Simplify the expression

the net pay (weekly) is $359.87

First, we know that the employee wins $11.50 per hour. The employee works 38 hours per week, then the weekly pay is:

38*$11.50  = $437

Now we need to discount the taxes:

There are 10% from federal tax, 6.2% from social services and 1.45% from medicare.

The total is:

10% + 6.2% + 1.45% = 17.65%

Then we need to discount a 17.65% of the weekly pay, we will get:

$437*(1 - 17.65%/100%) = $437*(1 - 0.1765) = $359.87

We conclude that the net pay (weekly) is $359.87

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Answer:

Number of orchestra tickets = 8

Number of balcony tickets = 22

Step-by-step explanation:

Let the number of orchestra tickets be x and number of balcony tickets be y.

So, cost of x orchestra tickets = $148 × x = $148x

Cost of y balcony tickets = $65 × x = $65y

Also given, total number of tickets = 30

Cost of 30 tickets = $2,614

So, the system of equations that can ne formed is :

x + y = 30     ....eq 1

148x + 65y = 2614    ....eq 2

Multiplying eq 1 by 148 and subtracting eq 2 by it,

148x + 148y - 148x - 65y = 4440 - 2614

83y = 1826

y = 1826/83 = 22

Substituting y = 22 in eq 1,

x + 22 = 30

So, x = 30 - 22 = 8

Thus,

Number of orchestra tickets = 8

Number of balcony tickets = 22

Answer: 89.21g

Step-by-step explanation:

36.81g + 52.40g = 89.21g

If the  unemployment rate has risen more than a percentage point to 8.9​% in February from 7.1% last​ November, then the relative change in the unemployment rate expressed as a  percentage is 25.35%

Given,

The unemployment rate in February = 8.9%

The unemployment rate in November = 7.1%

The relative change in unemployment rate = [(unemployment rate in February - unemployment rate in November )÷ unemployment rate in November ] ×100

The relative change in unemployment rate= [tex](\frac{8.9-7.1}{7.1} )100[/tex]

=25.35%

Hence, If the  unemployment rate has risen more than a percentage point to 8.9​% in February from 7.1% last​ November, then the relative change in the unemployment rate expressed as a  percentage is 25.35%

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Answer: 111 + 222 = 333

The exact side length of the rug is √35.

What is the area of parallelogram

Let take the area of length as =a

As per the given question

A square rug has an area=35 sq ft

Area is the quantity that measures the number of unit square

As we know

area of square= a^2

Square area formula: A = a²

As already given a^2=35

So a=√35

The exact side length of the rug is √35.

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Answer:

C i think

Step-by-step explanation:

Answer:

C.

Step-by-step explanation:

Answer A is the definition of literal language, while C is the definition of figurative. So, C is correct.

Answer:

for first question the answer will be .. 5

Step-by-step explanation:

easily,

[tex] \frac{1}{5} \times 5 = 1 [/tex]

There are [tex]6\frac{1}{4}[/tex] groups of 1/5 in 5.

The simplified value of  5 ÷ 4/5 is also [tex]6\frac{1}{4}[/tex].

A number line is a one-dimensional horizontal line where we can represent any real number. The origin of the number line is represented by zero left to it are all negative real numbers and right to it are all positive real numbers.

From the bar given above the number line, each group has subgroups of 4.

Therefore, The number of groups of (1/5) in 5 is [tex]6\frac{1}{4}[/tex] groups.

We know when a fraction is in the division we can reciprocate the fraction and the division changes to multiplication,

Therefore, 5 ÷ 4/5 = 5×5/4 = 25/4 = [tex]6\frac{1}{4}[/tex]

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Answer:

=============================

The missing statement is:

Justification is:

All the other options have mismatching or wrong statements or justifications:

The measure of ∠ DCE is equal to 57.5° by using the property of bisectors and vertically opposite angles.

We are given a ray diagram.

In the diagram, the measure of angles are given as:

∠ ACB = 25°

∠ ACG = 90°

We need to find the measure of angle ∠ DCE​ when CE bisects ∠ DCEF.

Now, we know that:

∠ BCG = ∠ DCF ( Vertically opposite angles.)      … (1)

Now,

∠ BCG = ∠ ACB + ∠ ACG

∠ BCG = 25° + 90°

∠ BCG = 115°     … (2)

∠ DCF = ∠ DCE + ∠ CEF

∠ DCF = ∠ DCE + ∠DCE ( as ∠ DCE = ∠ CEF since CE bisects them)

∠ DCF = 2 ∠ DCE   … (3)

Now, put (2) and (3) in (1), we get that:

115° = 2 ∠ DCE

∠ DCE = 115 / 2°

∠ DCE = 57.5°

Therefore, we get that, the measure of ∠ DCE is equal to 57.5° by using the property of bisectors and vertically opposite angles.

