Given the line segment CD = 30 cm and a point E on directed line segment CD that partitions CD into the ratio 4 to 1, find the lengths of line segments CE and ED.CE = 6 cm, ED = 24 cm

CE = 25 cm, ED = 5 cm

CE = 24 cm, ED = 6 cm

CE = 5 cm, ED = 25 cm

Answer:

15, 11, 60

Step-by-step explanation:

Let Joe's amount be x

x + x - 6 + 4x = 96

6x - 6 = 96

6x = 90

x = 15

Joe has 15

Kaitlin has 11

Miguel has 60

Using relations in a right triangle, it is found that the angle of depression is of θ = 56.31º.

The relations in a right triangle are given as follows:

For this problem, we have that:

Hence, considering θ as the depression angle, we have that:

tan(θ) = 9000/6000

tan(θ) = 1.5

θ = arctan(1.5)

θ = 56.31º.

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Answer:

Step-by-step explanation:

Set it up like an equation first:

[tex]\frac{\frac{-7}{25} }{x}=-\frac{1}{15}[/tex] and then bring up the lower fraction, flip it, and multiply:

[tex]-\frac{7}{25}*\frac{1}{x}=-\frac{1}{15}[/tex]

Simplify to

[tex]-\frac{7}{25x}=-\frac{1}{15}[/tex] and cross multiply to get

-25x = -105 and divide both sides by -25 to get that

[tex]x=\frac{105}{25}=\frac{21}{5}[/tex]

The coordinates of point r(-8,0).

The coordinates of the point r(x,y) which divides the line segment joining the points p([tex]x_{1}[/tex],[tex]y_{1}[/tex]) and q([tex]x_{2}[/tex],[tex]y_{2}[/tex]) internally in the ratio :[tex]m_{1}[/tex][tex]m_{2}[/tex] are

[tex]\left(\frac{m_{1}x_{2}+ m_{2}x_{1} }{m_{1}+m_{2} } ,\frac{m_{1}y_{2}+ m_{2}y_{1} }{m_{1}+m_{2} }\Rifgt)[/tex]

Given that,

Two end points on the line q(-14,0) and s(2,0). point r(x,y) is the partition of the line segment from q to s in a ratio 3:5

[tex]m_{1}[/tex] = 3 and [tex]m_{2}[/tex] = 5

By using the midpoint formula

[tex]\left(\frac{m_{1}x_{2}+ m_{2}x_{1} }{m_{1}+m_{2} } ,\frac{m_{1}y_{2}+ m_{2}y_{1} }{m_{1}+m_{2} }\Rifgt)[/tex]

[tex]\left(\frac{3(2)+ 5(-14) }{3+5 } ,\frac{3(0)+ 5(0) }{3+5 }\Rifgt)[/tex]

[tex]\left(\frac{-64}{8 } ,\frac{0 }{8 }\Rifgt)[/tex]

[tex]\left(-8 ,0\Rifgt)[/tex]

Hence, The coordinates of point r(-8,0).

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The total amount Aisha would pay to paint her room is $54.

The first step is to determine the area of the walls. The walls have the shape of a rectangle. Thus, the formula for the area of a rectangle would be used to determine the area of the wall.

Area of the three walls without a door : 3 x (length x width)

3 x (2.4 x 5.2) = 37.44 m²

Area of the fourth wall with a door : area of the wall - area of the door

(2.4 x 5.2) - (2 x 0.8) = 10.88m²

Total area of the walls : area of the three walls + area of the wall with the door

10.88m² + 37.44 m² = 48.32 m²

The next step is to determine the number of tins that would be needed:

48.32 /12 = 4.03

5 tins would needed

Cost of the 5 tins: (4 x 12) + (0.5 x 12) = $54

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A correlation is a causative association between two or more variables, whereas an autocorrelation is a correlation of variables in successive ranges.

A correlation is defined as a causative association between two or more variables (different variables), which can be used to make predictions about a given outcome.

