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Answer:

132 degrees

Step-by-step explanation:

Looking at angle A and angle B, they are alternate interior angles. That means they are congruent to one another. Knowing that, we can set up an equation A=B

We can now fill A and B with their given equations

5x-18=3x+42

Now we solve

2x=60

x=30

Now that we know x is 30, we can replace it in the equation for A

5x-18

5(30)-18

150-18

132 degrees

Answer:

132

Step-by-step explanation:

ANGLE A = ANGLE B

(INTERIOR ALTERNATE ANGLES)

5x - 18 = 3x  + 42

2x = 60

x = 30

angle a = 150 - 18

= 132

Answer:

a² + b² = c²

13² + 13² = c²

169 + 169 = c²

338 = c²

c = √338 or 18.385 or 13√2

Answer:

18.4

Step-by-step explanation:

13² + 13² = x²

169 + 169 = x²

338 = x²

x = 18.38477....

Answer:

Option (D)

Step-by-step explanation:

By applying Sine rule in the right ΔABD,

Sin(A) = [tex]\frac{\text{Opposite side}}{\text{Hypotenuse}}[/tex]

Sin(60)° = [tex]\frac{\text{BD}}{\text{AB}}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\text{BD}}{7\sqrt{3}}[/tex]

BD = [tex]7\sqrt{3}\times \frac{\sqrt{3} }{2}[/tex]

     = [tex]\frac{21}{2}[/tex]

Now by applying Cosine rule in the right ΔBDC,

Cos(45)° = [tex]\frac{\text{Adjacent side}}{\text{Hypotenuse}}[/tex]

[tex]\frac{1}{\sqrt{2}}=\frac{\frac{21}{2}}{x}[/tex]

x = [tex]\frac{21}{2}\times \sqrt{2}[/tex]

x = [tex]\frac{21\sqrt{2}}{2}[/tex]

Therefore, Option (D) is the correct option.

Answer:

[tex]V_2=305.14\ \text{inch}^3[/tex]

Step-by-step explanation:

The volume of a gas in a container varies inversely as the pressure on the gas.

[tex]V\propto \dfrac{1}{P}\\\\V_1P_1=V_2P_2[/tex]

If V₁ = 356 inch³, P₁ = 6 pounds/in², P₂ = 7 pounds/in², V₂ = ?

So, using the above relation.

So,

[tex]V_2=\dfrac{V_1P_1}{P_2}\\\\V_2=\dfrac{356\times 6}{7}\\\\V_2=305.14\ \text{inch}^3[/tex]

So, the new volume is [tex]305.14\ \text{inch}^3[/tex].

Answer:

Answer is D.

Step-by-step explanation:

do 4 times every value in the ( )

Answer:

[tex]\boxed{\sf D}[/tex]

Step-by-step explanation:

Answer:

5

Step-by-step explanation:

This shape is formed by two right triangles.

Let's start by the little one.

Let y be the third side.

Using the Pythagorian theorem we get:

y^2 = 6^2 + 3^2

y^2 = 36 + 9

y^2 = 45

y = 3√(5)

●●●●●●●●●●●●●●●●●●●●●●●●

Now let's focus on the second triangle. Let z be the third side.

The Pythagorian theorem:

6^2 + x^2 = z^2

Using the Pythagorian theorem on the big triangle :

[3√(5)]^2 + z^2 = (3+x)^2

45 + z^2 = 3x^2 + 6x + 9

36 +z^2 = 3x^2 +6x

So we have a system of equations.

36+ x^2 = z^2

36 +z^2 = 3x^2 +6x

We want to khow the value of x so we will eliminate z .

Add (36+x^2 -z^2 =0) to the second one.

36 + x^2-z^2+36+z^2 = 3x^2+6x

72 + x^2 = 3x^2 +6x

72 - 2x^2 -6x = 0

Multipy it by -1 to reduce the number of - signs

2x^2 + 6x -72 = 0

This is a quadratic equation

Let A be the discriminant

● a = 2

● b = 6

● c = -72

A = b^2-4ac

A = 36 -4*2*(-72) = 36 + 8*72 =612

So this equation has two solutions

The root square of 612 is approximatively 25.

