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The money that the employee earn by working 25 hours in one week is $400.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let f(x) represent the amount of money that the employee earn by working x hours. If the employee work more than 20 hours, then:

f(x) = 15x + 5(x - 20)

For 25 hours:

f(25) = 15(25) + 5(25 - 20) = 400

The money that the employee earn by working 25 hours in one week is $400.

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Answer:

I THINK ITS B

Step-by-step explanation:

CORRECT ME IF I'M WRONG!!!

Answer:

18

Step-by-step explanation:

[tex] \sf( \frac{ ∛125 \: - \: 14 }{2})(3 - 5)2[/tex]

Apply rule: ([tex] a[/tex]) = [tex] a[/tex]

[tex] \sf( \frac{ ∛125 \: - \: 14 }{2}) = \: \sf\frac{ ∛125 \: - \: 14 }{2}[/tex]

[tex] = \sf \frac{∛125 \: - \: 14 }{2}(3 - 5)2[/tex]

Substract the numbers (3-5) = (-2)

[tex] = \sf \frac{∛125 \: - \: 14 }{2}( - 2)2[/tex]

Change 2 to fraction: [tex] \sf \frac{2}{1} [/tex]

[tex] = \sf \frac{∛125 \: - \: 14 }{2} \times \frac{2}{1} ( - 2)[/tex]

Cross- cancel common factor 2

[tex] = \sf \frac{∛125 \: - \: 14 }{1} ( - 2)[/tex]

Let's calculate [tex] \sf \frac{∛125 \: - \: 14 }{1} [/tex]

[tex] = \sf \frac{∛125 \: - \: 14 }{1} [/tex]

Apply a fraction rule = a/1 = a

[tex] = \sf ∛125 \: - \: 14 [/tex]

[tex] \sf = - 9[/tex]

[tex] \sf \: = - 9( - 2)[/tex]

[tex] \sf \: = 18[/tex]

Values of x y z for system of equation are

x=[tex]\frac{565}{54}[/tex]

y=[tex]\frac{703}{54}[/tex]

z=[tex]\frac{-23}{6}[/tex]

we have system of equation as follows

x + 2y+3z = 25

z=(25-x-2y)/3..........(A)

3x - 3y + 4z = -23

z=(3y-3x-23)/4........(B)

-4x+y-z = -25

z=(25-4x+y)..............(C)

Equating we get

(25-x-2y)/3=(3y-3x-23)/4=(25-4x+y)

Solving this we get

x=[tex]\frac{565}{54}[/tex]

y=[tex]\frac{703}{54}[/tex]

z=[tex]\frac{-23}{6}[/tex]

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The number of more drapery pieces than upholstery pieces does Brandon have is 2.03 pieces

Number of pieces of upholstery fabric = 23 3/5 ÷ 2 5/6

= 118/5 ÷ 17/6

= 118/5 × 6/17

= 708 / 85

= 8 28/85

Number of pieces of drapery fabric = 18 1/2 ÷ 2 5/6

= 37/2 ÷ 17/6

= 37/2 × 6/17

= 222/34

= 6 18/34

= 6 9/17

How many more drapery pieces than upholstery pieces does Brandon have = 708 / 85 - 222/34

= (24072-18204) / 2890

= 5,868 / 2890

= 2.03 pieces

Therefore, the number of more drapery pieces than upholstery pieces does Brandon have is 2.03 pieces

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The equation that can model the situation will be 150 + 5n and 20n

Moving Company = 150 + 5n  

(n = the number of boxes)

Taxi = 20n

The break-even point is when these are the same.

150 + 5n = 20n

150 = 15n

n = 150/15 = 10

So with 10 boxes, it costs the same ($200) either way.

With 11 boxes, the Moving Company costs $205

With 11 boxes, the Taxi would cost $220.

It is cheaper to use the moving company with 11 or more boxes.  

Another example of a real life situation that would use systems of equations and write the system is that Company A charges a flat rate of $150, and an additional $5 for each box asn there is another taxi service that would charge $20 for each box.

