Simplify [4 -⅓ ( 64 ⅓ - 64 ⅔) ]​

Answer:

For b), the answer is 105mm

Step-by-step explanation:

(150+60)/2

An inequality that shows all of the ways that Nita will win if Erik chooses 17 is

This refers to the relation in mathematics that makes an unequal comparison between two numbers or expressions.

Therefore, according to the rules of the game, they would have to make random selections from 0 to 20 and if the difference between their two numbers is less than 10, then Erik wins, the inequality that would show all of the ways that Nita will win if Erik chooses 17 are listed above.

Hence, we can see that the complete question goes thus:

"Erik and Nita are playing a game with numbers. In the game, they each think of a random number from zero to 20. If the difference between their two numbers is less than 10, then Erik wins. If the difference between their two numbers is greater than 10, then Nita wins.

Use what you know about inequalities and absolute values to better understand the game."

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Answer:

[tex]sin\ x = \frac{15}{17}[/tex]

Step-by-step explanation:

So in this case, it's similar to the previous question you asked, except this time you know cosine, and as you may know cosine is defined as: [tex]\frac{adjacent}{hypotenuse}[/tex] and sine is defined as: [tex]\frac{opposite}{hypotenuse}[/tex]. So all we need to solve for is the opposite side, but since we know two sides, we can solve for the other using the Pythagorean identity: [tex]a^2+b^2=c^2[/tex]

Plug in known values:
[tex]8^2 + b^2 = 17^2[/tex]

Simplify:

[tex]64 + b^2 = 289\\b^2 = 225\\b = 15[/tex]

So the opposite side is 15, and the hypotenuse is already given, in this case it's 17 (the denominator of cosine). So plugging this into the definition of sin gives you: [tex]\frac{15}{17}[/tex]

Answer:

[tex] \frac{15}{17} [/tex]

Step-by-step explanation:

[tex] { \sin(x) }^{2} + { \cos(x) }^{2} = 1[/tex]

[tex] \cos(x) = 8 \div 17[/tex]

[tex] \sin(x) = \sqrt{1 - { \cos(x) }^{2} } [/tex]

[tex] \sin(x) = \sqrt{1 - {(8 \div 17)}^{2} } [/tex]

[tex] \sin(x) = 15 \div 17[/tex]

If Carol’s final grade is a 91, Carol's lowest test score was 82.

Given;

Average in 9 tests = 90

Average when the lowest score is removed = 91

The average or Mean is calculated as thus;

Average = [tex]\dfrac{\sum x}{n}[/tex]

Where the summation of x is the sum of test scores.

n = number of tests

n = 9 - 1 = 8 and average = 91

91 =  [tex]\dfrac{\sum x}{8}[/tex]

[tex]\sum x = 91 \times 8\\\\\sum x = 728[/tex]

Thus, Carol's test score in 8 subjects is 728

Let's consider the 9th subject represent the subject Carol had the least score.

n = 9

Average = 90

Average = 728 + the 9th subject

Multiply both sides by 9

90 x 9 = 728 + the 9th subject

810 = 728 + the 9th subject

Subtract 728 from both sides

810 - 728 = 728- 728 + the 9th subject

82 = the 9th subject

Hence, Carol's lowest test score was 82.

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The length of A"B" is 20 units

From the figure, we have:

A = (1, 4)

B = (4, 8)

The distance AB is:

[tex]AB = \sqrt{(x_2 -x_1)^2 + (y_2 -y_1)^2[/tex]

So, we have:

[tex]AB = \sqrt{(1 -4)^2 + (4-8)^2[/tex]

Evaluate

[tex]AB = \sqrt{25[/tex]

This gives

AB = 5

The scale factor of dilation is 4.

So, we have:

A'B' = 5 * 4

Evaluate

A'B' = 20

Hence, the length of A"B" is 20 units

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Answer:

Step-by-step explanation:

The lengths um, md and ud are 255 units, 215 units and 470 units respectively, when given that points u, m, and d are collinear with m between u and d and um=5x+30, md=3x+80, ud=10x+20. This can be obtained by adding um and md which is equal to ud. An equation is obtained and solve for x.

Given that, um=5x+30  

                  md=3x+80

                  ud=10x+20

We know that m is the midpoint of ud

Thus, um + md = ud

                 5x+30 + 3x+80 = 10x+20

                 5x+3x + 30+80 = 10x+20

                      8x + 110 = 10x+20 (taking terms with x to one side of equation)

                             10x - 8x = 110 - 20

                                     2x = 90

                                   ⇒ x = 45

Therefore the lengths,

um=5x+30 = 5(45)+30 = 225 + 30 = 255

md=3x+80 = 3(45)+80 = 135 + 80 = 215

ud=10x+20 = 10(45)+20 = 450 + 20 = 470

Hence the lengths um, md and ud are 255 units, 215 units and 470 units respectively, when given that points u, m, and d are collinear with m between u and d and um=5x+30, md=3x+80, ud=10x+20.