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Answer:

Equation: (14.71-4.48) [divided by] 3 =x

"Answer x": 3.41

Explanation:

14.71-4.48=10.23

10.23/3 = 3.41

3.41

76 degrees n 83 degrees it's div

As per the question statement, we are given a mathematical expression [tex]|a+5| -1[/tex] is [tex]$\left[\begin{array}{cc}\text { Solution: } & -\infty < a < \infty \\ \text { Interval Notation: } & (-\infty, \infty)\end{array}\right]$[/tex]

As this question involves a modulus function therefore let us suppose that [tex]b=|a+5|-1[/tex]

Find the absolute value vertex. In this case, the vertex for [tex]b=|a+5|-1[/tex] is [tex](-5,-1)[/tex]

Function range definition

The set of values of the dependent variable for which a function is defined

Domain of [tex]$|a-5|-1: \quad-\infty < a < \infty$[/tex]

Extreme Points of [tex]$|a-5|-1: \quad$[/tex] Minimum [tex]$(5,-1)$[/tex]

Find the range for the interval [tex]$-\infty < a < \infty: \quad-1 \leq f(x) < \infty$[/tex]

Combine the ranges of all domain intervals to obtain the function range [tex]$f(x) \geq-1$[/tex]

Axis interception points of [tex]$|a-5|-1: \quad X$[/tex]Intercepts: [tex]$(4,0),(6,0), Y$[/tex]Intercepts:[tex]$(0,4)$[/tex]

x - axis interception points of[tex]$|a-5|-1: \quad(4,0),(6,0)$[/tex]

y - axis interception point of [tex]$|a-5|-1: \quad(0,4)$[/tex]

X Intercepts: [tex]$(4,0),(6,0), Y$[/tex] Intercepts: [tex]$(0,4)$[/tex]

Asymptotes of [tex]$|a-5|-1$[/tex] : None

Vertical asymptotes of [tex]$|a-5|-1$[/tex] : None

Horizontal Asymptotes of [tex]$|a-5|-1$[/tex] : None

Extreme Points of[tex]$|a-5|-1$[/tex] : Minimum [tex]$(5,-1)$[/tex]

First Derivative Test definition

Suppose that [tex]$x=c$[/tex] is a critical point of [tex]$f(x)$[/tex] then,

If [tex]$f^{\prime}(x) > 0$[/tex] to the left of [tex]$x=c$[/tex]and [tex]$f^{\prime}(x) < 0$[/tex] to the right of [tex]$x=c$[/tex] then $x=c$ is a local maximum.

If [tex]$f^{\prime}(x) < 0$[/tex] to the left of [tex]$x=c$[/tex] and [tex]$f^{\prime}(x) > 0$[/tex] to the right of[tex]$x=c$[/tex] then $x=c$ is a local minimum.

If [tex]$f^{\prime}(x)$[/tex] is the same sign on both sides of [tex]$x=c$[/tex] then [tex]$x=c$[/tex]

is neither a local maximum nor a local minimum.

[tex]f^{\prime}(a)=\frac{a-5}{a-5}$$[/tex]

Plug [tex]$a=5$[/tex] into [tex]$|a-5|-1: \quad-1$[/tex]

Minimum [tex]$(5,-1)$[/tex]

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Answer:6x+3

Step-by-step explanation:

Answer:

  51%

Step-by-step explanation:

Given P(passing math) = 59%, P(passing physics) = 26%, and P(passing both) = 17%, you want to find the probability of passing only one of the courses.

We can record the given probabilities in a 2-way table (values shown in blue). The table is completed by making sure the totals add up (values shown in black).

The probability of passing one course and failing the other is the sum of the probabilities with a yellow background:

  42% +9% = 51%

The probability of passing one or the other is 51%.

__

Additional comment

We can also get there using the relation ...

  P(A+B) = P(A) +P(B) -P(AB)

The union of A and B also includes their overlap:

  P(A+B) = P(AB') +P(A'B) +P(AB)

In other words, the probability of interest is ...

  P(AB') +P(A'B) = P(A) +P(B) -2×P(AB) = 59% +26% -2(17%)

  P(AB') +P(A'B) = 51%

Answer:

21

Step-by-step explanation:

14 - (-20) + (-13) =

= 14 + 20 - 13

= 34 - 13

= 21

Work Shown:

34% of $242,020 = 0.34*242020 = 82,286.80

This rounds to 82,287 when rounding to the nearest whole number.

The vertex of f(x)=x²+10x+1 is found to be  (-5,-24).

The Vertex of the quadratic equation f(x)=ax²+bx+c is (h,k) where

h=-b/2a  and k=f(h).

In the given question

f(x)=x²+10x+1

On comparing with ax²+bx+c

we get a=1 , b=10 , c=1

So , h=-b/2a

h=-10/2(1)

h=-5     ...(i)

k=f(h)

Substituting the value of x=h in f(x) we get

k=(-5)²+10(-5)+1

k=25-50+1

k=-25+1

k=-24

Therefore , the vertex of f(x)=x²+10x+1 is (-5,-24).

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If out of 7 squares, 4 squares are in grey colour then if there are 15 white squares there will be 20 grey squares by solving through ratios.

Given that out of 7 squares, 4 squares are in grey colour.

We are required to find the number of grey colour squares if there are 15 white squares by using equivalent ratios.

Equivalent ratios are basically the ratios that are the same when we compare them. Two or more ratios can be compared with each other to find whether they are equivalent or not.

