Andrew drives through 5 intersections with stoplights on his way to work. He records that he is stopped by a red light in 2 of the 5 intersections. If he uses this as his general probability of getting stopped at a red light on his way to or from work, how many times can he expect to be stopped at a red light over the next 40 times he drives through an intersection on his route? Andrew should expect to be stopped times.

Answers

Answer 1

Answer:

He should expect to be stopped a total of 16 times

Step-by-step explanation:

What the question is saying is that out of every 5 intersections, he would be stopped at 2.

Now, for forty times in which he passes through an intersection , we want to know the number of times in which he would be stopped

In this case we only need to multiply the probability by the number of times in which he passes an intersection and that would be 2/5 * 40 = 16 times

Answer 2

Answer:

16

Step-by-step explanation:


Related Questions

30 POINTS!!!

Suppose f(x) = x2 and
g(x) = (1/3)^2. Which statement best compares the graph of g(x) with the graph of f(x)?

Image attached

Please help!!!

Answers

Answer:

A. The graph of g(x) is vertically compressed by a factor of 3.

Step-by-step explanation:

When there is a fraction, that means that there is a veritcal dilation.

Hope this helps! Good luck!

Find the area ratio of a regular octahedron and a tetrahedron regular, knowing that the diagonal of the octahedron is equal to height of the tetrahedron.

Answers

Answer:

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

Step-by-step explanation:

The area of a regular octahedron is given by:

area = [tex]2\sqrt{3}\ *edge^2[/tex]. Let a is the length of the edge (diagonal).

area = [tex]2\sqrt{3}\ *a^2[/tex]

Given that the diagonal of the octahedron is equal to height (h) of the tetrahedron i.e.

a = h, where h is the height of the tetrahedron and a is the diagonal of the octahedron. Let the edge of the tetrrahedron be e. To find the edge of the tetrahedron, we use:

[tex]h=\sqrt{\frac{2}{3} } e\\but\ h=a\\a=\sqrt{\frac{2}{3} } e\\e=\sqrt{\frac{3}{2} }a[/tex]

The area of a tetrahedron is given by:

area = [tex]\sqrt{3}\ *edge^2[/tex] = [tex]\sqrt{3} *(\sqrt{\frac{3}{2} }a)^2=\frac{3}{2}\sqrt{3} *a^2[/tex]

The ratio of area of regular octahedron to area tetrahedron regular is given as:

Ratio = [tex]\frac{2\sqrt{3}\ *a^2}{\frac{3}{2} \sqrt{3}*a^2} =\frac{4}{3}[/tex]

Please help i will mark brainliest

Answers

I think it’s 2/4x-6

Answer:

See below.

Step-by-step explanation:

To find the equation, we need to find the slope and the y-intercept. Afterwards, we can put the numbers into the slope-intercept form:

[tex]y=mx+b[/tex]

From the graph, we can see that the line crosses the y-intercept at y=-6. Thus, the y-intercept (b) is -6.

Now we need to find the slope. Pick any two points where the line crosses. I'm going to pick (0,-6) and (4,-7).

[tex]m=\frac{y_2-y_1}{x_2-x_1}=\frac{-7-(-6)}{4-0}= -1/4[/tex]

Therefore, the equation of the line would be:

[tex]y=mx+b\\y=-1/4x-6[/tex]

first correct answer gets best marks ​

Answers

Answer:

the answer would be x is less than 6.

Step-by-step explanation:

the reason why it would not be x is less than or equal to 6 is that the circle is not filled in.

Answer:

B

Step-by-step explanation:

x≤6

We can see from the graph that it starts from 6 and goes to 5, 4, 3, 2.

Hope this helps ;) ❤❤❤

Fill in the blank with a constant, so that the resulting expression can be factored as the product of two linear expressions: 2ab-6a+5b+___ Please include an explanation too!