The correlation coefficient enables the estimation of this association between different variables.

Moreover, an autocorrelation is a  special type of correlation between variables that can be found in successive ranges (e.g. range time intervals).

In conclusion, a correlation is a causative association between two or more variables, whereas an autocorrelation is a correlation of variables in successive ranges.

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The unit rate with the turtle crawling is 6/23 mile per hour.

According to the statement

we have given that the every 1/3 of an hour that the turtle is crawling, he can travel 2/23 of a mile. and

we have to find the unit rate with the turtle crawling

So, For this purpose,

we know that the

The unit rate of speed in miles per hour the number of miles in one hour.

So, to find the unit rate:

Divide distance in miles with the time which is in hours.

So,

unit rate of speed = distance / time

unit rate of speed = (2/23) / (1/3)

unit rate of speed = 6/23 mile per hour.

So, The unit rate with the turtle crawling is 6/23 mile per hour.

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The possible variable terms are a^5b^3, b^8, a^4b^4, a^8, ab^7

The expression is given as:

(2a + 3b)^8

The power of the above expression is 8

When expanded, the sum of powers on each variable must be 8.

Using the above explanation, we have the following possible variable terms:

a^5b^3, b^8, a^4b^4, a^8, ab^7

This is because the sum of powers equals 8

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The answer choice which represents the group which makes up the best sample for the researcher's survey as described in the task content is; Choice A; Parents of preschool children.

It follows from the task content that; Mrs. Greenaway has decided to open a new preschool in her town.

The need for a researcher therefore arises from the decision to either open the preschool or not.

Consequently, in a bid to choose a sample, Mrs. Greenaway's researcher must make sure the sample chosen is from the population space.

Therefore, answer choice A represents the correct sample.

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Answer:

D.

Step-by-step explanation:

Yes, because it passes the vertical line test.

Answer:

a = 3

Step-by-step explanation:

4(2a - 3) = -2(a - 9)

8a - 12 = -2a + 18

10a = 30

a = 3

Let's check if we got it correct by substituting 3 for a in the original equation.

4(2*3 - 3) = -2(3 - 9)

4(6 - 3) = -2(-6)

4(3) = 12

12 = 12

The values of trigonometric functions sin(x/2) is [tex]\sqrt{\frac{7+4\sqrt{3}}{14}}[/tex], cos(x/2) is [tex]\frac{1}{\sqrt{14(7+4\sqrt{3})}}[/tex] and tan(x/2) is 7+4√3.

Given that the value of trigonometric function csc(x)=7 for 90°<x<180°.

We are given:

csc(x)=7 Where: 90° < x < 180°

This interval indicates that the angle x in the second quadrant and we know that at that quadrant sin(x) and CSC(x) functions are positive and all other trigonometric functions sign are negative.

Now:

sin(x)=1/csc(x)

sin(x)=1/7

Using the trigonometric identity sin²x+cos²x=1, we get

cosx=±√(1-sin²x)

cosx=±√(1-(1/7)²)

cosx=±√((49-1)/49)

cosx=±(4√3)/7

As x is in second quadrant.

Therefore, cosx=-(4√3)/7

consider the inequality, 90°<x<180°

Divide by 2, we get

45°<x/2<90°

This is the angle x/2 in the first quadrant and in that quadrant all functions sign are positive.

Substitute the value of cosx in half angle formula for sine function,

[tex]\begin{aligned}\cos x&=1-2\sin^2\left(\frac{x}{2}\right)\\ -\frac{4\sqrt{3}}{7}&=1-2\sin^2\left(\frac{x}{2}\right)\\ 2\sin^2 \frac{x}{2}&=1+\frac{4\sqrt{3}}{7}\\\sin \frac{x}{2}&=\pm\sqrt{\frac{7+4\sqrt{3}}{14}}\end[/tex]

As x/2 lies in first quadrant then [tex]\sin \frac{x}{2}=\sqrt{\frac{7+4\sqrt{3}}{14}}[/tex]