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Answer:

2=A

Step-by-step explanation:

We know that F is directly proportional to A then the equation relating them is

F = kA ← k is the constant of proportion

To find k use the condition F = 36 when A = 9, then

36 = 9k ( divide both sides by 9 )

4 = k

F = 4A ← equation of proportion

When F = 8, then

8 = 8A ( divide both sides by 4 )

2 = A

Answer:

For plan B to save money cell phone user need to send 6000 texts per month as [tex]$\frac{x}{y}$[/tex] expresses the average texts sent per month by cell phone user and its obtained value is 6000.

Step-by-step explanation:

In the question it is given that a cell phone company offers two plans for minutes.

Plan A: $15 per month and $2 for every 300 texts.

Plan B: $25 per month and $0.50 for every 100 texts.

It is required to find that how many texts would be needed to send per month for plan B to save money. be needed to send per month for plan B to save money.

Step 1 of 6

In Plan A $15 per month and $2 for every 300 texts are costed so the cost of Plan [tex]$\mathrm{A}$[/tex] is given by following equation,

[tex]$A=15 y+\frac{2 x}{300}$[/tex]

In Plan B [tex]$\$ 25$[/tex] per month and [tex]$\$ 0 \cdot 50$[/tex]for every 100 texts are costed so the cost of Plan B is given by following equation,

[tex]$B=25 y+\frac{0 \cdot 50 x}{100}$[/tex]

Step 2 of 6

Now comparing the obtained equations [tex]$A=15 y+\frac{2 x}{300}$[/tex]

and [tex]$B=25 y+\frac{0 \cdot 50 x}{100}$[/tex]

[tex]$15 y+\frac{2 x}{300}=25 y+\frac{0 \cdot 50 x}{100}$[/tex]

Step 3 of 6

Subtract $15 y$ from both the sides of the obtained equation [tex]$15 y+\frac{2 x}{300}=25 y+\frac{0 \cdot 50 x}{100}$[/tex] and simplify using subtraction properties.

[tex]$$\begin{aligned}&15 y+\frac{2 x}{300}-15 y=25 y-15 y+\frac{0.50 x}{100} \\&\frac{2 x}{300}=10 y+\frac{1 x}{200}\end{aligned}$$[/tex]

Step 4 of 6

Subtract [tex]$\frac{x}{200}$[/tex] from both the sides of the obtained equation [tex]$\frac{2 x}{300}=10 y+\frac{1 x}{200}$[/tex] and simplify using subtraction properties.

[tex]$$\begin{aligned}&\frac{2 x}{300}-\frac{x}{200}=10 y+\frac{1 x}{200}-\frac{x}{200} \\&\frac{x}{600}=10 y\end{aligned}$$[/tex]

Step 5 of 6

Multiply both the sides of the obtained equation [tex]$\frac{x}{600}=10 y$[/tex] by 600 and simplify using multiplication properties.

[tex]$$\begin{aligned}&\frac{x}{600} .600=10 y .600 \\&x=6000 y\end{aligned}$$[/tex]

Step 6 of 6

Divide both the sides of the obtained equation x=6000 by y and simplify using division properties. As [tex]\frac{x}{y}[/tex] expresses the average texts sent per month by cell phone user. So, for plan B to save money cell phone user need to send 6000 texts per month.

[tex]$$\begin{aligned}&\frac{x}{y}=\frac{6000 y}{y} \\&\frac{x}{y}=6000\end{aligned}$$[/tex]

The inverse variation equation is y = -5/x

Equation is the defined as mathematical statements that have a minimum of two terms containing variables or numbers is equal

y varies inversely as x

Let variation constant = k

y = k/x, y = 25 and x = -15

25 = k/(-15),  k = -375 divded by 75 ,So k = -5

The inverse variation equation is y = -5/x

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In time period of 1980-1988 the rate of ticket price is $0.2 per year
Between time period 1989-1993 there is constant rate.
Between year 1994-2011 the increase in rate is same $0.2 per year

The graph could be divided up into three different periods of relatively consistent ticket price change: The years 1980 – 1988, 1989 – 1993 and 1994 – 2011.

The statistic is the study of mathematics which deal with relations between comprehensive data.

The graph is not available, in the question, so the graph could be as attached
For period 1980-1988
rate of change = 4.2-2.8/ 8 = 0.2
In time period of 1980-1988 the rate of ticket price is $0.2 per year

for period 1989-92 there is a straight line so,
Between time period 1989-1993 there at constant rate.