Ratio of white squares in 7 total squares=3:4

Total number of squares if there are 15 white squares=15*7/3=35

Number of grey squares=35-15=20

Hence if out of 7 squares, 4 squares are in grey colour then if there are 15 white squares there will be 20 grey squares by solving through ratios.

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The equation of the two-step problem is 2x = x + 2/3

The conditions are given as

Let the variable in the equation be x

So, a possible equation is as follows

2x = x + 2/3

The solution to the equation is as follows

2x = x + 2/3

Subtract x from both sides

x = 2/3

Hence, the equation of the two-step problem is 2x = x + 2/3

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Answer:

$28.00

Step-by-step explanation:

Set up an equation:

$22.40 = y(1 - 0.20)

Solve the contents of the parentheses first:

$22.40 = y(0.80)

Divide by 0.80 on both sides to isolate y:

($22.40)/0.80 = y

Solve:

$22.40 / 0.80 = $28.00

Answer:

  (1/3)log(5) +(4/3)log(x) +log(y)

Step-by-step explanation:

You want the expanded form of the expression log(∛5·x^4·y^3).

The applicable rules of logarithms are ...

  log(ab) = log(a) + log(b)

  log(a^b) = b·log(a)

In addition to these, you need to know that a root is a fractional power: the cube root is the 1/3 power. Of course, the rules of powers tell you ...

  (a^b)^c = a^(bc)

You can express the root as a fractional power before or after taking logs. Either way, you get the same result.

Before

  [tex]\log\left(\sqrt[3]{5\cdot x^4y^3}\right)=\log\left(5^{\frac{1}{3}}\cdot x^{\frac{4}{3}}y^{\frac{3}{3}}\right)=\boxed{\dfrac{1}{3}\log(5)+\dfrac{4}{3}\log(x)+\log(y)}[/tex]

After

  [tex]\log\left(\sqrt[3]{5\cdot x^4y^3}\right)=\dfrac{1}{3}\log\left(5}\cdot x^{4}y^{3}}\right)=\dfrac{1}{3}\left(\log(5)+4\log(x)+3\log(y)\right)\\\\=\boxed{\dfrac{1}{3}\log(5)+\dfrac{4}{3}\log(x)+\log(y)}[/tex]

__

Additional comment

Working a more complex problem is like working a simple problem, except it has more parts and may require the use of the rules more times. It may be tedious to keep track of all the parts, but it is not difficult.

Answer:

  x = 25

Step-by-step explanation:

The given angles, x° and (6x+5)° form a linear pair. You want the value of x.

Two angles that together form a straight line are called a linear pair. A linear pair of angles has a total measure of 180°.

  x° +(6x +5)° = 180°

  7x +5 = 180 . . . . . . . . . divide by °, collect terms

  7x = 175 . . . . . . . . . . subtract 5

  x = 25 . . . . . . . . . . divide by 7

The value of x is 25.

The height of the density curve is 1/5

From the graph, we have the following parameter

Base = 5 units

As a general rule, the area under the density curve must have a value of 1

This means that

Area = 1

So, we have

Area = Base * Height

So, we have

1 = 5 * Height

Evaluate

Height = 1/5

Hence, the height of the density curve is 1/5

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[tex]\begin{array}{ccll} \stackrel{recipe}{batch}&\stackrel{recipe}{flour}\\ \cline{1-2} 1&4\frac{2}{3}\\[1em] \frac{3}{4}&f \end{array}\implies \cfrac{1}{ ~~ \frac{3}{4}~~ }~~ = ~~\cfrac{ ~~ 4\frac{2}{3} ~~ }{f}\implies \cfrac{4}{3}~~ = ~~\cfrac{ ~~ \frac{4\cdot 3+2}{3}~~ }{f}\implies \cfrac{4}{3}=\cfrac{ ~~ \frac{14}{3} ~~ }{f} \\\\\\ \cfrac{4}{3}=\cfrac{14}{3f}\implies 12f=42\implies f=\cfrac{42}{12}\implies f=\cfrac{7}{2}\implies f=3\frac{1}{2}[/tex]

Answer:

I am pretty sure it is a i think.

Step-by-step explanation:

just subtract the fractions and you will get 6\10 and the you will simplify and get 3\5. so 3\4 is closest to 3\5.

"Premise: No loving persons are thoughtless persons. Premise: No loving persons are aggressive people. Conclusion: No aggressive people are thoughtless persons." This is a Valid Argument

This is further explained below.

Generally, In the context of deductive reasoning, an argument is considered to be sound if and only if it can be expressed in a way that makes it logically impossible for all of the premises to be true while still leading to an incorrect conclusion.

In conclusion, This is a Valid Argument

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[tex](\stackrel{x_1}{0}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{3}~,~\stackrel{y_2}{0}) ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{0}-\stackrel{y1}{9}}}{\underset{run} {\underset{x_2}{3}-\underset{x_1}{0}}} \implies \cfrac{ -9 }{ 3 }\implies -3 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{9}=\stackrel{m}{-3}(x-\stackrel{x_1}{0}) \\\\\\ y-9=-3x\implies y=-3x+9[/tex]