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Using breaking apart technique 4 X 23 = 92

To make the multiplication process simpler, you split up one or more of the numbers. It also explains why you would "carry" a number to the following column.

split apart definitions. verb: break anything down into its component parts. substitute words: separate, dismantle, take apart

According to the given question

4×23 = (20×4)+(3×4)

         =80+12

          =92

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The domain of the result f/g(x) is (7 /x ) + (5/x^2).

The two functions are:

f(x) = 7x + 5

and

g(x) = x^2

For calculating f/g(x)

f/g(x) is equivalent to f(x) / g(x)

Hence (f / g)(x) = f(x) / g(x)

                        = 7x + 5 / x^2

                        = (7 /x ) + (5/x^2)

the domain of the result f/g(x) is (7 /x ) + (5/x^2).

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Step-by-step explanation:

it is a ratio, and as such it is simply a fraction.

we can also write it

A/B = 6/7

and it gives us the same information.

and so,

30/B = 6/7

30 = 6B/7

210 = 6B

B = 210/6 = 35

Answer:

[tex](x+7)(x+2)[/tex]

Step-by-step explanation:

We have the equation:

[tex]x^{2} +9x+14[/tex].

The factors of 14 are 1, 2, 7 and 14.

Now, all we have to do is find two numbers that add up to 9.

We see that 2 and 7 add up to 9.

Therefore,

[tex]x^2+9x+14=(x+7)(x+2)[/tex]

Hope this helped! Have a great day!

Answer:

(x+7)(x+2)

Step-by-step explanation:

x^2+2x+7x+14           (Because 7 and 2 add to give 9 and multiply to give 14)

x(x+2) +7(x+2)            (Take out common factors, factorising)

(x+2)(x+7)                   (Put the things common in a bracket together and the bits that are outside in another bracket)

Answer:

d = 4

Step-by-step explanation:

the nth term of an AP is

[tex]a_{n}[/tex] = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

given a₄ = 27 , then

a₁ + 3d = 27 → (1)

Also

a₁ + a₄ = 42 ← substitute a₄ = 27

a₁ + 27 = 42 ( subtract 27 from both sides )

a₁ = 15

substitute a₁ = 15 into (1)

15 + 3d = 27 ( subtract 15 from both sides )

3d = 12 ( divide both sides by 3 )

d = 4

Answer: X^3 / Y^5

The first answer choice

Step-by-step explanation:

Answer:

-1.7 = rational

1.083 = rational

1.22345721 = rational

-1.090909 = rational

Step-by-step explanation:

Answer:

x = 3/5

Step-by-step explanation:

Isolate the x by subtracting 2x from both sides

2x + 3 = 7x

-2x        -2x

3 = 5x, Then divide by 5 on both sides

x = 3/5

Answer:

[tex] \sf \large \: x = \frac{3}{5} [/tex]

Step-by-step explanation:

Let's solve your equation step-by-step.

2x+3=7x

Step 1: Subtract 7x from both sides.

2x+3−7x=7x−7x

−5x+3=0

Step 2: Subtract 3 from both sides.

−5x+3−3=0−3

−5x=−3

Step 3: Divide both sides by -5.

−5x/−5 = −3/−5

x=3/5

The domain of given points are :

IR x ≠ ± √2

Points of discontinuity - x = ± √2 ( non removable / asymptote )

x - intercept =± √-5/2 , y -  intercept = 5/2

Any function that can be expressed as a polynomial divided by a polynomial is said to be rational. The domain of a rational function is the set of all numbers except the zeros of the denominator since polynomials are defined everywhere. First example: f(x) = x/ (x - 3).

y=2x²+5 / x²-2 x

= 2[tex]x^{2}[/tex]+5/(x-√2)(x+√2)

The domain of given points are :

IR x ≠  ± √2

Points of discontinuity - x = ± √2 ( non removable / asymptote)

x-intercept :

2√10x=2[tex]x^{2}[/tex]+5+2√10x

2[tex]x^{2}[/tex] = -5

[tex]x^{2}[/tex]=-5/2

x = ±√-5/2

y - intercept :

2 * [tex]0^{2}[/tex] + 5 / ( 0-√2)(0+√2)

= 5/2

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If the sprinter ran the race for 13.245 seconds, if we are to represent it in tenth then it would be written as 13.2

This is the term that is used to refer to the place value that would come immediately after the decimal point that we have in a group of digits that are a decimal.