● (-6-25)/4 = -31/4 = -7.75

● (-6+25)/4 = 19/4 = 4.75 wich is approximatively 5

A distance cannot be negative so x = 5

Answer:

work is shown and pictured

Answer:

-1

Step-by-step explanation:

Twice a number of 4 equals 8

4×2=8

less than 9 of the quotient (answer) of twice a number of 4, is -1

8-9= -1

The required expression for Nine less than the quotient of twice a number and four gives the equation, y =2x/4 - 9.

An expression Nine less than the quotient of twice a number and four. is to be determined.

The process in mathematics to operate and interpret the function to make the function simple or more understandableis called simplify and the process is called simplification.

The quotient is the result when one value gets divided by some other value.

Using simple arithmetic,
Let the number be x,
A number y is equal to nine less than the quotient of twice of x and four. So,
y = quotient of 2x and four - 9
y = 2x/4 - 9

Thus, the required expression for Nine less than the quotient of twice a number and four gives the equation, y =2x/4 - 9.

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Answer:

More number of words that can be made: [tex]\bold{2^n}[/tex]

Please refer to below proof.

Step-by-step explanation:

Given that:

The number of binary code words that can be made:

[tex]B(n) =2^n[/tex]

where n is the length of binary numbers.

Binary numbers means 2 possibilities either 0 or 1.

Here, suppose if we have 5 as the length of binary number.

And there are 2 possibilities for each digit.

So, total number of possibilities will be [tex]2\times 2\times 2\times 2\times 2 = 2^5[/tex]

If the length of binary number is 2.

The total words possible are [tex]2^2[/tex].

These numbers are:

{00, 01, 10, 11}

If the length of binary number is 3. (increasing the 'n' by 1)

The total words possible are [tex]2^3[/tex].

These words are:

{000, 001, 010, 100, 011, 101, 110, 111}

So, number of More binary words = 8 - 4 = 4 or [tex]2^2[/tex] or [tex]2^n[/tex].

So, the answer is [tex]2^n[/tex].

Let us try to prove in generic terms:

[tex]B(n) = 2^n[/tex]

Increasing the n by 1:

[tex]B(n+1) = 2^{n+1}[/tex]

Number of more words made by increasing n by 1:

[tex]B(n+1) -B(n)= 2^{n+1} -2^n\\\Rightarrow 2\times 2^{n} -2^n\\\Rightarrow 2^n(2-1)\\\Rightarrow \bold{2^n}[/tex]

Hence, proved that:

More number of words that can be made: [tex]\bold{2^n}[/tex]

Answer:

if he needs to walk, we can see that between the street and his house he must walk 4 times a distance of 0.5km, so this is a total of 4¨*0.5km = 2km.

Now he has a jet-pack, he can ignore the buildings and just travel in the shorter path, so we can draw a triangle rectangle, in such a way that the hypotenuse of this triangle is the distance between the home and the school.

One of the cathetus is the vertical distance, in this case, is 1km, and the other one is the horizontal distance, also 1km.

So the actual distance is given by the Pythagorean's theorem:

A^2 + B^2 = H^2

Where A and B are the cathetus, and H is the hypotenuse, then:

H^2 = 1km^2 + 1km^2

H = (√2)km = 1.41km.

Now, in the case that he has a jet-pack, he can actually go to the school using this hypotenuse line as his path, so in this case the distance and the displacement would be the same.

Distance: "how much ground an object has covered"

Displacement: "Difference between the final position and the initial position"

When he walks, the distance is 2km, but the displacement is 1.41km

When he uses the jet-pack, both the distance and the displacement are 1.41km

Answer and Step-by-step explanation:

The first thing is we can see in the image, when he walks, that between the house and his school he has to walk four times a distance of 0.5 km.  The result of this is a total of 4¨*0.5 km = 2 km.  The second thing is that he must walk 2 kilometers.  On the other hand, if he has a jetpack, he can simply take the shorter path by ignoring all the buildings.  This idea is where we can draw a triangular rectangle on the map in a way so that the hypotenuse of the triangle is the distance between the school and the home.  As for the Catheti, it is a vertical distance which in this case is two blocks of 0.5 km.  The result is that these catheti have a length of 2*0.5 km = 1 km.  The other is the distance of the horizontal line, which is 1 km.  The absolute distance of this path is given by Pythagorean's theorem, which is A^2 + B^2 = H^2.  Here, A and B are the cathetus, and H is the hypotenuse, then, H^2 = 1 km^2 + 1 km^2.  As well, H = (√2)km = 1.41 km.  Currently, in the situation where he has a jetpack, he can literally fly to the school utilizing this hypotenuse line for the path he would need to follow.  For this specific situation, the displacement, and the distance would be the exact same.  The reason for this is that the definitions of displacement and distance are displacement is the difference between the final position and the initial position and distance is how much area an item has covered.  Also, when he walks, the distance is 2 km and the displacement is 1.41 km.  Also, when he utilizes the jet pack, the distance is equal to the displacement.  Both of these are 1.41 km.