Again, Substitute the value of cosx in half angle formula for sine function,[tex]\begin{aligned}\sin x&=2\sin\left(\frac{x}{2}\right)\cos\frac{x}{2}\\ \frac{1}{7}&=2\sqrt{\frac{7+4\sqrt{3}}{14}}\cos\left(\frac{x}{2}\right)\\ \cos \frac{x}{2}&=\frac{1}{14\times\frac{\sqrt{7+4\sqrt{3}}}{\sqrt{14}}}\\\cos \frac{x}{2}&=\frac{1}{\sqrt{14(7+4\sqrt{3})}}\end[/tex]

Now find tan(x/2) as shown below, we get

[tex]\begin{aligned}\tan\frac{x}{2}&=\frac{\sin\frac{x}{2}}{\cos \frac{x}{2}}\\&=\frac{\sqrt{7+4\sqrt{3}}}{\sqrt{14}}\times \frac{\sqrt{14}\sqrt{7+4\sqrt{3}}}{1}\\&=7+4\sqrt{3}\end[/tex]

Hence, when trigonometric function csc(x)=7 for 90°<x<180° then sin(x/2) is [tex]\sqrt{\frac{7+4\sqrt{3}}{14}}[/tex], cos(x/2) is [tex]\frac{1}{\sqrt{14(7+4\sqrt{3})}}[/tex] and tan(x/2) is 7+4√3.

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Answer:  All-day ride passes cost $15 each

Step-by-step explanation:

It was $20 in total for everyone to get in, so subtract that from $65 to just focus on the all day passes

65-20=45

So the all day passes all together cost $45

Since your uncle bought it for you and 2 friends that's 3 people in total so divide 45 by 3

45 divided by 3 = 15

So each of the all day passes cost $15

Hope this helped  :)

The maximum and minimum values of f(x,y,z) = 2xyz are [tex]\frac{2}{\sqrt{3} } and \frac{-2}{\sqrt{3} }[/tex]

The value of the function at a maximum point is called the maximum value of the function and the value of the function at a minimum point is called the minimum value of the function. Differentiate the given function.

f(x,y,z)=2xyz

Computation:

Differentiating the given equation up to second order

[tex]\begin{gathered}f_x=2yz\Rightarrow 2yz=0 either y=0 or z=0\\f_y=0\Rightarrow 2xz=0 either x=0 or z=0\\f_z=0\Rightarrow 2xy=0 either x=0 or y=0\end{gathered}[/tex]

So, the critical point is (0,0,0)

Now, using the Lagrange's on the boundary,

[tex]\begin{gathered}g(x,y,z)=x^2+y^2+z^2-1=0\\g_x=2x\\g_y=2y\\g_z=2z\end{gathered}[/tex]

[tex]So, \bigtriangledown f=\lambda \bigtriangledown g\left < 5yz,5xz,5xy \right > =\lambda \left < 2x,2y,2z )[/tex]

By solving we get,

[tex]x^2=y^2=z^2[/tex] then,

[tex]\begin{gathered}x^2+y^2+z^2=1\\3z^2=1\\z=\pm \frac{1}{\sqrt{3} } \\y=\pm \frac{1}{\sqrt{3} } \\\\x=\pm \frac{1}{\sqrt{3} } \\\end{gathered} x 2+y 2+z 2=13z 2=1z=± 31[/tex]

[tex]So, the critical points are (0,0,0),(\frac{1}{\sqrt{3} } ,\frac{1}{\sqrt{3} } ,\frac{1}{\sqrt{3} } )and (\frac{-1}{\sqrt{3} }, \frac{-1}{\sqrt{3} },\frac{-1}{\sqrt{3} })[/tex]

