Find the perimeter or circumference and area of each figure. Round to the nearest hundredth.

Answer:

Step-by-step explanation:

From the question all the angles are on a a straight line which means the sum of their angles is 180 and one of the angles is 90°

So to find m \( \angle\) 6 , add

m \( \angle\) 5 and 90 and subtract the result from 180°

That's

m\( \angle\)5 + m\( \angle\)6 + 90 = 180

m\( \angle\)6 = 180 - 90 - m\( \angle\)5

m\( \angle\)6 = 180 - 90 - 22

We have the final answer as

Hope this helps you

In the question, a lines & angles diagram is provided, and we can see three angles labelled here. The angles are angle 5, angle 6 and a 90°.

GiveN,

We have to find the measure of the missing angle 6...

All the three angles are lying on a staright line, so their collective measure or the sum of these 3 angles is equals to 180°.

➝ Angle 5 + Angle 6 + 90° = 180°

➝ 22° + Angle 6 + 90° = 180°

➝ Angle 6 + 112° = 180°

➝ Angle 6 = 68°

So, the measure of Angle 6:

\( \huge{ \boxed{ \bf{68 \degree}}}\)

And we are done !!

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If Jesse can mow 3 yards in 8 hours. Jackson can mow twice as many yards per hour the constant of proportionality between the number of yards Jackson can mow and the number of hours is 3/4.

To find the constant of proportionality between the number of yards Jackson can mow and the number of hours, we can use the formula:

k = y/x

where k is the constant of proportionality, y is the number of yards, and x is the number of hours.

We know that Jesse can mow 3 yards in 8 hours, which means his rate of mowing is: 3 yards/8 hours = 3/8 yards per hour

We also know that Jackson can mow twice as many yards per hour as Jesse, which means his rate of mowing is:

2 * (3/8) yards per hour = 3/4 yards per hour

Now we can use the formula to find the constant of proportionality for Jackson:

k = y/x = (3/4) yards per hour / 1 hour = 3/4

Therefore, the constant of proportionality between the number of yards Jackson can mow and the number of hours is 3/4.

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The values of composite functions are,

(w о u)(2) = √7 + 2

(u о w)(2) = 9 + 4√2

Function composition is an operation о used in mathematics that takes two functions, f and g, and creates a function, h = g о f, such that h(x) = g. The function g is applied in this operation to the outcome of applying the function f to x.

Given:

u(x) = x^2 + 3

w(x) = √x+2

We have to find (w о u)(2) and (u о w)(2).

First to find (w о u)(x)

(w о u) = w(u(x))

         = w(x^2 + 3)

(w о u) = √(x^2 + 3) + 2

(w о u)(2) = √(2^2 + 3) + 2

 (w о u)(2) = √7 + 2

Now to find (u о w).

(u о w)(x) = u(w(x))

= u(√x+2)

= (√x+2)^2 + 3

= x + 4√x + 4 + 3

(u о w)(x) = x +  4√x + 7

(u о w)(2)  = 2 + 4√2 + 7

 (u о w)(2)  = 9 + 4√2

Hence,

The values of composite functions are,

(w о u)(2) = √7 + 2

(u о w)(2)  = 9 + 4√2

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Answer:

5 ft

Step-by-step explanation:

Area of a rectangle

length x width

(3x-2) * (x) = 65

3x^2 - 2x - 65 = 0

Δ/4 = 1 + 195 = 196

x1 = (1 + 14)/3 = 5

x2 (1-14)/3 = -13/3 (not acceptable because a length can’t be negative)

The present value of $102,000 in 10 years, given a long-term interest rate of 9 percent, is $43,229. This means that if Delia were to receive $102,000 in 10 years, it would be equivalent to receiving $43,229 today.

Therefore, Delia should choose to take the $38,000 today instead of waiting 10 years to receive $102,000. The present value of $102,000 is less than the amount offered today, so it is a better choice financially.
Explanation:
To calculate the present value, we use the formula: Present Value = Future Value / (1 + Interest Rate)^n, where n is the number of periods.