Answer:

[tex] - 24 + 16i + 45i + 15 = 9 + 61i[/tex]

Answer:

Hey there!

(2x+1)(x+1)

2x^2+1x+2x+1

2x^2+3x+1

The answer would be A.

Let me know if this helps :)

Answer:

The 99% confidence interval is  [tex]71.67 < \mu < 78.33[/tex]

Step-by-step explanation:

From the question we are told that

     The sample  size  is  [tex]n = 15[/tex]

      The  sample  mean is  [tex]\= x = 75[/tex]

        The  standard deviation is  [tex]s = 5[/tex]

 Given that confidence is  99%  then the level of significance is mathematically represented as

              [tex]\alpha = 100 - 99[/tex]

             [tex]\alpha = 1\%[/tex]

             [tex]\alpha = 0.01[/tex]

Next we obtain the critical values of  [tex]\frac{ \alpha }{2}[/tex] from the normal distribution table

   The  value is

                  [tex]Z_{\frac{ \alpha }{2} } = 2.58[/tex]

Generally the margin for error is mathematically represented as

            [tex]E = Z_{\frac{ \alpha }{2} } * \frac{ s}{ \sqrt{n} }[/tex]

=>         [tex]E = 2.58 * \frac{ 5}{ \sqrt{15} }[/tex]

=>         [tex]E = 3.3307[/tex]

   The  99% confidence interval is mathematically represented as

             [tex]\= x -E < \mu < \= x +E[/tex]

=>          [tex]75 - 3.3307 < \mu <75 + 3.3307[/tex]

=>          [tex]71.67 < \mu < 78.33[/tex]

Answer:

32  20  17  -57  13

-24  15  -31  31  -28

27  10  -7  18  22

Step-by-step explanation:

*Correct Question:

Which relationships have the same constant of proportionality between y and x as the equation 3y=27x?

Answer:

A, B, C

Step-by-step explanation:

Given that [tex] 3y = 27x [/tex] , we can simplify to get a proportionality statement that exists between X and y. Thus

[tex] 3y = 27x [/tex]

Divide both sides by 3

[tex] \frac{3y}{3} = \frac{27x}{3} [/tex]

[tex] y = 9x [/tex]

Thus, we can say, y would always be 9 times the quantity of x.

From the options given, examine which options have this proportionality statement as well.

Option A, y = 9x is same as the statement.

Option B, 2y = 18x, conforms to the same statement. If we simply further, we would have y = 9x

Option C, the graph shows that when x = 1, y = 9. This also confirms to the same constant of proportionality in the given equation.

Option D and E do not have same constant of proportionality.

The right options are A, B, and C.

Answer:

no of new computers sold : 8

no. of refurbished computer sold : 12

Step-by-step explanation:

Let the no. of new computers sold be x

let the no. of refurbished computers sold be y

Given

In March, the store sold 20 computers

x + y = 20

y = 20-x ----- equation 2

selling price of new computer = $500

selling price of x new computer = 500*x = 500x

selling price of refurbished computer = $200

selling price of y refurbished computer = 200*y = 200y

Total selling price of x new computer and y refurbished computers = 500x+200y

given that

total prioce of computer is $6400

thus

500x+200y = 6400

using y = 20-x  from equation 2

500x+200(20-x) = 6400

=> 500x+ 4000 - 200x = 6400

=> 300x = 6400 - 4000 = 2400

=> x = 2400/300 = 8

Thus,

no of new computers sold = 8

no. of refurbished computer sold = 20 -8 = 12

Answer:

6720 in ^3

Step-by-step explanation:

Volume = length * width * height

            = 30*16*14

                =6720 in ^3

Answer:

a,b and d

Step-by-step explanation:

●You can assign a probality based on your judgement and intuition.

●You can also assigni it based on the data of an histogram, in wich you see the frequency of the event you are interested in.