For period, 1994-2011

rate of change = 4.4-8/17 = 0.2
Between year 1994-2011 the increase in rate is same $0.2 per year

Thus, for the 3 Time period we have rate of change in ticket price is $0.2 per year, no change in ticket price, $0.2 per year respectively.

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Answer:

1, 4, 9, 16, 25, 36, 49, (....)

Answer:

Limit does not exist

Step-by-step explanation:

This is a piece-wise function, since we are finding the limit of f(x) when x is approaching 2 from the right, we know that x must be more than 2. However, based on the piece-wise function, there is no function that satisfy this condition, therefore the limit does not exist for this piece-wise function

Answer:

[tex]x=23, y=23\sqrt3[/tex]

Step-by-step explanation:

Since this triangle is half an equilateral triangle, x is half of 46 = 23

Use Pythagoras to solve y.

23^2 + y^2 = 46^2

[tex]y=23\sqrt3[/tex]

Answer: x∈∅ (there is no decision).

Step-by-step explanation:

[tex]\left \{ {{x+y=5} \atop {\frac{1}{x} +\frac{1}{y}=\frac{9}{14} }} \right.\ \ \ \ \ \left \{ {{x+y=5} \atop {\frac{y+x}{x*y}=\frac{9}{14} \ |*14*x*y\ (x\neq 0;\ y\neq 0) }} \right. \ \ \ \ \left \{ {{x+y=5} \atop {14*(x+y)=9*x*y}} \right. \ \ \ \ \left \{ {{x+y=5} \atop {14*5=9*x*y}} \right.\ \ \ \ \left \{ {{x+y=5} \atop {9*x*y=70}} \right. \\[/tex]

[tex]\left \{ {{y=5-x} \atop {9*x*y=70\ |:9}} \right. \ \ \ \ \left \{ {{y=5-x} \atop {x*(5-x)=\frac{70}{9} }} \right.\ \ \ \ \left \{ {{y=5-x} \atop {5x-x^2=\frac{70}{9} }} \right. \ \ \ \ \left \{ {{y=5-x} \atop {x^2-5x+\frac{70}{9} =0\ |*9}} \right.\ \ \ \ \left \{ {{y=5-x} \atop {9x^2-45x+70=0}} \right.[/tex]

[tex]\left \{ {{y=5-x} \atop {D=-495}} \right. \ \ \ \ \Rightarrow\ \ \ \ x\in\varnothing.[/tex]

Answer:

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

Step-by-step explanation:

Given Equation of line: y=mx+c,

where m = slope of line.

So, we will have to make y the subject to find the slope.

[tex]2x-3y=19\\2x-3y-2x=19-2x\\-3y=-2x+19\\y=\frac{-2x+19}{-3} \\y=\frac{2}{3} x-\frac{19}{3}[/tex]

From here, we can see the slope of the line is [tex]\frac{2}{3}[/tex].

Answer: 2/3

Step-by-step explanation:

2x-3y=19

minus 2x from both sides

-3y=-2x+19

divide both sides by -3

y=2/3x-19/3

slope is 2/3

36 cm²s⁻¹ is the rate at which the area of the square is increasing.

Given each side of the square is increasing at a rate of 6cm/s

Area, A = s², s is the side length

=> [tex]\frac{dA}{dt} =\frac{dA}{ds}* \frac{ds}{dt} =2s*(\frac{ds}{dt} )[/tex] {chain rule of differentiation}

[tex]\frac{ds}{dt}[/tex] = 6 cm/s

When A = 9 cm² => 9 = s²

=> s = √(9) => s = 3 cm

Hence, [tex]\frac{dA}{dt}[/tex] = 2 × 3 × 6 = 36 cm²s⁻¹

Hence, 36 cm²s⁻¹ is the rate at which the area of the square increases

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Answer:

15%

Step-by-step explanation:

$2325/$15500

Answer: 15

Step-by-step explanation: It’s given that the deductions reduce the original amount of taxes owed by $2,325.00. Since the deductions reduce the original amount of taxes owed by d%, the equation 2,32515,500=d100 can be used to find this percent decrease , d . Multiplying both sides of this equation by 100 yields 232,50015,500=d, or 15 = d. Thus, the tax deductions reduce the original amount of taxes owed by 15%.

The value of x is 14

From the question, the lines from the center of the circle to the chords have equal measures of 9 units

This line divides a chord into equal segments, where each segment is 7 units

So, we have:

x = 7 + 7

Evaluate

x = 14

Hence, the value of x is 14

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Answer:

488

Step-by-step explanation:

Calculate 20% of their take-home pay

3600 * 0.20 = 720

They already pay 122 and 110 per month

720 - (122 + 110)

720 - 232

488

So the largest monthly car payment they can afford is 488

Answer:

$7.76

Step-by-step explanation:

So you can express the price in the slope-intercept form: y=mx+b where m is the slope, but in this context it's the price per km,  and b is the y-intercept, but in this context it's the base price (how much it costs even if the length is 0 km)

So it gives you the slope of $0.22 per km. so you now have the equation

y = 0.22x + b

To find the base amount or y-intercept, you can take some of the given points, the price will be y, and the length will be x.

