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Considering the definition of an equation and the way to solve it, the girls will both pay the same if they buy six pairs of pants and one sweater.

An equation is the equality existing between two algebraic expressions connected through the equals sign in which one or more unknown values appear.

The solution of a equation means determining the value that satisfies it. In this way, by changing the unknown to the solution, the equality must be true.

To solve an equation, keep in mind:

Being "x" the number of pairs of pants, you know that:

The equation in this case is:

17.95x + 24 = 18.95x + 18

Solving:

24 -18= 18.95x - 17-95x

6= x

Finally, this means that Kelsey and Jeana pay the same if they buy six pairs of pants and one sweater.

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The number of quarters and dimes with Maria are 17 and 8 respectively.

Linear equations in mathematics are ones where the greatest power of the variable is one. A linear equation can only have one possible solution: a straight line. Thus, a simultaneous linear equation is a system of two linear equations in two or three variables that are solved concurrently to get a shared answer. Three methods are frequently used to solve simultaneous linear equations: substitution method, elimination method, and graphic method.

Let q be the number of quarters.

Let d be the number of dimes.

Given that the total amount is 5.05 in quarters and dimes

Therefore, 0.25 q + 0.10d = 5.05;

I once more, q = 2d + 1. — (ii) (ii)

Inferring from I and (ii), d = 4.80/0.60 = 8 and 0.25(2d + 1) + 0.10d = 5.05 and 0.60d = 4.80 respectively.

This is put into (ii), and the result is q = 2d + 1 = 2*8 + 1 = 17.

As a result, q = 17 and d = 8

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The geometric sequence has an explicit equation of f(n) = 3^(n - 1)

The given parameters are

r = 3

f(1) = 1

The above means that we have the following parameters:

The explicit equation of the geometric sequence is represented as

f(n) = f(1) * r^(n - 1)

Substitute the known values in the above equation

So, we have the following equation

f(n) = 1 * 3^(n - 1)

Evaluate the products

So, we have the following equation

f(n) = 3^(n - 1)

Hence, the explicit equation of the function is f(n) = 3^(n - 1)

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Answer:

4:3

Step-by-step explanation:

it will be 16:12 and when you divide both by 3 it will be 4:3

Part 1: The first five terms of the sequence are -1, 16, 69, 266, 1039

Part 2: The first five terms of the sequence are 12, 12, 130/9, 69/4, 126

Part 1:

For finding the first five terms we will substitute values in n from 1,2,3,4 and 5

an = 2n^2 + 5n-10

a1 = 2(1)^2 + 5(1)-10

=4+5-10 = -1

So, the first term is -1

a2 = 2(2)^2+5(2)-10

= 16+10-10 =16

The second term is 16

a3 = 2(3)^2+5(3)-10

=64+15-10

=69

The third term is 69

a4 = 2(4)^2+5(4)-10

=256+20-10

=266

The fourth term is 266

a5 = 2(5)^2+5(5)-10

= 1024+25-10

= 1039

The fifth term is 1039

Part 2:

an = 3n+4/n^2+5

a1 = 3(1)+4/1^2+5

= 3+4+5

= 12

The first term is 12

a2 = 3(2)+4/2^2+5

= 6+1+5

= 12

The second term is 12

a3 = 3(3)+4/3^2+5

= 9+4/9+5

= 81+4+45/9

=130/9

The third term is 130/9

a4 = 3(4)+4/4^2+5

= 12+4/16+5

=17+1/4

=69/4

The fourth term is 69/4

a5 = 3(5)+4/5^2+5

= 15+4/25+5

=20+4/25

=504/4

= 126

The fifth term is 126

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Yes, because 6 c+5=2(3 c+2)+1 and 3 c+2 is an integer. Option A

This is further explained below.

Generally, A natural number that is either positive or negative expressed as an integer, or the number zero itself are all examples of integers.

The additive inverses of the positive numbers that they relate to are represented by the negative numbers.