In the question, the rule I followed is the one that I have explained above. The number that is after the decimal value is given as 2.

Hence we can say that the the answer after it has been rounded to the nearest tenth is  13.2

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Answer:

THE CORRECT ANSWER'S ARE A,D AND F

Step-by-step explanation:

did the test on edge .

The simplest rational exponent form is [tex]x^\frac{3}{4}[/tex] .

We have given a rational exponent [tex]\sqrt{x} .\sqrt[4]{x}[/tex] .     ...(1)

By formula [tex]\sqrt[4]{x}[/tex] can be written as [tex]x^\frac{1}{4}[/tex] and [tex]\sqrt{x}[/tex] can be written as [tex]x^\frac{1}{2}[/tex] .

Put in equation (1) , we get

    [tex]\sqrt{x} .\sqrt[4]{x}=x^\frac{1}{2} .x^\frac{1}{4} \\[/tex]              ...(2)

Using formula [tex]a^\frac{m}{n} \times a^\frac{p}{q} = a^{\frac{m}{n} +\frac{p}{q} }[/tex]  

Equation (2) can be written as

                                       [tex]\sqrt{x} .\sqrt[4]{x}=x^{\frac{1}{2} +\frac{1}{4}}\\\\\sqrt{x} .\sqrt[4]{x}=x^{\frac{2+1}{4} }\\\\\sqrt{x} .\sqrt[4]{x}=x^{\frac{3}{4} }[/tex]

Hence , on simplification , we get  [tex]x^\frac{3}{4}[/tex] .

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Answer:

$50 = 5 days

$40 = 4 days

$30 = 3 days

$20 = 2 days

$10 = 1 day

Step-by-step explanation:

hope this helps = )

The mass of the bar of chocolate in kilogram, kg is calculated to be 1.8kg

Given data

mass of a bar of chocolate is 1800g

density of the chocolate is 9g/cm

density is defined the ratio of mass and volume. therefore the density of the chocolate will be the ratio of the mass of the chocolate bar to the volume of the chocolate. this represented mathematically as:

The mass of the bar of chocolate is given in grams as 1800g

converting the unit to kg is done as follows

if 1kg is equal to 1000g, then how many kg is 1800g

= 1800 / 1000

= 1.8 kg

hence the mass of the chocolate bar in kg is 1.8 kg

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The radius of the tank should increase √ 2 times.

We are given that:

To fill a pond whose volume is given as V₁ = 1 / 2 (4 / 3 π r₁³), we need a cylindrical tank whose volume is given by V₂ = 12 π r₂².

This means that the volume of the pond should be equal to the volume of the cylinder.

So, we get that:

1 / 2 (4 / 3 π r₁³) = 12 π r₂².

4 / 3 π r₁³ = 24 π r₂²

r₁³ = 18 r₂²

r₂ = √ ( r₁³ / 18)

r₂ = r₁ / 3 √r₁ / 2

Now, we will double the radius of the pond.

So, r₁ will become 2 r₁

Substituting it, we get that:

R₂ = 2r₁ / 3 √2r₁ / 2

R₂ = 2r₁ / 3 √r₁

So, we can see that:

R₂ / r₂ = 2r₁ / 3 √r₁ / (r₁ / 3 √r₁ / 2)

R₂ / r₂ = 2 / √2 = √ 2

Therefore, the radius of the tank should increase √ 2 times.

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The required interquartile range of the data set along the median is 300.

