What is the next number in the sequence? 55, 34, 21, 13, 8…?

Answer:

The given system of equations is inconsistent.

Step-by-step explanation:

We have been given a system of equations and we are asked to find whether our given system of equation is consistent, consistent and coincident, or inconsistent.

[tex]y=-3x+12...(1)[/tex]        

[tex]y=-6x+2...(2)[/tex]

Since we know that a system of equations is inconsistent if it has no solution. Parallel lines are inconsistent because they do not intersect.

Consistent system of equations can have one or infinitely many solution. As if two lines are same then they will have infinitely many solution. If lines intersect at only one point, then they will have only one solution.

Now let us check slope of given lines.

Upon dividing equation (2) by 2 we will get,

[tex]y=-3x+1...(2)[/tex]

We can see that both lines have slope of negative 3. Since y-intercept of both equations are different, so they will not intersect at any point.

Since our given equations represent parallel lines, therefore, our given system of equations is inconsistent.

The complete proof of the theorem in chronological order is;

The proof that when a transversal cut two parallel lines, alternate angles are equal.

1.

Statement: l ∥ m

Reason: Given

2.

Statement: m ∠ 1 = m ∠ 3

Reason: If lines are ∥, corresponding angles are equal.

3.

Statement: m ∠ 2 = m ∠ 3

Reason: Vertical angles are equal

4.

Statement: m ∠ 1 = m ∠ 2

Reason: Substitution

Therefore, alternate exterior angles are equal when a transversal cut two parallel lines,

Read more on vertical angles:

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Figure 1)

a) The area is A = 36( π - 2) ст²

b) The perimeter is P = 6 (π + 2√2) [tex]cm^2[/tex]

Figure 2)

a) The area is A = 576 cm²

b) The perimeter is P = 24(π+2) [tex]cm^2[/tex]

Figure 1)

a) To find the area of the figure 1 we have to do the substraction

The area of figure 1 is equivalent to the area of a triangle less the area of a quarter circle.

The area of quarter circle is equal to

A = 2

we have

r = 12 cm

Put the  value r as given

A = (12)2

A = 36 [tex]cm^2[/tex]

Area of a triangle formula is

A = (b)(h)

Given information,

b=12 cm

h = 12 cm

substitute

A = (12) (12)

A = 72 cm²

therefore

The area of the figure is

A = (36π - 72) [tex]cm^2[/tex]

Simplify

A = 36(π -2) [tex]cm^2[/tex]

b)

The perimeter of the figure 1 is equal to the circumference of a quarter circle plus the side AC of triangle

The perimeter of a quarter of circle is formula

C = 2πr

simplify

C = r

we have

r=12 cm

substitute

C = (12)

C = 67 cm

Find the length side AC

Applying the Pythagorean Theorem

[tex]AC^2 = 12^2 + 12^2[/tex]

[tex]AC^2 = 144+ 144\\AC^2 = 288\\AC = \sqrt{288}[/tex]

AC = 12√2 cm

P = (6π +12√2) cm

Taking 6 as a common multiplier we get

P = 6(π+2√2)

The perimeter of the figure 1 is

P = 6(π+2√2) cm

Part 2)

a) As we can see,

The area of a semicircle plus the area of a square less the area of a semicircle equals the area of figure 2.

The figure's area and the square's area are equal.

A = [tex]24^2[/tex]

A = 576 cm²

b) Find the perimeter of the figure 2

we know that

The perimeter of the figure 2 is equal to the length side AB plus the length side DC plus the circumference of two semicircles

The perimeter of the figure 2 is equal to two times the length side AB plus the circumference of one circle

P = 2(AB) + π D

P = 2(24) + π(24)

P = 48 +  π(24)

Take out 24

P = 24 (2 + π ) cm

Answer:

9 is an extraneous solution.

Step-by-step explanation:

Extraneous solution is a solution that when plugged into the equation do not hold true.

Here we are given an algebraic equation in terms of single variable x as:

[tex]\sqrt{x-5}-\sqrt{x}=5-----(1)[/tex]

Now, we will solve this equation to obtain the solution as:

on squaring both side of the equation we obtain:

[tex](\sqrt{x-5}-\sqrt{x})^2=5^2[/tex]

[tex]x-5+x-2\sqrt{x-5}\sqrt{x}=25\\\\2x-5-2\sqrt{x-5}\sqrt{x}=25\\\\2x-30=2\sqrt{x-5}\sqrt{x}\\\\on\ dividing\ both\ side\ by\ 2\ we\ obtain:\\\\x-15=\sqrt{x-5}\sqrt{x}[/tex]

Again on squaring both side of the equation we obtain:

[tex]x^2+225-30x=(x-5)x\\\\\x^2+225-30x=x^2-5x\\\\225=-5x+30x\\\\225=25x\\\\x=9[/tex]

Now when we plug x=9 back to the original equation i.e. equation (1) we get:

[tex]\sqrt{9-5}-\sqrt{9}=5\\\\\sqrt{4}-\sqrt{9}=5\\\\2-3=5\\\\-1=5[/tex]

Hence, the equation does not hold true.