Using Appendix B, we can find the present value factor for 10 years at a 9 percent interest rate, which is 0.423.
To calculate the present value, we multiply the future value ($102,000) by the present value factor (0.423): $102,000 * 0.423 = $43,229.
Comparing the present value of $43,229 to the $38,000 offered today, we can see that the latter is the better choice financially. Delia should choose to take the $38,000 today instead of waiting 10 years to receive $102,000.

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Answer:

f(x) is the blue graph, g(x) is the red graph.

x^2 - 3x + 17 - (2x^2 - 3x + 1) = 16 - x^2

16 - x^2 = 0 when x = -4, 4

So the area between these two graphs is (using the TI-83 graphing calculator):

fnInt (16 - x^2, x, -4, 4) = 85 1/3

Answer:

6+33j

Step-by-step explanation:

ok so most people would think to add first no y ok u would start with the 3 so I think as far as I remember you multiply or add the 3 to both numbers so I think that the answer would be 6+33j

When the input x is -18, the output h is 14. The answer is obtained using substitution method.

What is substitution method?

One of the algebraic techniques for solving simultaneous linear equations is the substitution approach. It entails changing any variable's value from one equation to the other by substituting it in. In this manner, a pair of linear equations are combined into a single, simple linear equation with just one variable.

Now, we are given h = 17+x/6

So, when the input x is -18, using substitution method,

h=17+(-18/6)

=17+(-3)

=17-3

=14

Hence, when the input x is -18, the output h is 14.

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let's recall that d = rt, namely distance = rate * time.

d = distance from one pier to another

c = speed of the current

b = speed of the boat

now, let's notice something, on the way over the boat travelled "d" kilometers in 4 hrs, and back the same "d" kilometers in 5 hours.

when going with the current, the boat is not really going at "b" km/s, is really going faster, at "b + c" km/s, because the current is adding to it.

now, when going against the current, the boat is not really going "b" fast, is really going slower, at "b - c", because the current is subtracting from it.

from the above, we can conclude that the 4hrs trip was with the current, since it took less time, and the 5hrs trip was against the current, and we also know that the boat does 70 km with the current in 3.5 hrs, ok hmmm let's put all that in a table.

\(\begin{array}{lcccl} &\stackrel{km}{distance}&\stackrel{km/h}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ \textit{on the way over}&d&b+c&4\\ \textit{on the way back}&d&b-c&5\\\cline{1-4} \textit{we also know}&70&b+c&3.5 \end{array}~\hfill \begin{cases} d=(b+c)4\\\\ d = (b-c)5 \end{cases} \\\\\\ \stackrel{\textit{substituting on the 2nd equation}}{(b+c)4~~ = ~~(b-c)5}\implies 4b+4c=5b-5c \\\\\\ 4c=b-5c\implies \boxed{9c=b} \\\\[-0.35em] ~\dotfill\)

\(\stackrel{\textit{we know that}}{70~~ = ~~(b+c)3.5}\implies 70=(b+c)\cfrac{7}{2}\implies \stackrel{\textit{substituting "b"}}{70=(9c+c)\cfrac{7}{2}} \\\\\\ 70=10c\cdot \cfrac{7}{2}\implies 70=\cfrac{70c}{2}\implies 140=70c\implies \cfrac{140}{70}=c\implies \blacktriangleright 2=c \blacktriangleleft \\\\\\ \stackrel{\textit{the speed of the boat in still water is then}}{9c=b\implies 9(2)=b}\implies \blacktriangleright 18=b\blacktriangleleft\)

Answer:

4.24

Step-by-step explanation:

3.64 + 3/5

3.64 + 0.6 <--- 3/5 = 0.6 as a decimal

= 4.24 <--- Final answer

The area of the domain enclosed by the curve with parametric equations x = tsin t, y = cost, t ∈ [0, 2π] is 2π + 2.

The parametric equations given are:

x = t sin t
y = cos t

To find the area of the domain enclosed by the curve, we can use the formula for the area of a region bounded by a curve given in parametric form:

A = ∫[a,b] y dx

where a and b are the limits of the parameter t that describe the domain of the curve.

In this case, we have:

a = 0
b = 2π

So, we need to compute:

A = ∫[0,2π] cos(t) (t sin(t)) dt

Using integration by parts with u = t and dv = sin(t) dt, we get:

A = [t cos(t)]|[0,2π] - ∫[0,2π] cos(t) dt + ∫[0,2π] sin(t) dt

The first integral evaluates to:

[t cos(t)]|[0,2π] = 2π

The second integral evaluates to:

∫[0,2π] cos(t) dt = [sin(t)]|[0,2π] = 0

The third integral evaluates to:

∫[0,2π] sin(t) dt = [-cos(t)]|[0,2π] = 2

Therefore, the area of the domain enclosed by the curve is:

A = 2π - 0 + 2 = 2π + 2

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is it 2x - 45 or is it 2x - 4 5 can you write the question again in comment

The Value of x and y are (x,y) = (7,3).

What is Algebraic expression ?

Algebraic expressions are numbers expressed using letters or alphabets without specifying their values. Algebra taught us how to express unknown values using letters such as x, y, and z. There can also be a single value that is placed before and multiplied by a variable in an algebraic expression in addition to these letters, which are called variables.

GIven equations are :

5x - 2y= 26.....(1)

y + 4 = x..........(2)

Solve by substitution :

substitute equation 2 in equation 1 and solve for y

5x - 2y= 26

5(y+4) - 2y = 26

5y + 20 -3y = 26

2y = 26 - 20

2y = 6

y = 3

now, the value of x will be :

x = y + 4

x = 3+4

x = 7

Therefore, The Value of x and y are (x,y) = (7,3).

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the answer i got was 18.12

Answer:

6+7√3  or  18.12

Step-by-step explanation:

7√3+2√9= 7√3+2*3= 6+7√3

6+7√3= 18.12

I'm sorry this isn't the answer, but I just want you to know that you are incredible and that I love you for you! You are special to everyone you meet, and should not change who you are. I know your life may be tough, but you are strong and can get through it!

Answer:

yes it is

Step-by-step explanation:

Answer:

x^2 + 14x + 49

Step-by-step explanation:

Complete the square:

x^2 + 14x

Take the coefficient of the x term, 14.

Divide by 2.

14/2 = 7

Square it.

7^2 = 49

Add this .

x^2 + 14x + 49

Answer:

2.16, 3.45

Step-by-step explanation:

Answer:

Hello!!! erz here ^^

Step-by-step explanation:

2.16< \(\sqrt{5}\) < 3.45

Hope this helps!! :D

Answer: 90 is a number which has 9 as the sum of its digits, 9+0=9 and if the digits are reversed, 09 is 81 less than 90.

Step-by-step explanation: Since the number must be a 2-digit number, it seems that 90 is the only number that qualifies.

If you need some equations, here is a way to set up a system to solve:

x+y= 9 rewrite to get a value for y to substitute:

y = 9-x

10x +y = 10y + x + 81 rewrite by moving (subtracting variable terms from both sides)

10x -x + y-10y = 81 combine like terms

9x - 9y = 81

Substitute for y and solve for x

9x - 9(9-x)= 81 distribute.

9x - 81 + 9x = 81

18x = 162. 162/18 = 9

x = 9

Substitute in the original equation to solve for y

9 + y = 9

y = 0

10(9) + 0 = 10(0) + 9 +81

90 - 81 = 09

The diameter of the small volcano 'Pico' would be = 8km

An isoline is defined as the line found on a map that has constant value of either distance, temperature or rainfall.

The isoline that is used in the given map above = 2000m.

The number of lines that surrounds the pico volcano = 4

Therefore the diameter of the volcano = 2000×4 = 8000m

But 1000m = 1km

8000m = 8km

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a) The pooled variance for your study is 171.38 (rounded to three decimal places).

b) The estimated standard error of the difference in sample means for your study is 1.526 (rounded to three decimal places).

c) The t statistic for your independent-measures t test, when the null hypothesis is that the two population means are the same, is 0.261 (rounded to three decimal places).

d) The degrees of freedom for this t statistic is 45.

a) The pooled variance for your study is calculated using the formula:

pooled variance \(= [(n1 - 1) * S1^2 + (n2 - 1) * S2^2] / (n1 + n2 - 2)\)

Using the given values:

pooled variance = \([(31 - 1) * (6.1^2) + (16 - 1) * (777.6)] / (31 + 16 - 2)\)

pooled variance = 171.38 (rounded to three decimal places)

b) The estimated standard error (SE) of the difference in sample means is calculated using the formula:

\(SE = \sqrt{(s1^2 / n1) + (s2^2 / n2)}\)

Using the given values:

\(SE = \sqrt {(6.1^2 / 31) + (171.38 / 16)}\)

SE = 1.526 (rounded to three decimal places)

c) The t statistic for the independent-measures t test is calculated as:

t = (M1 - M2) / SE

Using the given values:

t = (9.8 - 9.4) / 1.526

t = 0.261 (rounded to three decimal places)

d) The degrees of freedom for this t statistic is 45.

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Answer:

A. The function is Nonlinear.

1. The implementation that faster is P1 than P2.

2. The global CPI for each implementation are:

Global CPI = (10% x 3) + (30% x 2) + (40% x 1) + (20% x 4) = 1.9

Global CPI = (10% x 2) + (30% x 2) + (40% x 2) + (20% x 2) = 2

3. The clock cycles required in both cases are:

Clock cycles = 1.9 x 1.0E5 = 190,000

Clock cycles = 2 x 1.0E5 = 200,000

P1: 10% x 3 + 30% x 2 + 40% x 1 + 20% x 4 = 1.9 CPI

Execution time P1 = (1.9 x 1.0E5) / 3GHz = 63.33 microseconds.

P2: 10% x 2 + 30% x 2 + 40% x 2 + 20% x 2 = 2 CPI

Execution time P2 = (2 x 1.0E5) / 2.5GHz = 80 microseconds.

Therefore, P1 is faster than P2.

Global CPI = (Percentage of class A instructions x CPI of class A) + (Percentage of class B instructions x CPI of class B) + (Percentage of class C instructions x CPI of class C) + (Percentage of class D instructions x CPI of class D)

For P1:

Global CPI = (10% x 3) + (30% x 2) + (40% x 1) + (20% x 4) = 1.9

For P2:

Global CPI = (10% x 2) + (30% x 2) + (40% x 2) + (20% x 2) = 2

Clock cycles = Global CPI x Instruction count

For P1:

Clock cycles = 1.9 x 1.0E5 = 190,000

For P2:

Clock cycles = 2 x 1.0E5 = 200,000

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The solution to the system equation  is (x, y, z, w) = (23, 12, 3, 1).

What is equation?

An equation could be a formula that expresses the equality of 2 expressions, by connecting them with the sign =

Main body:

Here is a system of linear equations that represents the situation.

x +5y +10z +20w = 133 . . . total amount earned

x +y +z +w = 39 . . . . . . . . . total number of bills

y = 4z . . . . . . . . . . . . . . . . . . the number of 5s is 4 times the number of 10s

x = 2y -1 . . . . . . . . . . . . . . . . the number of 1s is 1 less than twice the number of 5s

_____

We can substitute for x and z in the first two equations:

... (2y-1) +5y +10(y/4) +20w = 133

... (2y-1) +y +(y/4) +w = 39

These simplify to

... 9.5y +20w = 134

... 3.25y +w = 40

Solving by your favorite method, you get

... y = 12

... w = 1

So the other values can be found to be

... x = 2·12 -1 = 23

... z = 12/4 = 3

hence ,The solution to the system is (x, y, z, w) = (23, 12, 3, 1).

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The three conditions for a function to be continuous at a point x=a are:

a) The function is defined at x=a.

b) The limit of the function as x approaches a exists.

c) The limit of the function as x approaches a is equal to the value of the function at x=a.

The continuity of a function can be analyzed by observing its graph. However, as the graph is not provided, a specific discussion about its continuity cannot be made without further information. It is necessary to examine the behavior of the function around the point in question and determine if the three conditions for continuity are satisfied.

The function f(x) = { *** is not defined in the question. In order to discuss its continuity, the function needs to be provided or described. Without the specific form of the function, it is impossible to analyze its continuity. Different functions can exhibit different behaviors with respect to continuity, so additional information is required to determine whether or not the function is continuous at a particular point or interval.

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Answer:

1/2

Step-by-step explanation:

Slope is the rise over the run.  This means the difference in the y values over the difference in the x values.

The two points on the graph are (2,-2) and ( -6, -6 ).

The difference in the y values is a total of 4 units.

The difference in the x values is a total of 8 units.

The slope would then be 4/8 which simplifies to 1/2.

The line slants upward from left to right so the slope is positive.

Answer:

9 months

Step-by-step explanation:

Answer:

it will be 9 months like the first person said

Answer:

b

Step-by-step explanation:

a) 12 x 5 = 60

doesnt work

b) 0 x 5 =  0

works

c) 7 x 5 = 35

doesnt work

d) 5 x 5 = 25

doesnt work

e) 8 x 5 = 40

doesnt work

f) 9 x 5 =45

doesnt work

the particle's position at any time l is given by: x(t) = (21/2)t^2 - (17/2) y(t)  (7/2)t^3 - (5/2) z(t) = (1/2)t^2 - (1/2) w(t) = (1/4L)t^2 - (1/4L)

To find the particle's position at any time l, we can integrate its velocity vector with respect to time. Given that v(1) = (21, 21, 1, 2/4L), let's perform the integration.

(a) Position at any time l:

Integrating the velocity vector, we have:

∫(v(t)) dt = ∫((21t, 21t^2, t, (2/4L)t)) dt

To find the position, we integrate each component of the velocity vector separately:

∫(21t) dt = (21/2)t^2 + C1

∫(21t^2) dt = (7/2)t^3 + C2

∫(t) dt = (1/2)t^2 + C3

∫((2/4L)t) dt = (1/4L)t^2 + C4

Adding the constant terms, we get:

x(t) = (21/2)t^2 + C1

y(t) = (7/2)t^3 + C2

z(t) = (1/2)t^2 + C3

w(t) = (1/4L)t^2 + C4

Now, we need to determine the values of the constants C1, C2, C3, and C4. To do so, we'll use the initial conditions provided.

Given that the particle starts at the point (2, 1, 0) when t = 1, we substitute these values into the position equations:

x(1) = (21/2)(1)^2 + C1 = 2

y(1) = (7/2)(1)^3 + C2 = 1

z(1) = (1/2)(1)^2 + C3 = 0

w(1) = (1/4L)(1)^2 + C4 = 0

From these equations, we can solve for the constants C1, C2, C3, and C4.

C1 = 2 - (21/2) = -17/2

C2 = 1 - (7/2) = -5/2

C3 = 0 - (1/2) = -1/2

C4 = 0 - (1/4L) = -1/4L

Therefore, the particle's position at any time l is given by:

x(t) = (21/2)t^2 - (17/2)

y(t) = (7/2)t^3 - (5/2)

z(t) = (1/2)t^2 - (1/2)

w(t) = (1/4L)t^2 - (1/4L)

(b) To find the cosine of the angle between the velocity vector v(1) and the position vector at t = 1, we can calculate their dot product and divide it by the product of their magnitudes.

Let's calculate the cosine:

cosθ = (v(1) · r(1)) / (|v(1)| |r(1)|)

Substituting the values:

v(1) = (21, 21, 1, 2/4L)

r(1) = (2, 1, 0, 0)

|v(1)| = √((21)^2 + (21)^2 + (1)^2 + (2/4L)^2) = √(882 + 882 + 1 + (1/2L)^2) = √(1765 +

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