● Then there is the classical method based on mathematical calculations of equaly likely outcomes.

The three methods of assigning probabilities are:

b. intuition

e. relative frequency

d. equally likely outcomes

Probabilities may occasionally be determined by a person's subjective opinion or personal conviction. This approach depends on the individual's perception of an event's probability or intuition. It is crucial to remember that probabilities based on intuition may not always be precise or trustworthy.

With this approach, probabilities are calculated based on the relative frequencies of historical events that have been observed. Probabilities can be calculated based on the relative frequency of different outcomes by gathering data and calculating their frequencies.

This approach makes the supposition that each potential result has an equal likelihood of happening.

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Answer:

f(1/3) = 6

Step-by-step explanation:

f(x) =-3x+7

Let x = 1/3

f(1/3) =-3*1/3+7

        = -1 +7

       = 6

Answer:

f(1/3) = 6

Step-by-step explanation:

The function is:

● f(x) = -3x+7

Replace x by 1/3 to khow the value of f(1/3)

● f(1/3) = -3×(1/3) +7 = -1 +7 = 6

Answer:

151 9/19

Step-by-step explanation:

Step-by-step explanation:

Option A is the correct answer because it is equal to 151.47

Answer:

75 yards long and 90 yards wide.

Step-by-step explanation:

Let's first find the perimeter of the main rectangle:

100x2 + 65x2 =

330

_________________________________________

Next we need to find two numbers that match:

75 and 90

75x2 + 90x2 =

330

_________________________________________

75x90 is 6750 (More Area)

100x60 is 6500 (Less Area)

Answer:

106.68 centimeters

Step-by-step explanation:

40.64 cm/16 in = x cm/42 in

Cross multiplying, we get:

16x = 40.64(42)

16x = 1706.88

x = 106.68 centimeters

Answer:

86.28 ft²

Step-by-step explanation:

The figure given consists of a rectangle and a semicircle.

The area of the figure = area of rectangle + area of semicircle

Area of rectangle = [tex] l*w [/tex]

Where,

l = 10 ft

w = 8 ft

[tex] area = l*w = 10*8 = 80 ft^2 [/tex]

Area of semicircle:

Area of semicircle = ½ of area of a circle = ½(πr²)

Where,

π = 3.14

r = ½ of 8 = 4 ft

Area of semi-circle = ½(3.14*4) = 6.28 ft²

Area of the figure = area of rectangle + area of semi-circle = 80 + 6.28 = 86.28 ft² (nearest hundredth)

Answer:

the area of the figrue is 105.12

Step-by-step explanation:

Answer:

The  number is  [tex]N =1147[/tex] students

Step-by-step explanation:

From the question we are told that

    The population mean is  [tex]\mu = 281[/tex]

     The standard deviation is  [tex]\sigma = 34.4[/tex]

    The sample size is  n = 2000

percentage of the would you expect to have a score between 250 and 305 is mathematically represented as

      [tex]P(250 < X < 305 ) = P(\frac{ 250 - 281}{34.4 } < \frac{X - \mu }{\sigma } < \frac{ 305 - 281}{34.4 } )[/tex]

Generally  

             [tex]\frac{X - \mu }{\sigma } = Z (Standardized \ value \ of \ X )[/tex]

So  

         [tex]P(250 < X < 305 ) = P(-0.9012< Z<0.698 )[/tex]

       [tex]P(250 < X < 305 ) = P(z_2 < 0.698 ) - P(z_1 < -0.9012)[/tex]

From the z table  the value of  [tex]P( z_2 < 0.698) = 0.75741[/tex]

                                         and  [tex]P(z_1 < -0.9012) = 0.18374[/tex]

     [tex]P(250 < X < 305 ) = 0.75741 - 0.18374[/tex]

      [tex]P(250 < X < 305 ) = 0.57[/tex]

The  percentage is  [tex]P(250 < X < 305 ) = 57\%[/tex]

The  number of students that will get this score is

           [tex]N = 2000 * 0.57[/tex]

           [tex]N =1147[/tex]

Answer:

7). x = 3.2, AC = 42.8

8). a = 6, YZ = 38

Step-by-step explanation:

It is given in the question that a point B is between A and C of the line segment AC.