Answers

Answer:

[tex]2ab - 6a + 5b - 15[/tex]

Step-by-step explanation:

Given

[tex]2ab - 6a + 5b + \_[/tex]

Required

Fill in the gap to produce the product of linear expressions

[tex]2ab - 6a + 5b + \_[/tex]

Split to 2

[tex](2ab - 6a) + (5b + \_)[/tex]

Factorize the first bracket

[tex]2a(b - 3) + (5b + \_)[/tex]

Represent the _ with X

[tex]2a(b - 3) + (5b + X)[/tex]

Factorize the second bracket

[tex]2a(b - 3) + 5(b + \frac{X}{5})[/tex]

To result in a linear expression, then the following condition must be satisfied;

[tex]b - 3 = b + \frac{X}{5}[/tex]

Subtract b from both sides

[tex]b - b- 3 = b - b+ \frac{X}{5}[/tex]

[tex]- 3 = \frac{X}{5}[/tex]

Multiply both sides by 5

[tex]- 3 * 5 = \frac{X}{5} * 5[/tex]

[tex]X = -15[/tex]

Substitute -15 for X in [tex]2a(b - 3) + 5(b + \frac{X}{5})[/tex]

[tex]2a(b - 3) + 5(b + \frac{-15}{5})[/tex]

[tex]2a(b - 3) + 5(b - \frac{15}{5})[/tex]

[tex]2a(b - 3) + 5(b - 3)[/tex]

[tex](2a + 5)(b - 3)[/tex]

The two linear expressions are [tex](2a+ 5)[/tex] and [tex](b - 3)[/tex]

Their product will result in [tex]2ab - 6a + 5b - 15[/tex]

Hence, the constant is -15

Can someone help me with these two questions I don’t know how to do it and it’s due at 11 I would really appreciate it

Answers

Answer:

6. Unit rate = 1.3 yards per second

7. Unit rate = 0.8 page per minute

Step-by-step explanation:

The unit rate is simply the comparison of 2 quantities, whereby dividing both, the denominator must be 1.

For example, in the graph given comparing distance walked over time, when x (time in s) = 3, y (distance in yd) = 4.

Unit rate represented by the slope is the yards covered per second.

Unit rate = [tex] \frac{4}{3} = 1.33 [/tex]

Unit rate ≈ 1.3 yards per second

For the second graph given, unit rate of the slope is the number of pages read per minute.

From the graph, 4 pages is read at 5 minutes.

Thus,

Unit rate = [tex] \frac{4}{5} = 0.8 [/tex]

Unit rate = 0.8 page per minute

writing linear equations

Answers

Answer:

The graph in blue: y=4/3x-2

Step-by-step explanation:

Slope-intercept form is y=mx+b, m being the slope and b being the y-intercept.

To find the slope, you just put rise/run of the line.

Hope this helped :)

Solve by the quadratic formula: x^2= 6x-4

Answers

Answer:

3 [tex]\pm[/tex] [tex]\sqrt{5}[/tex].

Step-by-step explanation:

x^2 = 6x - 4

x^2 - 6x + 4 = 0

Now, we can use the quadratic formula to solve.

[tex]\frac{-b\pm\sqrt{b^2 - 4ac} }{2a}[/tex], where a = 1, b = -6, and c = 4.

[tex]\frac{-(-6)\pm\sqrt{(-6)^2 - 4 * 1 * 4} }{2 * 1}[/tex]

= [tex]\frac{6\pm\sqrt{36 - 4 * 4} }{2}[/tex]

= [tex]\frac{6\pm\sqrt{36 - 16} }{2}[/tex]

= [tex]\frac{6\pm\sqrt{20} }{2}[/tex]

= [tex]\frac{6\pm2\sqrt{5} }{2}[/tex]

= 3 [tex]\pm[/tex] [tex]\sqrt{5}[/tex]

x = 3 [tex]\pm[/tex] [tex]\sqrt{5}[/tex].

Hope this helps!

Pls help I need help with 12

Answers

Answer:

B. 14

Step-by-step explanation:

22/x = 11/(21-x)

462 - 22x = 11x

462 = 33x

x = 14

Answer: The value of x is 14, answer choice B

Let y be the other line segment connected to x

Using proportions:

[tex]\dfrac{11}{22}=\dfrac{y}{x}[/tex]

Cross multiply and simplify

[tex]22y=11x[/tex]

[tex]y=\dfrac{1}{2}x[/tex]

We know that x and y add to 21, so we can create the following equation:

[tex]x+y=21[/tex]

Substitute y=(1/2)x

[tex]x+\dfrac{1}{2}x=21[/tex]

Simplify by adding like terms

[tex]\dfrac{3}{2}x=21[/tex]

Divide both sides by 3/2

[tex]x=14[/tex]

Let me know if you need any clarifications, thanks!

A cube has a side length of 5 cm. Determine the surface area of the largest pyramid that will fit inside the cube. Round if necessary.

Answers

Answer:

The surface area of the pyramid is 80.9 cm²

Step-by-step explanation:

The side length, s of the cube is given as 5 cm

Therefore, the largest pyramid that can fit into the cube will have a base side length, s = The side length of the cube = 5 cm

The height, h of the largest pyramid = The height of the cube = 5 cm.

