Geometry: complete this proof, ASAP!!!!!!!!

The option that explains the point (1, 1400) is (c) the person surveyed took a day trip

From the graph, we have the following representations:

The point (1,1400) implies that

x = 1 and y = 1400

This means that

The number of days on vacation is 1 and the total cost of vacation is 1400

Hence, the option that explains the point (1, 1400) is (c) the person surveyed took a day trip

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Based on the information, there were initially 1200 ferns; the monthly percent rate of change is 68% and this situation is modeled by an exponential growth function

The given parameters from the question are:

P(n) = 1200(1.68)^n

Where

n represent the number of months

An exponential growth function is represented as:

P(n) = a * (1 + r)^n

Where

a represents the initial value

r represents the rate

By comparison, we have

a = 1200

1 + r = 1.68

This gives

r = 0.68 = 68%

Hence, there were initially 1200 ferns; the monthly percent rate of change is 68% and this situation is modeled by an exponential growth function

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1,200

68

growth

Got it right on Edmentum

Answer:

[tex]x=\frac{P^2t}{(m-t^3)}[/tex]

Step-by-step explanation:

[tex]P^2=\frac{mx}{t} -t^2x\\\\P^2+t^2x=\frac{mx}{t}\\\\t(P^2+t^2x)=mx\\\\P^2t+t^3x=mx\\\\P^2t=mx -t^3x\\\\P^2t=x(m-t^3)\\\\\frac{P^2t}{(m-t^3)} =\frac{x(m-t^3)}{(m-t^3)} \\\\\frac{P^2t}{(m-t^3)}=x[/tex]

Concluding, the end behavior is:

as x ⇒ ∞, f(x) ⇒ -∞

as x ⇒ -∞, f(x) ⇒ ∞

Here we have the function:

[tex]f(x) = -2*\sqrt[3]{x}[/tex]

When we evaluate the function in x that tends to positive infinity, the cubic root also tends to infinity, while because of the negative factor that multiplies it, we conclude that the end behavior is:

as x ⇒ ∞, f(x) ⇒ -∞

When x tends to negative infinity, the opposite happens:

as x ⇒ -∞, f(x) ⇒ ∞

Concluding, the end behavior is:

as x ⇒ ∞, f(x) ⇒ -∞

as x ⇒ -∞, f(x) ⇒ ∞

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Answer:

here's your answer step by step explanation

Step-by-step explanation:

page 6 is missing comment me i will send you page 6

The probability that out of 8 newly hired people is 3.2.

According to the statement

we have given that the some conditions and based on these conditions we have to find the probability that out of 8 newly hired people.

So, For this purpose,

we have given that the

new employee in a fast-food chain still being with the company at the end of the year is  0.5

Now, we See that is probability of remaining with the company.

There is only one is 0.5.

According to the probability law

P + Q = 1

and this become

Q = 1 - 0.60.

And the probability of one people hired is Q.

Then

Q = 0.4

And probability that out of 8 newly hired people is

8Q = 0.4 * 8

it become 3.2.

So, The probability that out of 8 newly hired people is 3.2.

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the train travels at 45 meters per each second

this question can be rewritten into the equation d = 45t, where d represents the distance traveled and t represents the time elapsed.

we are given two seperate times, so we can replace the variable t with each respective time. this leaves us with only one missing variable, so we can successfully isolate to find the other.

for i), we substitute t for 30 seconds, shown as follows

d = 45t

d = 45(30)

d = 1350m

for ii), we know that there are 60 seconds in every minute, so in multiplying 60 seconds by two we get the total amount of seconds in two minutes, which is 120.

we can now use 120 to substitute t in our equation

d = 45t

d = 45(120)

d = 5400m

hope this helps!!

Answer:

i) 1350 m

ii) 5400 m

Step-by-step explanation:

To calculate the distance traveled by the train, we can use the formula:

[tex]\large\boxed{\sf Distance=Speed \times Time}[/tex]

[tex]\hrulefill[/tex]

Given values:

Substitute the given values into the formula for distance:

[tex]\begin{aligned}\sf Distance&= \sf 45\; m/s \times 30\;s\\& = \sf 1350\:m\end{aligned}[/tex]

Therefore, the train travelled a distance of 1350 meters in 30 seconds.

[tex]\hrulefill[/tex]

Given values:

As the time is given in a different unit of time than the speed, we must first convert the time into seconds. As 1 minute = 60 seconds, then:

[tex]\sf 2\; minutes = 60 \;s \times 2 = 120 \;s[/tex]

Substitute the given values into the formula for distance:

[tex]\begin{aligned}\sf Distance&= \sf 45\; m/s \times 120\;s\\& = \sf 5400\:m\end{aligned}[/tex]

Therefore, the train travelled a distance of 5400 meters in 2 minutes.