So, by substituting the critical points we get,

[tex]\begin{gathered}f(0,0,0)=0\\f(\frac{1}{\sqrt{3} }, \frac{1}{\sqrt{3} },\frac{1}{\sqrt{3} })=2(\frac{1}{\sqrt{3} })(\frac{1}{\sqrt{3} })(\frac{1}{\sqrt{3} })\\=\frac{2}{3\sqrt{3} } \\f(-\frac{1}{\sqrt{3} },-\frac{1}{\sqrt{3} },-\frac{1}{\sqrt{3} })=2(-\frac{1}{\sqrt{3} })(-\frac{1}{\sqrt{3} })(-\frac{1}{\sqrt{3} })\\=-\frac{2}{3\sqrt{3} }\end{gathered}[/tex]

Hence the maximum and minimum values of f(x,y,z) are [tex]\frac{2}{\sqrt{3} } and \frac{-2}{\sqrt{3} }[/tex]

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I'll assume the ODE is

[tex]y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}[/tex]

Solve the homogeneous ODE,

[tex]y'' - 3y' + 2y = 0[/tex]

The characteristic equation

[tex]r^2 - 3r + 2 = (r - 1) (r - 2) = 0[/tex]

has roots at [tex]r=1[/tex] and [tex]r=2[/tex]. Then the characteristic solution is

[tex]y = C_1 e^x + C_2 e^{2x}[/tex]

For nonhomogeneous ODE (1),

[tex]y'' - 3y' + 2y = e^x[/tex]

consider the ansatz particular solution

[tex]y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x[/tex]

Substituting this into (1) gives

[tex]a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1[/tex]

For the nonhomogeneous ODE (2),

[tex]y'' - 3y' + 2y = e^{2x}[/tex]

take the ansatz

[tex]y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}[/tex]

Substitute (2) into the ODE to get

[tex]b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1[/tex]

Lastly, for the nonhomogeneous ODE (3)

[tex]y'' - 3y' + 2y = e^{-x}[/tex]

take the ansatz

[tex]y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}[/tex]

and solve for [tex]c[/tex].

[tex]ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16[/tex]

Then the general solution to the ODE is

[tex]\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}[/tex]

The equation can be used to model the number of subscribers, y, in the city t years after 2000 is  y = 200,000(1+60)t.

An equation is a mathematical statement that is made up of two expressions connected by an equal sign. For example, 3x – 5 = 16 is an equation. Solving this equation, we get the value of the variable x as x = 7.

The first step is to calculate the percentage increase for each year.

If the subscribers increase by 60%, it means that the subscribers will be (100+60)% of the previous year.

(100+60)% = (100/100 + 60/100)

                 =  (1 + 0.6)

After 2,000, the subscribers will be 200,000×(1+0.6).

For the following years the 200,000×(1+0.6)t

So, the value of y will be given by,

Y=200,000(1+0.6)t

Hence,The equation Y=200,000(1+0.6)t .

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Step-by-step explanation:

[tex] \sf \: T_n = a+(n-1)d[/tex]

[tex] \sf \: T_n = \frac{1}{4} +(n-1) \frac{1}{4} [/tex]

[tex]\sf \: T_n = \frac{1}{4} + \frac{n}{4} - \frac{1}{4} [/tex]

[tex]\sf \: T_n = \frac{n}{4}[/tex]

an equation for the nth term of the arithmetic sequence. is n/4

Step-by-step explanation:

TN=n/4 is the nth term of arithmetic sequence.

Answer:

Step-by-step explanation:

x=100 bc like 180-80=100

y=80 bc vertical angles

Answer:

y = 80*

x = 100*

Step-by-step explanation:

y is the same to the 80* angle and therefore x is 100*

180* - 80 = 100*

Answer:

See picture.

C is the point that does not move because it is on the line of reflection
Yes, a reflection is a rigid motion.  It does not change size.  A dilation changes the size.

Step-by-step explanation:

In the picture, you will see that I traced the triangle on tracing paper and then folded (flipped) the drawing over the line of reflection.

