Pls help 10pts which model shows 33 1/3?

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.

Answer:

gg

Step-by-step explanation:

Answer:

Three consecutive integers are 43, 44, 45

Step-by-step explanation:

Let three consecutive integers be x, (x + 1), (x + 2)

According to the question, if the third of three consecutive integers is subtracted from 132, the result is the sum of the first and second.

[tex] \therefore [/tex]

[tex] \implies \rm 132 - (x + 2) = x + (x + 1) \\ \\ \implies \rm 132 - x - 2 = x + x + 1 \\ \\ \implies \rm 130 - x = 2x + 1 \\ \\ \implies \rm 130 - 1 = 2x + x \\ \\ \implies \rm 129 = 3x \\ \\ \implies \rm 3x = 129 \\ \\ \implies \rm x = \frac{129}{3} \\ \\ \implies \rm x = 43[/tex]

So,

The integers are as follows:

x = 43

x + 1 = 44

x + 2 = 45

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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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:

  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:

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

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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Answer:

46

Step-by-step explanation:

13a+4b

Substitute

13(2)+4(5)

Multiply

26+20

Add

46

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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Work Shown:

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

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

Answer:

For Ai - P = 4 , Q = 4 , R = 4

For Aii - P = 3 , Q = 4 , R = 5

For Aiii - b^13 × b^-2 / b^1 = b^10

Step-by-step explanation:

I hope this is correct answer.

Answer:

The 1st option is correct.

We are not clear that WX = XY.

3rd is is not correct.

4th option is also incorrect.

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.

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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......................

[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]

Step-by-step explanation:

Across

1.26

2.59

3.19

5.38

6.82

7.62

8.48

9.72

11.40

12.64

Down

1.24

2.59

3.18

4 42

5.32

6.88

7.65

8.42

9.70

10.34

Answer:6x+3

Step-by-step explanation:

Hence , the coordinate of point B is (7 , -3) .

We have given two coordinate A (-3,3) and M (2,0) .

Let the coordinate of B (x, y) .

Using midpoint formula , we get

[tex](2,0) = ( \frac{-3+x}{2} ,\frac{3+y}{2} )[/tex]

On comparing , we get

[tex]\frac{-3+x}{2} =2\\\ \\-3+x = 4\\x = 7[/tex]

[tex]\frac{3+y}{2} =0\\3+y=0\\y=-3[/tex]

We get coordinate (7 , -3) .

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"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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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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Based on the time that it took Kenny and his family to reach the campsite and set it up, the time when they finished setting up the campsite was 8:22 pm

First, find the time that Kenny and his family reached their campsite:
= 3:48 pm + 2 hours 59 minutes

= 6:47 pm

After getting to the campsite at 6:47 pm, Kenny and his family took 1 hour 35 minutes to finish setting up their campsite:

= 6:47 pm + 1 hour 35 minutes

= 8:22 pm

In conclusion, Kenny and his family finished setting up at 8:22 pm.

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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]

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:

21

Step-by-step explanation:

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

= 14 + 20 - 13

= 34 - 13

= 21

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.

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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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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