Answer:

y = - 2x + 8

Step-by-step explanation:

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

y = [tex]\frac{1}{2}[/tex] x - 9 ← is in slope- intercept form

with m = [tex]\frac{1}{2}[/tex]

given a line with slope m then the slope of a line perpendicular to it is

[tex]m_{perpendicular}[/tex] = - [tex]\frac{1}{m}[/tex] = - [tex]\frac{1}{\frac{1}{2} }[/tex] = - 2

y = - 2x + c ← is the partial equation

to find c substitute (3, 2 ) into the partial equation

2 = - 6 + c ⇒ c = 2 + 6 = 8

y = - 2x + 8 ← equation of perpendicular line

[tex]{ \qquad\qquad\huge\underline{{\sf Answer}}} [/tex]

Let's solve ~

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {a}^{2} - 64 }{ {a}^{2} - 10a + 24} \sdot \cfrac{ {a}^{2} - 12a + 36 }{ {a}^{2} + 4a - 32} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{( {a}^{} + 8)(a - 8) }{ {a}^{2} - 6a - 4a+ 24} \sdot \cfrac{ {a}^{2} - 6a - 6a + 36 }{ {a}^{2} + 8a - 4a- 32} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{( {a}^{} + 8)(a - 8) }{ {a}^{} (a - 6) - 4(a - 6)} \sdot \cfrac{ {a}^{} (a - 6) - 6(a - 6) }{ {a}^{}(a + 8) - 4(a + 8)} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{( {a}^{} + 8)(a - 8) }{(a - 6) (a -4)} \sdot \cfrac{ (a - 6) (a - 6) }{ {}^{}(a - 4)(a + 8)} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{(a - 8) }{(a -4)} \sdot \cfrac{ (a - 6) }{ {}^{}(a - 4)} [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{(a - 8)(a - 6) }{(a -4) {}^{2} } [/tex]

Or [ in expanded form ]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {a}^{2} - 8a - 6a + 48 }{ {a}^{2} - 8a + 16 } [/tex]

[tex]\qquad \sf  \dashrightarrow \: \cfrac{ {a}^{2} -14a + 48 }{ {a}^{2} - 8a + 16 } [/tex]

Answer:

[tex]\dfrac{(a-8)(a-6)}{(a-4)^2}[/tex]

Step-by-step explanation:

Given expression:

[tex]\dfrac{a^2-64}{a^2-10a+24} \cdot \dfrac{a^2-12a+36}{a^2+4a-32}[/tex]

Factor the numerator and denominator of both fractions:

[tex]\textsf{Apply the Difference of Two Squares formula} \:\:\:x^2-y^2=(x-y)(x+y):[/tex]

[tex]\begin{aligned} a^2-64 & =a^2+8^2 \\ & =(a-8)(a+8)\end{aligned}[/tex]

[tex]\begin{aligned}a^2-10a+24 & =a^2-4a-6a+24\\& = a(a-4)-6(a-4)\\ & = (a-6)(a-4) \end{aligned}[/tex]

[tex]\begin{aligned}a^2-12a+36 & =a^2-6a-6a+36\\& = a(a-6)-6(a-6)\\ & = (a-6)(a-6) \end{aligned}[/tex]

[tex]\begin{aligned}a^2+4a-32 & =a^2+8a-4a-32\\& = a(a+8)-4(a+8)\\ & = (a-4)(a+8) \end{aligned}[/tex]

Therefore:

[tex]\dfrac{(a-8)(a+8)}{(a-6)(a-4)} \cdot \dfrac{(a-6)(a-6)}{(a-4)(a+8)}[/tex]

[tex]\textsf{Apply the fraction rule}: \quad \dfrac{a}{b} \cdot \dfrac{c}{d}=\dfrac{ac}{bd}[/tex]

[tex]\dfrac{(a-8)(a+8)(a-6)(a-6)}{(a-6)(a-4)(a-4)(a+8)}[/tex]

Cancel the common factors (a + 8) and (a - 6):

[tex]\dfrac{(a-8)(a-6)}{(a-4)(a-4)}[/tex]

Simplify the numerator:

[tex]\dfrac{(a-8)(a-6)}{(a-4)^2}[/tex]

The expression to figure out today’s ticket sales is 7x+5y.

Today's ticket sale is less than that of yesterday's.