⇒ AB + BC = AC

By substituting the values of segments,

5x + (9x - 2) = (11x + 7.6)

14x - 2 = 11x + 7.6

14x - 11x = 7.6 + 2

3x = 9.6

x = 3.2

Therefore, AC = 11x + 7.6

                        = 11(3.2) + 7.6

                        = 35.2 + 7.6

                        = 42.8

Question (8).

If Y is a point between X and Z,

⇒ XY + YZ = XZ

By substituting the measures of these segments given in the question,

(3a - 4) + (6a + 2) = (5a + 22)

9a - 2 = 5a + 22

9a - 5a = 22 + 2

4a = 24

a = 6

Since, length of segment YZ = 6a + 2

                                                = 6(6) + 2

                                                = 38

Answer:

38 = JKL

Step-by-step explanation:

JKM = JKL + LKN + NKM

Substituting what we know

104 = JKL + LKN +33

KN  bisects LKM so NKM =  LKN

33 =  LKN

104 = JKL + 33 +33

104 = JKL + 33 +33

Combine like terms

104 = JKL +66

104 - 66 = JKL

38 = JKL

Answer:  ∡JKL=38°

Step-by-step explanation:

KN bidects ∡LKM => ∠KLN=∡NKM=33°

=> ∡LKM=∠KLN+∡NKM=33°+33°=66°

=>∡JKL= ∡JKM-∡LKM= 104°-66°=38°

∡JKL=38°

Answer:

Reserve rate = 9%

Step-by-step explanation:

Reserve ratio/rate is the percentage of deposits which commercial banks are required to keep as cash, as directed by the central banks.

first, let us calculate the reserve amount as follows:

Reserve = Deposit - (free amount to lend out)

Reserve = 28,000 - 25,480 = $2,520

[tex]Reserve\ rate = \frac{Reserves}{Deposits} \times100\\Reserve\ rate = \frac{2520}{28000} \times100\\=\frac{252000}{28000} =9\%[/tex]

Therefore the reserve rate = 9%

a. The first part asks for how many ways they can be seated together in a row. Therefore we want the permutations of the set of 6 people, or 6 factorial,

6! = 6 [tex]*[/tex] 5

= 30 [tex]*[/tex] 4

= 360 [tex]*[/tex] 2 = 720 possible ways to order 6 people in a row

b. There are two cases to consider here. If the doctor were to sit in the left - most seat, or the right - most seat. In either case there would be 5 people remaining, and hence 5! possible ways to arrange themselves.

5! = 5 [tex]*[/tex] 4

= 20 [tex]*[/tex] 3

= 120 [tex]*[/tex] 1 = 120 possible ways to arrange themselves if the doctor were to sit in either the left - most or right - most seat.

In either case there are 120 ways, so 120 + 120 = Total of 240 arrangements among the 6 people if the doctor sits in the aisle seat ( leftmost or rightmost seat )

c. With each husband on the left, there are 3 people left, all women, that we have to consider here.

3!  = 3 [tex]*[/tex] 2 6 ways to arrange 3 couples in a row, the husband always to the left

The area is given by the integral

[tex]\displaystyle A=2\pi\int_Cx(t)\,\mathrm ds[/tex]

where C is the curve and [tex]dS[/tex] is the line element,

[tex]\mathrm ds=\sqrt{\left(\dfrac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt[/tex]

We have

[tex]x(t)=t+\sqrt 2\implies\dfrac{\mathrm dx}{\mathrm dt}=1[/tex]

[tex]y(t)=\dfrac{t^2}2+\sqrt 2\,t+1\implies\dfrac{\mathrm dy}{\mathrm dt}=t+\sqrt 2[/tex]

[tex]\implies\mathrm ds=\sqrt{1^2+(t+\sqrt2)^2}\,\mathrm dt=\sqrt{t^2+2\sqrt2\,t+3}\,\mathrm dt[/tex]

So the area is

[tex]\displaystyle A=2\pi\int_{-\sqrt2}^{\sqrt2}(t+\sqrt 2)\sqrt{t^2+2\sqrt 2\,t+3}\,\mathrm dt[/tex]

Substitute [tex]u=t^2+2\sqrt2\,t+3[/tex] and [tex]\mathrm du=(2t+2\sqrt 2)\,\mathrm dt[/tex]:

[tex]\displaystyle A=\pi\int_1^9\sqrt u\,\mathrm du=\frac{2\pi}3u^{3/2}\bigg|_1^9=\frac{52\pi}3[/tex]