The question has given, the median of a data set is 760, the third quartile is 950, and the first quartile is 650,
To determine the interquartile range.

Statistics is the study of mathematics that deals with relations between comprehensive data.

Here,
The first quartile is 650,
The third quartile is 950,
Since, the interquartile range is given as,

IQR = third quartile  - first Quartile
IQR = 950 - 650
IOR = 300

Thus, the required interquartile range of the data set along the median is 300.

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Answer:

Step-by-step explanation:

We have

[tex]f(x) = 6x + 5[/tex]

[tex]g(x) = 3x[/tex]

To find the value of either of these functions for a specific value of x, simply substitute that value into the corresponding function equation.

[tex]\bold{f(2)}[/tex]
Substitute x with 2 in 6x + 5
==> [tex]6\cdot \:2+5=17[/tex]   (Answer)

[tex]\bold{g(4)}[/tex]
Substitute x with 4 in g(x) = 3x
==> [tex]3 \cdot4=12[/tex]  (Answer)

[tex]\bold{f(-2)}[/tex]
Substitute x with -2 in 6x + 5
[tex]6\left(-2\right)+5=-7[/tex]  (Answer)

[tex]\bold{g(0)}[/tex]
Substitute x with 0 in g(x) = 3x
==> [tex]3\cdot0=0[/tex]  (Answer)

[tex]\bold{f(\frac{1}{2})}[/tex]
Substitute x with [tex]\frac{1}{2}[/tex] in 6x + 5
==>  [tex]6\cdot \frac{1}{2}+5 = 3 + 5 = 8[/tex]   (Answer)

[tex]\bold{g(-4)}[/tex]
Substitute x with -4 in g(x) = 3x
==> [tex]3\left(-4\right) = -12[/tex]  (Answer)

The average rate of change or the slope is found as 39.

The slope of the line is given by;

m = (y₂ - y₁)/(x₂ - x₁)

The function is given as;

g(x) = 3x³ + 1

The points are; -1 and 4.

Put x = -1

g(-1) = 3(-1)³ + 1

g(-1) = -2

Put x = 4

g(4) = 3(4)³ + 1

g(4) = 193

The points are;

(x₁ ,y₁) = (-1, -2)

(x₂, y₂) = (4, 193)

Put the values in the slope formula;

m = (193 + 2)/(4 + 1)

m = 195/5

m = 39

Thus, the value of the average rate of change is 39; which is also the slope of the function.

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Answer:

1.3

Step-by-step explanation:

(−2 3/5)(−0.5)

=(−13/5)(−0.5)

=−13/5(−0.5)

=1.3

=1.3

[tex]\begin{array}{llll} 10x=5y-8\\\\ 15x=-5y-2\\\\ \end{array}\implies \stackrel{\textit{using elimination}}{\begin{array}{llll} 10x-5y=-8\\\\ 15x+5y=-2\\\cline{1-1} 25x+0=-10 \end{array}} \\\\[-0.35em] ~\dotfill[/tex]

[tex]25x=-10\implies x=\cfrac{-10}{25}\implies x=-\cfrac{2}{5} \\\\\\ \stackrel{\textit{substituting "x" in the 1st equation}}{10\left(-\frac{2}{5} \right)=5y-8\implies }-4=5y-8\implies 4=5y\implies \cfrac{4}{5}=y \\\\[-0.35em] ~\dotfill\\\\ ~\hfill {\LARGE \begin{array}{llll} \left( -\frac{2}{5}~~,~~\frac{4}{5} \right) \end{array}}~\hfill[/tex]

Answer: Izzy's speed is [tex]6\frac{9}{10}[/tex] mph.

Step-by-step explanation: You have to divide the number of miles by the number of hours.

[tex]4\frac{3}{5} / \frac{2}{3}\\[/tex]

Convert the mixed fraction [tex]4\frac{3}{5}[/tex] to an improper fraction.