In conclusion, Because 6c+5=2(3c+2)+1 and 3c+2 is an integer, the answer to this question is yes.

is correct, Integer

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CQ

[tex](b) Is\ $6 c+5$ \an \odd \integer?\\\\Yes, because $6 c+5=2(3 c+2)+1$ and $3 c+2$ is \ an \ integer.\\\\Yes, because $6 c+5=2(3 c+2)$ and $3 c+2$ is \ an integer.\\\\No, because $6 c+5=2(3 c+2)+1$ and $3 c+2$ is an integer.\\\\No, because $6 c+5=2(3 c+2)$ and $3 c+2$ is an integer.[/tex]

Answer:

x = (1/2)

x = -3

Step-by-step explanation:

Quadratic:

2x² + 5x = 3

           -3     -3

----------------------

2x² + 5x - 3 = 0

-b ± √b² - 4(a)(c)

      -------------------

             2(a)

-5 ± √5² - 4(2)(-3)

---------------------------      

             2(2)

-5 ± √25 + 24

---------------------------

             4

-5 ± √49

----------------

      4

-5 ± 7

----------

    4

-5 + 7         2         1

---------- = ------- = ------

   4             4         2

-5 - 7        -12

---------- = --------- = -3

   4              4

----------------------------------------------------------------------------------------------------------

Factored (for fun):

(2x² + 6x) (-1x - 3) = 0

2x(x + 3) -1 (x + 3)

(2x - 1)(x + 3) = 0

2x - 1 = 0,    x + 3 = 0

     +1   +1         -3    -3

---------------    ------------------

  2x = 1             x = -3

 ÷2   ÷2

-------------

x = (1/2)

I hope this helps!

The solution to the equations are n(-2) = 4/3, n(0) = 2/3 and n(5) = -1

The equation of the function is given as:

n(x) = -1 -1/3x + 1 2/3

The replacement set is given as

x = -2, 0, 5

So, we replace the variables using the elements in the replacement set

When x = -2, we have

n(-2) = -1 -1/3 x -2 + 1 2/3

Evaluate the equation

So, we have the following equation

n(-2) = 4/3

When x = 0, we have

n(0) = -1 -1/3 x 0 + 1 2/3

Evaluate the equation

So, we have the following equation

n(0) = 2/3

When x = 5, we have

n(5) = -1 -1/3 x 5 + 1 2/3

Evaluate the equation

So, we have the following equation

n(5) = -1

Hence, the values of the functions are 4/3, 2/3 and -1

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The capture of the forts at Vincennes, Kaskaskia and Cahokia help end the war because it helped the Americans control the Appalachian Mountains.

It was the War frontier battle that was fought in present-day Vincennes in which the Indiana won by the militia led by George Rogers Clark over a British garrison led by Henry Hamilton.

The battle happened on 1778 where George Rogers Clark & his men reached Kaskaskia, seized it from the British men and bringing the colonies' battle for independence to the western edge of British territory in North America.

It was one of the battle that happened in the West as the leader of the secret expeditionary forces captured Kaskaskia, Cahokia and Vincennes in 1778 to 1779.

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X   Y
3    1.333
4    10.667
5    12
6    13.333
8    48

The inequality that represents the possible number of hours the chemistry club could rent the meeting room is 13 + 4t ≤ 33 or t ≤ 5.

The possible numbers of hours the chemistry club could rent the meeting room are  1, 2, 3, 4 and 5 hours.

Given that:-

Reservation fee charged to rent the meeting room by college = $ 13

Additional fee per hour charged by the college = $ 4

The maximum amount the chemistry club want to spend on meeting room = $ 33

Let t be the number of hours.

Hence,

13 + 4t ≤ 33

4t  ≤ 33 - 13

4t  ≤ 20

t  ≤ 5 hours.

Hence, the possible number of hours the meeting room can be booked by chemistry club are 1, 2, 3, 4 and 5 hours.

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The function of the  area of the forest in terms of time is A(t) = π(2t + 1)²

The given parameters are

Radius function,  r(t) = 2t + 1

From the question, we understand that:

The pattern is circular

The area of a circle is

A = πr²

Express as a function

A(r) = πr²

Substitute r(t) = 2t + 1

A(t) = π(2t + 1)²

Hence, the area function in terms of time is A(t) = π(2t + 1)²

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The division of  8 1/2 by the positive difference between 4 4/5 and 7 7/10 is 85/29.

We can find the difference between 4 4/5 and 7 7/10, but we will need to convert the mixed number into the improper fraction as

Then we have 24/5 and 77/10, then the difference between the values are:(77/10 - 24/5) = 29/10

Then we can proceed to find the division of this number that we got from the difference as ( 29/10)/  8 1/2

But we can make the  8 1/2 as improper fraction from the mixed fraction it was 17/2

Then the division of the values will be   ( 17/2)/ ( 29/10)  = 85/29

Therefore the division is 85/29.