The surface area of a pyramid =  Area of base, A + 1/2 × Perimeter of base, P × Slant height, S

The slant height of the pyramid = √(h² + (s/2)²) = √(5² + (5/2)²) = (5/2)×√5

The perimeter of the base = 4×5 = 20 cm

The area of the base = 5×5 = 25 cm²

The surface area of a pyramid = 25 + 1/2×20×(5/2)×√5 = 80.9 cm².

The surface area of a pyramid =  80.9 cm².

Calcule o valor dos produtos a) (-4). (-7/8) b) (-4). (-7/8) n sei pq tá repetido tá aqui na folha ;-; c) (-4).(+ 3,5) d) (-2).(-3/4). (-1/7) PRA HOJEE

Answers

Answer:

a) (-4). (-7/8) = 28/8

b) (-4). (-7/8)= 28/8

c) (-4).(+ 3/5) = -12/5

d) (-2).(-3/4). (-1/7) = (6/4)(-1/7)= -6/28

Step-by-step explanation:

a) (-4). (-7/8) = 28/8

b) (-4). (-7/8)= 28/8

c) (-4).(+ 3/5) = -12/5

d) (-2).(-3/4). (-1/7) = (6/4)(-1/7)= -6/28

É repetido talvez com sinal alterado para mostrar a diferença na resposta. Se o sinal for alterado, a resposta também se tornará negativa. Seria - 28/8.

Quando um número negativo é multiplicado por um número negativo, obtém um número positivo. Mas quando um número positivo é multiplicado por um número negativo, dá uma resposta negativa.

Sempre se lembre

negativo * negativo = positivo

positivo * negativo = negativo

negativo * positivo = negativo

positivo * positivo = positivo

Em palavras simples, dois sinais diferentes dão um sinal negativo e dois sinais semelhantes dão um sinal positivo na multiplicação.

English

a) (-4). (-7/8) = 28/8

b) (-4). (-7/8)= 28/8

c) (-4).(+ 3/5) = -12/5

d) (-2).(-3/4). (-1/7) = (6/4)(-1/7)= -6/28

Its repeated maybe with a changed sign to show the difference in the answer. If the sign is changed the answer would also become negative . It would become - 28/8.

When a negative number is multiplied with a negative number it gives a positive number. But when a positive number is multiplied with a negative number it gives a negative answer.

Always remember  

negative * negative= positive  

positive *negative=negative

negative *positive =negative

positive * positive =  positive

In simple words two unlike signs give a negative sign and two similar signs give a positive sign in multiplication.

ASAP!! Please help me. I will not accept nonsense answers, but will mark as BRAINLIEST if you answer is correctly with solutions.

Answers

Answer:

The function has two real roots and crosses the x-axis in two places.

The solutions of the given function are

x = (-0.4495, 4.4495)

Step-by-step explanation:

The given quadratic equation is

[tex]G(x) = -x^2 + 4x + 2[/tex]

A quadratic equation has always 2 solutions (roots) but the nature of solutions might be different depending upon the equation.

Recall that the general form of a quadratic equation is given by

[tex]a^2 + bx + c[/tex]

Comparing the general form with the given quadratic equation, we get

[tex]a = -1 \\\\b = 4\\\\c = 2[/tex]

The nature of the solutions can be found using

If [tex]b^2- 4ac = 0[/tex] then we get two real and equal solutions

If [tex]b^2- 4ac > 0[/tex] then we get two real and different solutions

If [tex]b^2- 4ac < 0[/tex] then we get two imaginary solutions

For the given case,

[tex]b^2- 4ac \\\\(4)^2- 4(-1)(2) \\\\16 - (-8) \\\\16 + 8 \\\\24 \\\\[/tex]

Since 24 > 0

we got two real and different solutions which means that the function crosses the x-axis at two different places.

Therefore, the correct option is the last one.

The function has two real roots and crosses the x-axis in two places.

The solutions (roots) of the equation may be found by using the quadratic formula

[tex]$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$[/tex]

[tex]x=\frac{-(4)\pm\sqrt{(4)^2-4(-1)(2)}}{2(-1)} \\\\x=\frac{-4\pm\sqrt{(16 - (-8)}}{-2} \\\\x=\frac{-4\pm\sqrt{(24}}{-2} \\\\x=\frac{-4\pm 4.899}{-2} \\\\x=\frac{-4 + 4.899}{-2} \: and \: x=\frac{-4 - 4.899}{-2}\\\\x= -0.4495 \: and \: x = 4.4495 \\\\[/tex]

Therefore, the solutions of the given function are

x = (-0.4495, 4.4495)

A graph of the given function is also attached where you can see that the function crosses the x-axis at these two points.