Hence, 9 is a extraneous solution

Answer:

The answer is the option C

IV, III,I,II

Step-by-step explanation:

The factored form of the expression 15a + 25b with respect to the Greatest Common Factor (GCF) is 5 (3a + 5b). The GCF of 15 and 25 is 5.

To factor the Greatest Common Factor (GCF) from 15a + 25b, we first have to identify the GCF. The GCF of 15 and 25 is 5. Hence, we divide each term in the expression by this GCF.

We get: 5 (3a + 5b). So, the expression 15a + 25b factored by the GCF is 5 (3a + 5b).

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Answer:

Compare your personal records to the bank's statement

we have been asked to simplify

[tex] (3x-5)-(5x + 1) [/tex]

To Simplify this, we will first open the parenthesis, as below

[tex] (3x-5)-(5x + 1)=3x-5-5x-1\\
\\
\text{take the like terms together and simplify we get}\\
\\
(3x-5)-(5x + 1)=3x-5x-5-1\\
\\
(3x-5)-(5x + 1)=-2x-6\\
\\

[/tex]

Hence the simplified expression is -2x-6

Answer:

-2x-6

Step-by-step explanation:

Answer:

r=C/2*pie

Step-by-step explanation:

I got it correct on TTM

Answer:

x = -4 and x = -16.

Step-by-step explanation:

We have the equation,

[tex]x^{2}+20x+100=36[/tex]

i.e. [tex]x^{2}+20x+64=0[/tex]

As the solution of a quadratic equation [tex]ax^{2}+bx+c=0[/tex] is given by [tex]x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}[/tex].

On comparing, a = 1, b = 20, c = 64.

So, the solution is,

[tex]x=\frac{-20\pm \sqrt{(20)^{2}-4\times 1\times 64}}{2a\times 1}[/tex].

i.e. [tex]x=\frac{-20\pm \sqrt{400-256}}{2}[/tex].

i.e. [tex]x=\frac{-20\pm \sqrt{144}}{2}[/tex].

i.e. [tex]x=\frac{-20\pm 12}{2}[/tex].

i.e. [tex]x=\frac{-20+12}{2}[/tex] and [tex]x=\frac{-20-12}{2}[/tex].

i.e. [tex]x=\frac{-8}{2}[/tex] and [tex]x=\frac{-32}{2}[/tex].

i.e. x = -4 and x = -16

Thus, the solution of the given equation is x = -4 and x = -16.

Answer:

The scale factor is 2.5 and the coordinates of point C are (–2.4, –0.8).

Step-by-step explanation:

Final answer:

The John Street Barber Shop pays a flat rate of $25 for the first 8,000 gallons of water, with an additional charge of $3.50 per thousand gallons above this threshold. Accordingly, the total costs for using 7,000 gallons is $25, for 10,000 gallons is $32, and for 13,000 gallons is $42.50.

Explanation:

The John Street Barber Shop pays $25 per month for the first 8,000 gallons of water and $3.50 per thousand gallons above 8,000 gallons. To calculate the total water cost when the barber shop uses 7,000 gallons, 10,000 gallons, and 13,000 gallons, we follow this methodology:

Thus, the total costs for using 7,000 gallons, 10,000 gallons, and 13,000 gallons are $25, $32, and $42.50 respectively.

5x-3(2x+7)=2x+9

This is the equation and Raul had to solve it for x

So he started with simplifying the parenthesis first

To simplify the parenthesis -3 gets multiplied with 2x and 7 , to get rid of parenthesis

So it came out to be

5x-3(2x+7)=2x+9

STEP-1

5x-6x -21 = 2x+9 , So the STEP-1 is correct,

After this in Step -2 , he has combined the like terms on left side, the like terms are 5x and -6x , 5x-6x should combine to -x , because 5-6 = -1

So

STEP-2

-x -21 = 2x+9 , So STEP -2 is INCORRECT

So he made a mistake in STEP-2

After that subtract 2x from both sides

-3x -21 = 9

add 21 to both sides

-3x = 30

divide both sides by -3

x=-10

Hence Option B is correct , he made a mistake in STEP -2

The formula to find change in entropy is given by :

[tex] Change in entropy = \frac{Q}{T} [/tex]

where Q is change in heat and T is temperature.

We are given Q = 500 joule and T = 20 degree C

Plugging these values in the formula,

[tex] Change in entropy = \frac{500}{20} [/tex]

We get change in entropy = 25 J/Celsius.