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depreciated value of truck after 1year is $16863.2  

% is a relative number that is used to represent hundredths of any quantity. Since one percent (1%) equals one tenth of something, 100 percent denotes the entire amount, and 200 percent denotes twice the amount mentioned.

new truck valued depreciates 14.4% then

if old truck value be 100 % , depreciated value will  be 85.6 %

so depreciated value will be = 19700* 85.6 /100

                                                 = $16863.2

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The area of the figure is 144 square centimeters

The area of a shape is the amount of space on the shape

For most regular quadrilaterals, you multiply the side lengths to determine the area

The given parameters are

Shape = parallelogram

Base = 9 cm

Height = 16 cm

The area of a parallelogram shape is

Area = Base * Height

So, we have

Area = 9 * 16

Evaluate the expression

Area = 144

Hence, the area  is 144 square centimeters

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Answer:

48

Step-by-step explanation:

Given:

[tex]\textsf{Maximize}: \quad z=6x+12y[/tex]

[tex]\begin{aligned}&\textsf{Subject to}: \quad &4x+5y & \leq 20\\&&8x+y &\leq 20\\&&x\geq 0, y&\geq 0\end{aligned}[/tex]

Graph the lines:

[tex]\textsf{Draw the line } \;\;4x+5y=20 \;\;\textsf{and shade under the line}.[/tex]

[tex]\textsf{Draw the line } \;\;8x+y=20 \;\;\textsf{and shade under the line}.[/tex]

[tex]\textsf{Draw the line } \;\;x=0\;\;\textsf{and shade above the line}.[/tex]

[tex]\textsf{Draw the line } \;\;y=0\;\;\textsf{and shade above (to the right of) the line}.[/tex]

Therefore, the feasible region is bounded by the corner points:

Determine the value of z at the corner points by substituting the x and y values of the points into the equation for z:

[tex]\textsf{Value of $z$ at $A(0,0)$}: \quad 6(0)+12(0)=0[/tex]

[tex]\textsf{Value of $z$ at $B(0,4)$}: \quad 6(0)+12(4)=48[/tex]

[tex]\textsf{Value of $z$ at $C\left(\dfrac{5}{2},0\right)$}: \quad 6\left(\dfrac{5}{2}\right)+12(0)=15[/tex]

[tex]\textsf{Value of $z$ at $D\left(\dfrac{20}{9},\dfrac{20}{9}\right)$}: \quad 6\left(\dfrac{20}{9}\right)+12\left(\dfrac{20}{9}\right)=40[/tex]

Hence, the maximum value of z is 48 at B(0, 4).

The given equation is 1-degree linear equation in one variable.

What is linear equation?

A linear equation is one that may be written as a1x1+a2x2+......+anxn in mathematics, where a1, a2,...., an are the coefficients, which are frequently real integers. The coefficients, which may be any expressions as long as they don't contain any of the variables, can be thought of as the equation's parameters. The coefficients a1 to a must not all be 0 in order for the equation to have any sense. An alternative method for creating a linear equation is to equalize a linear polynomial over a field, from which the coefficients are drawn, to zero. The numbers that, when used to replace the unknowns in such an equation, result in the equality, are the solutions.

x+1 = x+1 is a 1-degree linear equation in one variable (x).

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The equation of line in slope-intercept form is found as y = -3x + 7.

For the given question;

The two passing point of the line are;

(x1, y1) = (2,1)

(x2, y2) = (1,4)

Slope = m = (y2 - y1)/(x2 - x1)

Put the values.

m = (4 - 1)/(1 - 2)

m = -3

Then, the equation of the line in slope intercept form is found using two point equation.

y - y1 = m(x - x1)

y - 1 = -3(x - 2)

Simplifying,

y = -3x + 7

-3 is the slope and 7 is the y intercept.

Thus, the equation of line in slope-intercept form is found as y = -3x + 7.

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There are 14 students in set R.

There are 10 students in set R but not in set S.

There are 11 students in set S.

There are 21 total students in sets R and S.

There are 4 students common to sets R and S.

The music teacher will get $280 for the piano students.

The music teacher will get $140 for the saxophone students.