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

Work Shown:

(x hours strength)/(4 hours endurance) = 5/3

x/4 = 5/3

3x = 4*5

3x = 20

x = 20/3

x = (18+2)/3

x = 18/3 + 2/3

x = 6 + 2/3

x = 6 & 2/3

He needs to spend 6 full hours, plus an additional 2/3 of an hour, on strength training.

1 hr = 60 min

2/3 hr = (2/3)*60 min

2/3 hr = 40 min

6 hr + 2/3 hr = 6 hr + 40 min

Therefore, he needs to spend 6 hours, 40 minutes on strength training.

This is equivalent to 6*60+40 = 360+40 = 400 minutes.

See below for the solution to the systems of equations

System of equations 1

Here, we have

y = -5x/3 + 3

y = x/3 -3

We start by plotting the graph of both equations

Next, we write out the point of intersection of the equations

See attachment for the equation of the system

From the attached graph, the point of intersection of the equations is (3, -2)

Hence, the solution to the system is (3, -2)

System of equations 2

Here, we have

y = 4x + 3

y = -x -2

We start by plotting the graph of both equations

Next, we write out the point of intersection of the equations

See attachment for the equation of the system

From the attached graph, the point of intersection of the equations is (-1, -1)

Hence, the solution to the system is (-1, -1)

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4 andfirst find x by finding its smallest factor

The solution to the given system of equation is -7 and 7

A linear equation is an algebraic equation of the form y=mx+b. involving only a constant and a first-order (linear) term, where m is the slope and b is the y-intercept. Occasionally, the above is called a "linear equation of two variables," where y and x are the variables.

Given:

4x² = 196

4x²/4= 196/4

x² = 49

x=√49

x= ±7

Hence, the solution to the given system of equation is -7 and 7

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The rounding off a given number implies approximating the number to the required value. Thus the answer to the given question is 6 000 mm.

Rounding off a given number implies approximating the number to the required value. This can be done in two major ways: rounding up or rounding down.

Rounding up implies considering the required digit if it is up to or greater than 5, which turns to 1. Rounding down requires no consideration of digits less than 5 which turns to 0.

Thus considering the given question, Since the second digit is 4 which is lesser than 5, then it turns to a zero. So that;

rounding off 6 400 mm would give 6 000 mm.

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Since Point R divides PQ in the ratio 1:3. the x-coordinate of Q is 3

The question has to do with line division

This is the division of a line into a given ratio.

Since Point R divides PQ in the ratio 1:3. If the x-coordinate of R is -1 and the x-coordinate of P is -3.

Let

Since R divides PQ in the ratio, 1:3, we have that

PR/RQ = 1/3

(x₂ - x₁)/(x₃ - x₂) = 1/3

Substituting the values of the variables into the equation, we have

(x₂ - x₁)/(x₃ - x₂) = 1/3

(-1 - (-3))/(x₃ - (-3)) = 1/3

(-1 + 3)/(x₃ + 3) = 1/3

2/(x₃ + 3) = 1/3

x₃ + 3 = 2 × 3

x₃ + 3 = 6

x₃ = 6 - 3

x₃ = 3

So, the x-coordinate of Q is 3

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The difference of 32 · x² / (x² + 8 · x + 15) - 14 · x² / (x² - 9) is equal to [18 · x³ - 166 · x²] / [(x + 3) · (x + 5) · (x - 3)]. (Correct choice: B)

In this question we have a subtraction between two rational functions, which have to be simplified by algebra properties. The complete procedure is presented below:

32 · x² / (x² + 8 · x + 15) - 14 · x² / (x² - 9)        Given

32 · x² / [(x + 3) · (x + 5)] - 14 · x² / [(x + 3) · (x - 3)]         Factorization

[x² / (x + 3)] · [32 / (x + 5) - 14 / (x - 3)]      Distributive and associative properties

[x² / (x + 3)] · [32 · (x - 3) - 14 · (x + 5)] / [(x + 5) · (x - 3)]      Subtraction of rational numbers with distinct denominators