When deriving the quadratic formula by completing the square, what expression can be added to both sides of the equation to create a perfect square trinomial?

Answers

Answer:

According to steps 2 and 4. The second-order polynomial must be added by [tex]-c[/tex] and [tex]b^{2}[/tex] to create a perfect square trinomial.

Step-by-step explanation:

Let consider a second-order polynomial of the form [tex]a\cdot x^{2} + b\cdot x + c = 0[/tex], [tex]\forall \,x \in\mathbb{R}[/tex]. The procedure is presented below:

1) [tex]a\cdot x^{2} + b\cdot x + c = 0[/tex] (Given)

2) [tex]a\cdot x^{2} + b \cdot x = -c[/tex] (Compatibility with addition/Existence of additive inverse/Modulative property)

3) [tex]4\cdot a^{2}\cdot x^{2} + 4\cdot a \cdot b \cdot x = -4\cdot a \cdot c[/tex] (Compatibility with multiplication)

4) [tex]4\cdot a^{2}\cdot x^{2} + 4\cdot a \cdot b \cdot x + b^{2} = b^{2}-4\cdot a \cdot c[/tex] (Compatibility with addition/Existence of additive inverse/Modulative property)

5) [tex](2\cdot a \cdot x + b)^{2} = b^{2}-4\cdot a \cdot c[/tex] (Perfect square trinomial)

According to steps 2 and 4. The second-order polynomial must be added by [tex]-c[/tex] and [tex]b^{2}[/tex] to create a perfect square trinomial.

Answer: D

Step-by-step explanation:

EDGE 2023

What fraction is equal to six-sevenths times eight-fifths?

Answers

Answer:

1 13/35 (mixed number) or 48/35 (simplified)

Step-by-step explanation:

6/7 times 8/5

= (6 times 8) / (7 times 5)

= 48/35 or 1 13/35

hope this helped :)

Answer:

48/35

Step-by-step explanation:

6/7*8/5=48/35

48 - 8x equivalent expression

Answers

Answer:

8(6-x)

Step-by-step explanation:

Both 48 and 8 can be divisible by 8.

48 ÷ 8 = 6

8 ÷ 8 = 1

Therefore you get the answer 8(6-x)

as the simplest form.

Hope this helps.

Points A(-l, y) and B(5,7) lie on a circle with centre 0(2, -3y). Find the values of y. Hence, find the radius of the circle

Answers

Answer:

The answer is below

Step-by-step explanation:

Points A(-l, y) and B(5,7) lie on a circle with centre O(2, -3y). This means that AB is the diameter of the circle and OA = OB = radius.

For two points X([tex]x_1,y_1[/tex]) and Y([tex]x_2, y_2[/tex]), the coordinates of the midpoint (x, y) between the two points is given as:

[tex]x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}[/tex].

For A(-l, y) and B(5,7) with center O(2, -3y), the value of y can be gotten by:

[tex]For\ x\ coordinate:\\2=\frac{-1+5}{2}\\ 2=2.\\For\ y\ coordinate:\\-3y=\frac{y+7}{2}\\ -6y=y+7.\\-6y-y=7\\-7y=y\\y=-1[/tex]

The value of y is -1. Therefore A is at (-1, -1) and O is at (2, -3(-1))= (2, 3)

The radius of the circle = OA. The distance between two points X([tex]x_1,y_1[/tex]) and Y([tex]x_2, y_2[/tex]) is given as:

[tex]|OX|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\\\Therefore\ the\ radius \ |OA|\ is :\\|OA|=\sqrt{(2-(-1))^2+(3-(-1))^2}=\sqrt{25}=5[/tex]

The radius of the circle is 5 units

the diagram above shows a rectangle inscribed in a circle AB=10 and AC =12 caculate the total surface area of the shaded part

Answers

Answer:

[tex]71.63 \: \: \mathrm{cm^2 }[/tex]

Step-by-step explanation:

Once we know the diameter of the circle, we can figure out the problem.

The diameter of the circle = The diagonal of the rectangle inscribed in the circle

To find the diagonal of the rectangle, we can use a formula.

