming 4 digit numbers / 2

8*7*6*5 = 1680 / 2 =  840 ways

Answer:

The number to be multiplied is the "multiplicand"

Step-by-step explanation:

a base when it is written in exponential notation

9514 1404 393

Answer:

  125.6 cm²

Step-by-step explanation:

The relevant area formula is ...

  Area = (1/2)ac·sin(B)

  Area = (1/2)(13 cm)(21 cm)·sin(67°) ≈ 125.6 cm²

Answer: $5650

Step-by-step explanation:

El precio de la carrera es:

y = ($50/km)*x + $4500.

Donde x representa la cantidad recorrida en Km.

Ahora se nos pregunta:

¿ Cuanto cuesta una carrera si la distancia recorrida fue de 23 kilómetros?

Para esto, debemos reemplazar la variable en la equacion por 23km:

x = 23km

y = ($50/km)*23km + $4500 = $5650

Answer:

distance from the flying object to

the ground

= 7.2 melo(unit of measurement)

Step-by-step explanation:

The distance between the robot and Jo is 5 melo( unit Of measurement)

Let the distance between the flying object and the ground= y

Let's the remaining length of the closest between robot and Jonny and the ground be x.

Y/(x+5)= tan 29.... equation 1

Y/x= tan 42.... equation 2

Equating the value of y

Tan 29(x+5) = tan42(x)

Tan29/tan 42 = x/(x+5)

0.61562(x+5)= x

3.0781= x- 0.61562x

3.0781= 0.38438x

3.0781/0.38438= x

8.008= x

8= x

Y/x= tan 42

Y/8= 0.9004

Y= 7.203

Y= 7.2 melo (unit of measurement )

Answer:

V = 2143.57 cm^3

Step-by-step explanation:

The volume of a sphere is

V = 4/3 pi r^3

The diameter is 16 so the radius is 1/2 of the diameter or 8

V = 4/3 ( 3.14) (8)^3

V =2143.57333

Rounding to the nearest hundredth

V = 2143.57 cm^3

Answer:

2143.57 cm^3

Step-by-step explanation:

V = 4/3 * 3.14 * r^3

r = 1/2 * 16 = 8

So V = 4/3 * 3.14 * 8^3

= 2143.57 cm^3.

Answer: 55958180.97

Step-by-step explanation:

9.3 x 9.3 x 9.3 x 9.3 x 9.3 x 9.3 x 9.3 x 9.3

86.49 x 86.49 x 86.49 x 86.49

= 55958180.97

Answer:

The number is 11.

Step-by-step explanation:

[tex]3(8+x)=57\\24+3x=57\\3x=33\\x=11[/tex]

Answer:

Step-by-step explanation:

g(x) has the smallest minimum value. All three functions share the same domain and the y-intercept is also the same for all three thus options (C),(D), and (E) is correct.

A graph is a diametrical representation of any function between the dependent and independent variables.

For example y = x² form a parabola now by looking at only the graph we can predict that it has only a positive value irrespective of the interval of x.

As per the given table,

The minimum value among all functions is -204 which is for g(x).

The domain is the set of x since all function defines the same x thus they have the same domain.

At y-intercept x = 0

Since at x = 0 all function is 3 thus all three will have the same y-intercept.

Hence "The smallest minimal value is for g(x). The y-intercept for all three functions is the same, and all three functions have the same domain".

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