[x² / (x + 3)] · [32 · x - 96 - 14 · x - 70] / [(x + 5) · (x - 3)]     Distributive property / (- 1) · a = - a

[x² / (x + 3)] · (18 · x - 166) / [(x + 5) · (x - 3)]      Distributive property / Definitions of addition and subtraction

[18 · x³ - 166 · x²] / [(x + 3) · (x + 5) · (x - 3)]         Mutiplication between rational numbers / Multiplication between powers / Distributive property

The difference of 32 · x² / (x² + 8 · x + 15) - 14 · x² / (x² - 9) is equal to [18 · x³ - 166 · x²] / [(x + 3) · (x + 5) · (x - 3)]. (Correct choice: B)

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Answer:

  8

Step-by-step explanation:

The given ratios can be combined to a composite ratio using least common multiples.

We can identify each of the entities with a 2-letter designation:

  TT - Tumtum tree; TW - tulgey wood; BS - frumious Bandersnatch;

  ST - slithy tove; BG - borogoves; MR - mome raths; JW - Jabberwocky;

  JB - Jubjub birds.

Then the given ratios are ...

The path from BS to JW is unique, so we can work our way there one ratio at a time. Having done this once, we can go back and do it again with the least possible multiplier.

Here, that turns out to be 125 Bandersnatches. We can combine these to the composite ratio ...

  125 BS : 125 TW : 2500 TT : 25 JB : 5 ST : 2 BG : 400 MR : 200 JW

Then the ratio of Bandersnatches to Jabberwocks is 125 : 200 = 5 : 8.

If there are 5 frumious Bandersnatches, there should be 8 Jabberwocks.

__

Additional comment

We know 1 JW : 2 MR and 200 MR : 1 BG. In order to get a common value of 200 MR in these ratios, we need to multiply the first by 100:

  100 JW : 200 MR : 1 BG

Next, we notice the other ratio involving BG is 2 BG : 5 ST, which means we need to multiply by 2 again:

  200 JW : 400 MR : 2 BG : 5 ST

The number of JB is 5 times the number of ST, so this extends to ...

  ... 5 ST : 25 JB

The number of TT is 100 times the number of JB, so we have ...

  ... 5 ST : 25 JB : 2500 TT

Continuing in like fashion, we get the composite ratio shown above. We originally worked this starting from BS. Reducing the JB/TT ratio from 2/200 to 1/100 helped eliminate an extra factor of 2.

Step-by-step explanation:

it's not really a math problem per se, it is more like a regular riddle with numbers.

true, everything involving numbers is essentially math, but then what will you do in life with things like earning a salary, going shopping, checking a calendar and schedules for bus, train and flights or business meetings, sharing a pizza with friends, playing computer games, ...

life is for a big part numbers and therefore math.

so, don't be afraid, just use your brain and common sense !

let's think this through (just be strict, concentrated, and follow each line literally) :

5 fr. Band. means 5 tulgey woods.

which means 20×5 = 100 tumtum trees.

as there are 2 jubjub birds in 200 tumtum trees, that means that there is 1 jubjub bird in the given 100 tumtum trees.

as there are 5 jubjub birds in 1 slithy toves, but we have only 1 bird, that means we have only 1/5 slithy toves.

there are 5 slithy toves in 2 borogoves. that means 1 slithy toves in 2/5 borogoves. and 1/5 slithy toves in 2/5 / 5 = 2/25 borogoves.

there are 200 mome raths in each borogove. and we have 2/25 borogoves. that means we have

200 × 2/25 = 400/25 = 16 mome raths.

and there are 2 mome raths per jabberwocky. we have 16 raths. how many pairs of raths does that make ?

16/2 = 8 pairs of raths = jabberwocks.

so, there should be 8 jabberwocks.

you see ? that's all that was needed. this was like a labyrinth riddle, where you need to find the way from an inside starting point to the exit without crossing a line (or wall).