[tex]\frac{4 * 5 + 3}{5} / \frac{2}{3} = \frac{23}{5} / \frac{2}{3}[/tex]

Multiply the first fraction by the reciprocal of the second fraction (the second fraction upside down).

[tex]\frac{23}{5} / \frac{2}{3} = \frac{23}{5} * \frac{3}{2} = \frac{23 * 3}{5*2} = \frac{69}{10}[/tex]

Convert the result into a mixed fraction.

69 / 10 = 6 with a remainder of 9

A: [tex]6\frac{9}{10}[/tex]

Answer: x=5/4

Step-by-step explanation:

−6−(−4x+2)−x=-(5x-2)

−6+4x-2−x=-5x+2

-x+4x-6-2=-5x+2

3x-8=-5x+2

3x-8+8=-5x+2+8

3x=-5x+10

3x+5x=-5x+5x+10

8x=10

x=10/8

x=5/4

Hi there,

I hope you and your family are staying safe and healthy during this anomalous time.

[tex]-6 -(-4x + 2) - x = -(5x-2)[/tex]

The first step is to distribute

[tex]-6 + 4x - 2-x=-(5x-2)[/tex]

Now let's subtract the numbers

[tex]-8 +3x = -(5x-2)[/tex]

[tex]3x - 8 = -(5x-2)[/tex]

[tex]3x - 8 = -5x+2[/tex]

[tex]3x = -5x + 2 +8\\3x = -5x +10[/tex]

[tex]3x + 5x = 10 \\8x = 10\\x = \frac{10}{8}[/tex]

Once we simplify that, we get:

[tex]x = \frac{5}{4}[/tex] as the answer

Happy to help! All the best this semester.

Check the picture below.

[tex]~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ A(\stackrel{x_1}{5}~,~\stackrel{y_1}{-1})\qquad B(\stackrel{x_2}{21}~,~\stackrel{y_2}{11}) ~\hfill AB=\sqrt{(~~ 21- 5~~)^2 + (~~ 11- (-1)~~)^2} \\\\\\ ~\hfill AB=\sqrt{( 16 )^2 + ( 12)^2}\implies \boxed{AB=20}[/tex]

[tex]B(\stackrel{x_1}{21}~,~\stackrel{y_1}{11})\qquad C(\stackrel{x_2}{26}~,~\stackrel{y_2}{-1}) ~\hfill BC=\sqrt{(~~ 26- 21~~)^2 + (~~ -1- 11 ~~)^2} \\\\\\ ~\hfill BC=\sqrt{( 5)^2 + ( -12)^2}\implies \boxed{BC=13} \\\\\\ C(\stackrel{x_1}{26}~,~\stackrel{y_1}{-1})\qquad D(\stackrel{x_2}{2}~,~\stackrel{y_2}{-8}) ~\hfill CD=\sqrt{(~~ 2- 26~~)^2 + (~~ -8- (-1)~~)^2} \\\\\\ ~\hfill CD=\sqrt{( -24)^2 + ( -7)^2}\implies \boxed{CD=25}[/tex]

[tex]D(\stackrel{x_1}{2}~,~\stackrel{y_1}{-8})\qquad A(\stackrel{x_2}{5}~,~\stackrel{y_2}{-1}) ~\hfill DA=\sqrt{(~~ 5- 2~~)^2 + (~~ -1- (-8)~~)^2} \\\\\\ ~\hfill DA=\sqrt{( 3)^2 + ( 7)^2}\implies \boxed{DA=\sqrt{58}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\LARGE Perimeter}}{20~~ + ~~13~~ + ~~25~~ + ~~\sqrt{58} ~~ \approx ~~ \text{\LARGE 65.6}}[/tex]

now, as far as the area goes, let's check the picture below, hmmm, there are two triangles, both have a base of 21, and you can see their heights there, well, let's simply get the area of each triangle and sum them up.

[tex]\cfrac{1}{2}(21)(12)~~ + ~~\cfrac{1}{2}(21)(7)\implies 126~~ + ~~73.5\implies \text{\LARGE 199.5}[/tex]