[tex]d=\sqrt{w^2 + l^2}[/tex]

The width is 10 cm and the length is 12 cm.

[tex]d=\sqrt{10^2 + 12^2}[/tex]

[tex]d \approx 15.62[/tex]

The diagonal of the rectangle inscribed in the circle is 15.62 cm.

The diameter of the circle is 15.62 cm.

Find the area of the whole circle.

[tex]A=\pi r^2[/tex]

The [tex]r[/tex] is the radius of the circle, to find radius from diameter we can divide the value by 2.

[tex]r = \frac{d}{2}[/tex]

[tex]r=\frac{15.62}{2}[/tex]

[tex]r=7.81[/tex]

Let’s find the area now.

[tex]A=\pi (7.81)^2[/tex]

[tex]A \approx 191.625[/tex]

Find the area of rectangle.

[tex]A=lw[/tex]

Length × Width.

[tex]A = 12 \times 10[/tex]

[tex]A=120[/tex]

Subtract the area of the whole circle with the area of rectangle to find area of shaded part.

[tex]191.625-120[/tex]

[tex]71.625 \approx 71.63[/tex]

The graph of g(x) resembles the graph of f(x)=x^2, but it has been changed. Which of these is the equation of g(x)?

Answers

Answer:

A.

Step-by-step explanation:

Anwer A has the following equation:

[tex]g(x)=\frac{3}{5}x^2-3[/tex]

In this equation, we can calculated the intercept replacing x by 0, as:

[tex]g(x)=\frac{3}{5}0^2-3=-3[/tex]

if this is the answer, the graph of g(x) should be through the point (0,-3) and that happens.

Additionally, the roots of the equations are calculated replacing g(x) by 0 and solving for x, so:

[tex]0=\frac{3}{5}x^2-3\\x_1=\sqrt{5}=2.236\\x_2=-\sqrt{5}=-2.236[/tex]

It means that the graph of g(x) should be through the points (2.236,0) and (-2.236,0) and that happens too.

So, the answer is A, [tex]g(x)=\frac{3}{5}x^2-3[/tex]

Helppppppppppppppppp

Answers

Answer/Step-by-step explanation:

Surface area of cylinder = 2πrh + 2πr²

Surface area of the cylinder with,

height (h) = 4 ft

radius (r) = 5 ft

Surface area = 2*3.14*5*4 + 2*3.14*5²

= 125.6 + 157

= 282.6 ft²

Surface area of the cylinder with,

height (h) = 12 yd

radius (r) = 3 yd

Surface area = 2*3.14*3*12 + 2*3.14*3²

= 282.74 yd²

Identifying relationships from diagrams

Answers

Answer: <CED is the right angle, which measures 90 degrees. Since the measure of a straight angle is 180 degrees. <CEA must also be 90 degrees by the Definition of Right Angle. A bisector cuts the angle measure in half. m<AEB is 45 degrees.

What is the following product?

Answers

Answer:

[tex]\boxed{6\sqrt{6} }[/tex]

Step-by-step explanation:

[tex]\sqrt{12} \sqrt{18}[/tex]

Multiply square roots.

[tex]\sqrt{12 \times 18}[/tex]

[tex]\sqrt{216}[/tex]

Simplify square root.

[tex]\sqrt{36} \sqrt{6}[/tex]

[tex]6\sqrt{6}[/tex]

Find the area of the shape shown below

Answers

Answer: 28

Step-by-step explanation:

I can't really think of a way to explain this well without visuals and idk how to add images on my answer. But, what I normally do is draw out the shape on paper divide the shape into different sections. Solve the area of the separate sections. It simplifies the more complex figure and turns them into basic shapes. After solving each shape, add all of them together and that leaves you with the area. Hopefully you understand what I mean. I hope this sort of helped:)

x=y-y and 2x+4y=10 solve using substitution​

Answers

Answer:

(0, 2.5)

Step-by-step explanation:

Well we substitute y-y into x in the following equation,

2x + 4y = 10

2(y-y) + 4y = 10

2y - 2y + 4y = 10

Combine like terms

2y - 2y = 0

4y = 10

10/4

y = 2.5

If y is 2.5 we can plug those into y.

2x + 4(2.5) =10

2x + 10 = 10

-10

2x = 0

0/2

x = 0

I NEED HELP WITH THIS! I need to pass...

Answers

Answer:  A) The log parent function has negative values in the range.