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Answer:

5 miles in 50 minutes

Substitute [tex]x=\sin(t)[/tex] and [tex]dx=\cos(t)\,dt[/tex], and recall the definition of the elliptic integral of the second kind,

[tex]\displaystyle E(k) = \int_0^{\pi/2} \sqrt{1 - k^2 \sin^2(\theta)} \, d\theta[/tex]

Then the integral has a value of

[tex]\displaystyle \int_0^1 \sqrt{\frac{2-x^2}{1-x^2}} \, dx = \int_0^{\pi/2} \sqrt{2-\sin^2(t)} \, dt \\\\ ~~~~~~~~~~~~~~~~~~~~~ = \sqrt2 \int_0^{\pi/2} \sqrt{1 - \frac12 \sin^2(t)} \, dt \\\\ ~~~~~~~~~~~~~~~~~~~~~= \boxed{\sqrt2 E\left(\frac1{\sqrt2}\right)}[/tex]

Answer:

2/9

Step-by-step explanation:

The possibilities that satisfy the condition are 4 and 8.

This is 2 numbers out of the possible 9, so the probability is 2/9.

Answer: That is the answer

Step-by-step explanation:

Answer:

v = 4

Step-by-step explanation:

4(2v - 1) = 28

4(2v - 1)/4 = 28/4

2v - 1 = 7

2v - 1 + 1 = 7 + 1

2v = 8

2v/2 = 8/2

v = 4

Answer:

c.

Step-by-step explanation:

13+6.2+5.8

= 13+(6.2+5.8)

- the sum in the parentheses is easy to calculate (it = 12).

Answer: Yes  

Step-by-step explanation: The real number, which most exceeds its cube, is A 21

Answer:

Proofs provided below

Step-by-step explanation:

[tex]\bold {\text{Prove } (a + b)^2 = a^2 + 2ab + b^2}}[/tex]

[tex]\\\\\implies (a + b)^2 = (a+b) \times (a+b)\\\\\implies (a + b)^2 = a \times (a+b) +b \times (a+b)\\\\\implies (a + b)^2 = a \times a + a \times b + b \times a + b \times b\\\\\implies (a + b)^2 = a^2+ab+ba+b^2\\\\\implies (a + b)^2 = a^2+ab+ab+b^2 \;\;\;\;\text{ since ab = ba}\\\\\implies (a+b)^2 = a^2+2ab+b^2\\\\[/tex]

[tex]\bold{\text{Prove } a^2-b^2 \,=\, (a+b)(a-b)\\\\}[/tex]

1. Add and subtract ab to LHS
[tex]\implies a^2-b^2 = a^2-b^2-ab+ab\\\\\implies a^2-b^2 = a^2-ab+ab-b^2\\\\[/tex]

2. Factorize the above expression
[tex]\implies a^2-b^2 = a(a-b)+b(a-b)\\\\\implies a^2-b^2 = (a-b)(a+b) \;\;\;\; \text{since (a-b) is a common factor in RHS }[/tex]

∴  (a² - b²) = (a - b) (a + b)

[tex]\bold{\text{Prove $\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$= $2+\sqrt{\ensuremath{3}}$}}\\[/tex]

[tex]\text{1. Multiply LHS by $\dfrac{\sqrt{\text{}3}-1}{\sqrt{3}-1}$}\\\\\implies $\dfrac{\sqrt{\text{}3}+1}{\sqrt{3}-1}$ \times $\dfrac{\sqrt{\text{}3}-1}{\sqrt{3}-1}$ \\\\[/tex]

[tex]\implies \dfrac{ (\sqrt{3} + 1)(\sqrt{3}-1) }{(\sqrt{3} - 1)(\sqrt{3}-1) }\\\\[/tex]

Numerator is
[tex](\sqrt{3} + 1)(\sqrt{3}-1) = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2\\\\[/tex]

Denominator is
[tex](\sqrt{3}-1)^2 = (\sqrt{3})^2 - 2\cdot \sqrt{3} \;\cdot 1 + (-1)^2\\\\= 3 - 2 \sqrt{3} + 1\\\\= 4 - 2 \sqrt{3}\\\\[/tex]

So LHS becomes
[tex]\dfrac{2}{4 - 2\sqrt{3}} \\\\[/tex]

Dividing numerator and denominator by 2 yields

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

Multiply numerator and denominator by [tex]{2-\sqrt {3}}[/tex]

[tex]$\dfrac{1\cdot(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3)}}$[/tex]

[tex]=$\dfrac{\ensuremath{2}+\sqrt{3}}{2^{2}-(\sqrt{3)^{2}}}=$$\dfrac{\ensuremath{\ensuremath{2}+\sqrt{3}}}{4-3}=\ensuremath{2}+\sqrt{3}$[/tex]

Hence Proved

Answer:

oh god what da heck is this math :')

Step-by-step explanation: