In this problem, you will investigate the location of the points of concurrency for any equilateral triangle.
a. Concrete Construct three different equilateral triangles on tracing paper and cut them out. Fold each triangle to locate the circumcenter, incenter, centroid, and orthocenter.

Answer:

1 hour has 3600 seconds

to get how many dollars Jack earns we have to do this

3600 sec x $0.02 = $72

40 weeks x $72 = $2880

Answer:

$149,760

Step-by-step explanation:

As:

⇒ 1 hour = 60 × 60 seconds = 3600 seconds

As:

⇒ 52 × 40 = 2080 hours per year

Therefore, the number of seconds Jack works per year is:

⇒ 2080 × 3600 = 7,488,000 seconds

If Jack earns $0.02 per second, then the number of dollars per year he earns is:

⇒ $0.02 × 7,488,000 = $149,760

The company profit approximately increased by E) 70%.

An approximation is an equivalent value or number obtained by rounding off to the required significant figure.

The result obtained through approximation is not an exact value but is the nearest rounded value.

For instance, 71.4% can be approximated to 71% or 70%.  58 minutes can be rounded off or approximated to 1 hour.

Profit in 3 months = $3,500

Profit in 6 months = $6,000

Increase in profit = $2,500 ($6,000 - $3,500)

Percentage increase in profit = 71.43% ($2,500/$3,500 x 100)

71.43% ≈ 70%

Thus, the company profit approximately increased by E) 70%.

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The given sequence 100,117,134,151,168, ........... is in AP. The next three terms are 185, 202, 219..

The difference between two numerical orders is a constant value in Arithmetic Progression (AP). Another name for it is Arithmetic Sequence.

We'd emerge across a few fundamental words in AP that have been labeled as:

The AP can also be described in terms of common differences, as shown below.

Now, the given sequence is; 100,117,134,151,168, ...........

The series comprises of five given terms.

Let the initial term be 'a₁' = 100.

The second term be 'a₂' = 117.

The third term be 'a₃' = 134.

And, the fourth term is 'a₄' = 151.

The AP must have the equal common difference. So,

d = a₃ - a₂

Substitute the values.

d = 134 - 117

d = 17

Thus, the computed common difference is 17.

or d = a₄ - a₃ (Substitute the values)

d = 151 - 134

d = 17

Since, a₃ - a₂ = a₄ - a₃

Thus, we can get to know that the given sequence is in AP.

Now, the 6th term is estimated as; a₆ = a₅ + d = 168 + 17

a₆ = 185

Similarly, the 7th term will be; a₇ = a₆ + d = 185 + 17

a₇ = 202

And, the 8th term will be calculated as; a₈ = a₈ + d = 202 + 17

a₈ = 219.

Therefore, it can be said that the given sequence is in AP next three terms be 185, 202, 219.

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Answer: In geometry, when you translate an object, it means that you have turned that object in a different direction. Hence a translated angle is an angle that has been turned in a different direction.

From the first option in part A, it is clear from the image that the angle didn't change. It moves 7 units from the positive to the negative part of the x-axis.

From Part B, the angles also do not change. Note that when an angle is rotated or translated, it does not modify the angles.

From Part C, what we have is the reflection of the angles. This means that both angles are equal because they are a reflection of each other.

From Part D, the two sets of parallel lines will remain parallel to each other also long as they are both translated as the same time.

From part E, although in each set of parallel lines, the two sets of lines remain equidistant to each other, the parallel line that is at an angle to the Y-axis will not intersect that which is at an angle to the x-axis if they are extended infinitely.

Step-by-step explanation: hope this helps alittle

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

They are congruent triangles.

What do you mean by congruent?

In the triangle JKL, the sides can be calculated as following:

J(2;5); K(1;1)

            => JK = [tex]\sqrt{( ( 1 - 2 )^{2} + ( 1 - 5 )^{2} } = \sqrt{( -1 )^{2} + ( -4)^{2} } \\= \sqrt{ 1 + 16 } = \sqrt{17}[/tex]

J(2;5); L(5;2)

            JL = [tex]\sqrt{( 5 - 2 )^{2} + ( 2 - 5 )^{2} } = \sqrt{3^{2} + ( -3)^{2} } \\[/tex]

                 = [tex]\sqrt{ 9 + 9 } = \sqrt{18} = 3\sqrt{2}[/tex]

K(1;1); L(5;2)

            =>  KL = [tex]\sqrt{ ( 5 - 1 )^{2} + ( 2 - 1 )^{2} } = \sqrt{4^{2} + 1^{2} } = \sqrt{ 1 + 16} = \sqrt{17}[/tex]

In the triangle QNP, the sides can be calculate as following:

Q(-4;4); N(-3;0)

            => QN  = [tex]\sqrt{[ -3 - ( -4)^{2} ] + ( 0-4)^{2} } = \sqrt{1^{2} + 16 } = \sqrt{17}[/tex]

Q (-4;4); P(-7;1)

  => QP =  [tex]\sqrt{ [ -7 - ( -4 )^{2} + ( 1 - 4 )^{2} } = \sqrt{( -3)^{2} + ( -3)^{2} } = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}[/tex]

N(-3;0); P(-7;1)

      =>  NP = [tex]\sqrt{[-7 -( -3)]^{2} + ( 1 - 0 )^{2} } = \sqrt{ - 4^{2} + 1 } = \sqrt{ 16 + 1} = \sqrt{ 17}[/tex]

It can be seen that QPN and JKL have: JK = QN; JL = QP; KL = NP

=> They are congruent triangles

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1)The neighborhood park is 60m by 50 m. The park is reduced in size by 321m2 as a portion of it is ripped up to create a uniform-width sidewalk around it. What quadratic equation best captures the circumstances?

2)A rectangle's length is 1 m less than twice its width. The rectangle has a 120 m2 area. What quadratic equation best describes the circumstances?

Answer for the first question is the quadratic equation is [tex]4x^{2} -220x+321=0\\[/tex].

Answer for the second question is The quadratic equation is [tex]x^{2} -x-120=0[/tex].

1)Let 'x' stand in for the sidewalk's width.

The area of the park is: [tex]A_{1} =60\times50[/tex]

The reduced area of the park measures

60-2x by 50-2x with area of [tex]A_{2} =(60-2x)(50-2x)[/tex]

[tex]A_{1} -A_{2} =321[/tex]

[tex]60\times50-(60-2x)(50-2x)=321[/tex]

[tex]3000-(60-2x)(50-2x)=321\\3000-(3000-120x-100x+4x^{2} )=321\\220x-4x^{2} =321\\4x^{2} -220x+321=0\\[/tex]

Therefore, the quadratic equation is [tex]4x^{2} -220x+321=0\\[/tex].

2) Given that,

Width=x.

length=2x-1

Area=120[tex]m^{2}[/tex]

We know area=length[tex]\times[/tex]width

120=x(x-1)

[tex]x^{2} -x=120\\x^{2} -x-120=0[/tex]

Therefore, The quadratic equation is [tex]x^{2} -x-120=0[/tex].

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The value of fraction is 6 1/4 inches are equal to 19+6/425mm.

Given that the precipitation in March was 4 3/4 inches below average and the precipitation in April was 1 1/2 inches below average.

We must determine how many inches the precipitation in March and April fell short of typical.

Add 4 3/4 inches and 1 1/2 inches to get the answer. So,

4 3/4 + 1 1/2

19+6 / 4

= + 4 312

6 1/4 inches are equal to 19+6/425mm.

six inches Therefore, the precipitation was 4 less than usual for both March and April.

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Here are some steps in the photo.

Mark me Brainliest. :)

Source of the photo: Calculator Soup

If a red string of holiday lights blinks every 3 seconds. a blue string of holiday lights blinks every 4 seconds, then the number of times the red and blue string of holiday lights will blink in one minute is equal to 20 and 15 respectively.

The number of times each string of holiday lights will blink can be determined by using division.

As we have to calculate the number of times they blink in one minute, we can convert one minute into seconds.

One minute = 60 seconds

Now, the number of times each string of light blinks can be calculated by dividing the total seconds by the seconds after which the light blinks

For the red string of holiday lights :

The number of times the light will blink in one minute = 60 ÷ 3 = 20

For the blue string of holiday lights :

The number of times the light will blink in one minute = 60 ÷ 4 = 15

Hence the red string of holiday lights will blink 20 times and the blue string of holiday lights will blink 15 times in one minute.

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Step-by-step explanation:

let this number is x

The value of the expression R = x^2/y when y = 3.8 x 10^5 and x = 5.9 x 10^4 is 9.2 x 10^3

Expressions are mathematical statements that are represented by variables, coefficients and operators

The expression is given as

R = x^2/y

Where

y = 3.8 x 10^5

x = 5.9 x 10^4

Substitute y = 3.8 x 10^5 and x = 5.9 x 10^4 in the equation R = x^2/y

So, we have

R = (5.9 x 10^4)^2/3.8 x 10^5

Evaluate the exponent in the above equation

So, we have

R = 34.81 x 10^8/3.8 x 10^5

Evaluate the quotient in the above equation

So, we have

R = 9.16052631579 x 10^3

Approximate the above expression

So, we have

R = 9.2 x 10^3

Hence, the value of the expression R = x^2/y when y = 3.8 x 10^5 and x = 5.9 x 10^4 is 9.2 x 10^3

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Answer:

B. 4/5

Glad to help :)

The ninth and tenth term of the given arithmetic sequence are: -18 and -21.

A series of numbers called an arithmetic progression or arithmetic sequence (AP) has a constant difference between the terms.

Now,

Given: 6, 3, 0, -3, ....; it is a decreasing AP.

Here, a = 6 (initial term) and d = -3 (common difference)

Thus, in the formula: [tex]a_n[/tex] = a + (n - 1) d, substituting these values:

=>  [tex]a_n[/tex] = 6 + (n - 1)(-3)

For ninth term, n = 9

Thus,  [tex]a_n[/tex] = [tex]a_9[/tex] = 6 + (9 - 1)(-3)

=> [tex]a_9[/tex] = 6 + (8)(-3)

=> [tex]a_9[/tex] = 6 -24

=> [tex]a_9[/tex] = -18

Similarly, for the tenth term, n = 10

Thus, [tex]a_n=a_{10}[/tex] = 6 + (10 -1) (-3)

=> [tex]a_{10}[/tex] = 6 + (9) (-3)

=> [tex]a_{10}[/tex] = 6 - 27

=> [tex]a_{10}[/tex] = -21

Hence, The ninth and tenth term of the given arithmetic sequence are: -18 and -21.

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Number of songs that will be downloaded in 2020 = 2600 million

Given that in 2004, 200 million songs were downloaded and in 2006, 500 million songs were downloaded from the internet.

When we graph this trend, these data corresponds to the points (4,200) and (6, 500).

Also it will be a straight line. So its slope can be found.

Slope = [tex]\frac{y_2-y_1}{x_2-x_1} = \frac{500-200}{6-4}[/tex] = 300/2 = 150

This means that there is an increase of 150 million songs downloads pe year.

Also 2020-2006 = 14 years

So the number of songs downloaded in 2020 = Number of songs downloaded in 2006 + 14 x 150 = 500 + 14 x 150

                                                     = 500 +2100

                                                     = 2600 million

So 2600 million songs will be downloaded legally from the internet by 2020.

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The advanced trail is 4.32 times as long as the beginner's trail.

A decimal is a number that is made up of integers and non-integers. The whole numbers are separated from numbers that are not whole number by a decimal point. An example of a decimal is 1.2.

Let:

The equation that represents the length of the medium trail: 1.2 x b = 1.2b

The equation that represents the length of the advanced trail: 3.6 x 1.2b = 4.32b

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Answer:

Step-by-step explanation:

Answer: x=-3/4

Step-by-step explanation:

12-x=15+3x

12-x+x=15+3x+x

12=15+4x

12-15=15-15+4x          ==> isolate the variable x.

-3=4x

4x/4=-3/4

x=-3/4

Answer: Large Box: 18.75 kilograms

Small Box: 15.75 kilograms

Step-by-step explanation:

l=large box. s=small box

5l+3s=141

7l+9s=273

(5l+3s=141)*3

15l+9s=423    ==> subtract this equation

7l+9s=273   ==> subtract this equation

8l=150 ==> subtract the two above equations

l=150/8

l=18.75

5(18.75)+3s=141

93.75+3s=141

93.75+3s=141.00

3s=47.25

s=15.75

Large Box: 18.75 kilograms

Small Box: 15.75 kilograms

An expression can be used in algebra to represent a value, and that value may rely on the values given to the variables that appear in the expression.

If the expression be (-3)(-9) then the value of (-3)(-9) is 27.

Let (-3)(-9) be the given expression

We know the rule , (-) × (-) = (+)

(-3) × (-9) = +27

Therefore, the value of (-3)(-9) is 27.

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In mathematics, the logarithm is the reciprocal of a power. This means that the logarithm of a given number x is the exponent by which another fixed number (the base b) must be increased to get that number x. The simplest logarithm counts the number of times the same element occurs when multiplying repeatedly. The logarithm of x to base b is denoted by log b (x), or log b x without parentheses. Also, if confusion is impossible or the base doesn't matter, log x shows that even without an explicit base: B. In big O notation.

Logarithm base 100 of 100 is 1.

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Answer: the length of the 1st side is 4 m

Step-by-step explanation:

P=a+b+c

b=a+3

c=4a

Hence,

27=a+(a+3)+4a

27=a+a+3+4a

27=6a+3

27-3=6a+3-3

24=6a

Divide both parts of the equation by 6:

4=a

Thus,

a=4 m

Answer:

Step-by-step explanation:

In all three pictures, the data are correctly matching except this one..

~~

Answer:

7.2

Step-by-step explanation:

7.4-0.2=7.2

Answer:

6 is the y intercept and 1 is the slope, so it goes up 1 unit right 1 unit

Step-by-step explanation:

Binary relation  is a function .

Functions are relationships where each input has a single output. There is one output for each input in a relation that passes the vertical line test.

How can I tell if a relationship serves a purpose?

A relation is considered to be a function if each element in the domain is mapped to a distinct element in the codomain.

What are the characteristics of a binary relation?

Binary relationship A subset of P x Q from the set P to Q is defined as R. If (a, b) R and R P x Q, then a and b are connected by R, or aRb. When sets P and Q are equal, we say that R is a relation on P, for example.

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The formula to find the sum of an arithmetic sequence: Sn = n/2[2a + (n - 1)d], where a is the first term, n is the number of terms and d is the common difference.

1) Let's substitute the values into the formula.

S17 = 17/2[2(11) + (17 - 1)6]

S17 = 17/2[22 + (16)6]

S17 = 17/2[22 + 96]

S17 = 17/2[118]

S17 = 17/2 × 118

S17 = 1003

Therefore, the sum of the first 17 terms of this arithmetic sequence is 1003.

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Answer:

See proof below

Step-by-step explanation:

[tex]4cos^2(x) +4sin(x) = 5\\\\\sf Divide\; both \; sides\; by\; 4\\== > cos^2(x) + sin(x) = \dfrac{5}{4} .................(1)\\\\cos^2(x) + sin^2(x) = 1\\\\cos^2(x) = 1- sin^2(x) \\\\\textsf {Substituting into equation (1) }\\1-sin^2(x) + sin(x) = \dfrac{5}{4}\\\\\\sf Subtract\; 1\; from\; both \; sides\\\mathsf{-\sin^2(x) + \sin(x) = \dfrac{5}{4} - 1}\\\\ \dfrac{5}{4} - 1 = \dfrac{5}{4} - \dfrac{4}{4} = \dfrac{1}{4}\\[/tex]

[tex]\sf {-\sin^2(x) + \sin(x) = \dfrac{1}{4}\\[/tex]

[tex]\sf Subtract\; \dfrac{1}{4}\;from\;both\;sides\\[/tex]
[tex]{-\sin^2(x) + \sin(x) - \dfrac{1}{4} = 0\\\\[/tex]

Multiply throughout by -1
[tex]\sf \sin^2(x) - \sin(x) + \dfrac{1}{4} = 0\\\\[/tex]

Let u = sin(x). Substituting we get
[tex]\sf u^2 - u + \dfrac{1}{4} = 0[/tex]

This is a quadratic equation which can be solved using sum of squars

Note that

[tex]\sf (u - \dfrac{1}{2})^2 = u^2 -2\cdot \dfrac{1}{4}u + (\dfrac{1}{2})^2 \\= \sf u^2 - u + \dfrac{1}{4} = 0[/tex]

So
[tex]\sf (u - \dfrac{1}{2})^2 = 0\\Solution\; is \;u = \dfrac{1}{2}\\\\Substituting\sin(x) \;for\;u \;gives\\sin(x) = \dfrac{1}{2}[/tex]

PROVED

Answer:

Step-by-step explanation:

Answer:

A: D:(-6,6)

B: D:(-7,5]

C: Not a function D: -5

D: Not a function D:(-4,4)

E: Not a function D:(0,∞)

F:  D:[0,∞)

G: D:[-3,5)

H: D:(-4,2)

I: D:[-3,4]

J: Not a function D:(-3.-4)

K: D:(-∞,∞)

L: D:(-∞,∞)

The equation that we need to solve and the solution is:

A = 11.7mi/3 = 3.9 mi

We conclude that he needs to ride an average of 3.9 miles per day.

Here we know that Carter wants to ride his bicycle 28.7 miles this week, and he has already ridden 17 miles, so the distance left is:

D = 28.7 mi - 17mi = 11.7mi

Now, if he has 3 days to complete that distance, then the average distance that he needs to travel each day is given by the quotient:

A = 11.7mi/3 = 3.9 mi

So that is the equation that we need to solve and the solution, we conclude that he needs to ride an average of 3.9 miles per day.

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Answer:

The first one would be -11x + 3

The second on would be 29x + 3

Step-by-step explanation:

For the first equation you add -2 to 5 so you have 9x+3-20x. Then you add the 9x to -20x which will leave you with -11x+3.

For the second equation you add the -2 to 5 and have 9x+3+20x. You add 9x to 20x which will leave you with. 29x+3

[tex]format \: \: y = mx + c[/tex]

[tex]m = \frac{5 - 0}{10 - 0} = \frac{5}{10} = \frac{1}{2} [/tex]

[tex]c = 0 \: since \: the \: line \: passes \: through \: the \: origin \\ [/tex]

[tex](d) \: \: \: y = \frac{x}{2} [/tex]