Step-by-step explanation:

The domain of y = ln (x)  is D: x > 0

The domain of y = [tex]\sqrtx[/tex][tex]\sqrt x[/tex]  is  D: x ≥ 0

The range of y = ln (x)  is: R: -∞ < y < ∞

So the only valid option is A because the range of a log function contains negative y-values when 0 < x < 1.

Solve the following 2 + 8 ÷ 2 x 3

Answers

Answer:

14

Step-by-step explanation:

Solution,

Use the BODMAS Rule:

B = Bracket

O = Of

D = Division

M= Multiplication

A = Addition

S = Subtraction

Now,

Let's solve,

[tex]2 + 8 \div 2 \times 3[/tex]

First we have to divide 8 by 2

[tex] = 2 + 4 \times 3[/tex]

Calculate the product

[tex] = 2 + 12[/tex]

Calculate the sum

[tex] = 14[/tex]

Hope this helps...

Good luck on your assignment..

Answer:

14

Step-by-step explanation:

2 + 8 ÷ 2 x 3 =

There is an addition, a division, and a multiplication. According to the correct order of operations, we do first the multiplications and divisions in the order they appear from left to right.

= 2 + 4 x 3

= 2 + 12

Now we do the addition.

= 14

asap help !!

How much additional interest is earned if $8000 is invested for 7 years at 6.5% when
interest is compounded annually, as compared with simple interest paid at the same
rate?

Answers

Answer:

put it in a calculator, 8,000 times whatever number u need

The following box plot shows the number of years during which 40 schools have participated in an interschool swimming meet: A box and whisker plot is drawn using a number line from 0 to 10 with primary markings and labels at 0, 5, 10. In between two primary markings are 4 secondary markings. The box extends from 1 to 6 on the number line. There is a vertical line at 3.5. The whiskers end at 0 and 8. Above the plot is written Duration of Participation. Below the plot is written Years. At least how many schools have participated for more than 1 year and less than 6 years?

Answers

Answer:

Step-by-step explanation:

The box encloses data between the two quartiles, namely at least half of the data.  If there are 40 schools, then half of them would be in the box, between 1 and 6.

see attached plot.

Answer:

really hard to tell what the box plot is like without an attachment so im gonna help u find it out anyway

Step-by-step explanation:

basically when u look at a  box plot and the range the line in the middle is the median  and then the max the lowest range the lower quartile and then the higher quartile you can find ur anser, simply find the median first, find where the lower quartile is and then the lowest number in the group thats in betweeen 1 and 6

simplify: m–(a–k–b) HElp!

Answers

Answer: m-a+k+b

Step-by-step explanation:

m-(a-k-b)= m+-1*(a-k-b). Then, simply use distributive property to simplify the equation into m-a+k+b

Hope it helps <3

Simplify (2^3)^–2. PLEASE I NEED HELP U WILL GET 10 POINTS

Answers

Answer:

I won't give you the answer straight away so you take the time to read my answer and understand

Step-by-step explanation:

We knoe that 2 to the third is 8. when you square to a negative power, you do squaring normally, and then take the reciprocal of that number. so 8 to the second power is 64, and we flip it over, sp the answer is 1/64

Please help me to solve this . Thank you so much .
And if possible , could you explain the answer too ?

Base on the diagram , state

a) The point which is 2 cm from R and 4 cm from P

b) The point which is more than 2 cm from R and 4 cm from T

c) The location of a moving point X in the diagram such that it is less than 4 cm from P and more than 2cm from R

d) The location of a moving point Y in the diagram such that YR < 2 cm and YP < 4 cm

e)The location of a moving point Z in the diagram such that ZT > 4 cm , ZP > 4 cm and ZR > 2 cm

Answers

Answer:

a) N

b) L

c) area I

d) area II

e) area VI

Step-by-step explanation:

a) the points that are 2cm from R are Q, N, M, S. Then, points that are 4cm from P are K, N, R. So, the only one point that works for both is N.

b) the points that are >2cm from R are P, K, L, T. We do not count those are exactly 2cm from R. Then, points that are 4cm from T are R, M, L. Ans is L.

c) <4cm from P, are area I and II. Then area that are >2cm from R are I, VI, and V. So, the only area that works for both is I.

d) <2cm from R, are areas II, III, and IV. Then, <4cm from P, are areas I and II. So, the only one works for both is area II.

e) >4cm from T, are areas I, II, III, VI. Then, >4cm from P, are III, IV, V, VI. Finally, >2cm from R, are areas I, VI, V. The only one that works for all three conditions is area VI.

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