I'll use the point (6, 4.68)

Plugging these values into the equation you get:

4.68 = 0.22(6) + b

multiply

[tex]4.68 = 1.32 + b[/tex]

subtract 1.32 from both sides

[tex]3.36 = b[/tex]

So now we have the full equation:

y = 0.22x + 3.36

Now to find the price if the length is 20km, simply plug in 20 as x

y = 0.22(20) + 3.36

Multiply

y = 4.4 + 3.36

Add

y=7.76

Answer:

y = (2/5) OR y = (6/5)

Step-by-step explanation:

The first step is isolating the expression within the absolute value bars. The first thing we can do is subtract both sides by 8. If we do that, we get -2|4-5y| = -4. Now, to completely isolate the absolute value, we would have to divide by -2. This yields |4 - 5y| = 2. Finally, we can remove the absolute value bars. However, to do this, we need to first understand what an absolute value bar does to an equation. Lets say that |x| = 2. Absolute value describes the DISTANCE of some quantity from 0 (on the number line). Therefore, x (which is inside the absolute value bars) can be either positive or negative 2 (they are BOTH two units away from 0). Similarly, in this case, (4 - 5y) can either be 2 or -2 (because the absolute value of both is 2). Now we have two possible solutions to solve for:

4 - 5y = 2 OR 4 - 5y = -2

5y = 2 OR 5y = 6

y = (2/5) OR y = (6/5)

If we plug both of these answers back into the equation we can see that they both check out.

Scale factor = (image)/(preimage)

[tex]QR=3\\\\Q'R'=1[/tex]

So, the scale factor is 1/3

The horizontally compress the exponential function f(x)=2^x by a factor of 3 is f(3x)=2^3x

When we compress a function y=f(x) by a scale factor of 'k', where k >1, then the function after compression will become :

y=f(x), where k>1

We have given the Exponential function.

The scale factor is a measure for similar figures, that look the same but have different scales or measures. Suppose, two circle looks similar but they could have varying radii. The scale factor states the scale by which a figure is bigger or smaller than the original figure.

The scale factor of the compression (k): 5

Then after horizontal compression, the exponential function will become

f(3x)=2^3x

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Answer:

Answer is Similar-AA :D

Reason:

We're given these two angle congruences:

[tex]\angle P \cong \angle S\\\\\angle Q \cong \angle T[/tex]

These are the blue and red angles marked in the diagram below. Since we have two pairs of congruent angles, we can use the AA similarity theorem to prove the triangles are similar.

Similar triangles have the same shape, but are likely different sizes. Triangles with the same shape indicates the corresponding angles are congruent.

The info that PR = 12 and SU = 3 isn't used or needed. Your teacher may have put that in there as a distraction or filler.

The value of the function operation f(a + h) - f(a) is; h² + 5h + 2ah

We are given the function;

f(x) = x² + 5x

We want to find f(a + h) - f(a)

Let us first find f(a + h) to get;

f(a + h) = (a + h)² + 5(a + h)

⇒ a² + 2ah + h² + 5a + 5h

⇒ a² + 5a + h² + 5h + 2ah

Let us now find f(a) to get;

f(a) = a² + 5a

Thus, f(a + h) - f(a) = a² + 5a + h² + 5h + 2ah - a² - 5a

f(a + h) - f(a) = h² + 5h + 2ah

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All the statements are true about the triangle except

1. a diagonal line extends from angle 8 to form angle.

According to the statement

There is a triangle has angles 6, 7, 8 and there are also some statements are given that matched with the triangle.

But the statement

1. a diagonal line extends from angle 8 to form angle.

is not true because it is not possible to extend a diagonal line from angle 8.

So, All the statements are true about the triangle except

1. a diagonal line extends from angle 8 to form angle.

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Step-by-step explanation:

[tex] \frac{2 {}^{4} }{3 \sqrt{xy} } [/tex]

[tex] \frac{2 { }^{4} }{ \sqrt{9xy} } [/tex]

[tex] \frac{1}{2 {}^{ - 4} \sqrt{9xy} } [/tex]

[tex] \frac{1}{ \frac{ \sqrt{9xy} }{16} } [/tex]

[tex] \frac{1}{ \frac{ \sqrt{9xy} }{ \sqrt{256} } } [/tex]

[tex] \frac{1}{ \sqrt{ \frac{9xy}{256} } } [/tex]

[tex]( \sqrt{ \frac{9xy}{256} } ) {}^{ - 1} [/tex]