According to the question,

Adult tickets cost $7 each.

Child tickets cost $5 each.

Let the number of adult and child tickets sold be x and y respectively.

Expression for ticket sales= $(7x+5y)

Dexter has sold 125 kid's tickets and 100 adult tickets today.

Ticket sales = $(7*100+5*125) = $ 1325

He sold 120 adult tickets and 120 kid tickets yesterday.

Ticket sales = $(7*120+5*120) = $ 1440

Thus, from the expression it can be seen that yesterday' s ticket sale is greater than today's ticket sale.

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Answer:

5 un^2

Step-by-step explanation:

the liens from (0, 0) to (1, 3) and (1, 3) to (4, 2) are perpendicular meaning that the angle at (1, 3) is a right angle

to find the lengths of the sides we must use the pythagorean theorem

a^2 + b^2 = c^2

for the leftmost side

we have 1^2 + 3^2 = c^2

1 + 9 = 10

c^2 = 10

c = sqrt(10)

for the top side

we have

the same thing

1^2 + 3^2 = c^2

1 + 9 = 10

c^2 = 10

c = sqrt(10)

you must multiple sqrt(10) by sqrt(10) and then by 1/2

sqrt(10) * sqrt(10) is 10

10 * 1/2 is 5

the area is 5 un^2

Answer:

[tex]\textsf{Midpoint rule}: \quad \dfrac{2\pi}{\sqrt[3]{2}}[/tex]

[tex]\textsf{Trapezium rule}: \quad \pi[/tex]

[tex]\textsf{Simpson's rule}: \quad \dfrac{4 \pi}{3}[/tex]

Step-by-step explanation:

Midpoint rule

[tex]\displaystyle \int_{a}^{b} f(x) \:\:\text{d}x \approx h\left[f(x_{\frac{1}{2}})+f(x_{\frac{3}{2}})+...+f(x_{n-\frac{3}{2}})+f(x_{n-\frac{1}{2}})\right]\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]

Trapezium rule

[tex]\displaystyle \int_{a}^{b} y\: \:\text{d}x \approx \dfrac{1}{2}h\left[(y_0+y_n)+2(y_1+y_2+...+y_{n-1})\right] \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]

Simpson's rule

[tex]\displaystyle \int_{a}^{b} y \:\:\text{d}x \approx \dfrac{1}{3}h\left(y_0+4y_1+2y_2+4y_3+2y_4+...+2y_{n-2}+4y_{n-1}+y_n\right)\\\\ \quad \textsf{where }h=\dfrac{b-a}{n}[/tex]

Given definite integral:

[tex]\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x[/tex]

Therefore:

Calculate the subdivisions:

[tex]\implies h=\dfrac{2 \pi - 0}{4}=\dfrac{1}{2}\pi[/tex]

Midpoint rule

Sub-intervals are:

[tex]\left[0, \dfrac{1}{2}\pi \right], \left[\dfrac{1}{2}\pi, \pi \right], \left[\pi , \dfrac{3}{2}\pi \right], \left[\dfrac{3}{2}\pi, 2 \pi \right][/tex]

The midpoints of these sub-intervals are:

[tex]\dfrac{1}{4} \pi, \dfrac{3}{4} \pi, \dfrac{5}{4} \pi, \dfrac{7}{4} \pi[/tex]

Therefore:

[tex]\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2}\pi \left[f \left(\dfrac{1}{4} \pi \right)+f \left(\dfrac{3}{4} \pi \right)+f \left(\dfrac{5}{4} \pi \right)+f \left(\dfrac{7}{4} \pi \right)\right]\\\\& = \dfrac{1}{2}\pi \left[\sqrt[3]{\dfrac{1}{2}} +\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}+\sqrt[3]{\dfrac{1}{2}}\right]\\\\ & = \dfrac{2\pi}{\sqrt[3]{2}}\\\\& = 4.986967483...\end{aligned}[/tex]

Trapezium rule

[tex]\begin{array}{| c | c | c | c | c | c |}\cline{1-6} &&&&&\\ x & 0 & \dfrac{1}{2}\pi & \pi & \dfrac{3}{2} \pi & 2 \pi \\ &&&&&\\\cline{1-6} &&&&& \\y & 0 & 1 & 0 & 1 & 0\\ &&&&&\\\cline{1-6}\end{array}[/tex]

[tex]\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{2} \cdot \dfrac{1}{2} \pi \left[(0+0)+2(1+0+1)\right]\\\\& = \dfrac{1}{4} \pi \left[4\right]\\\\& = \pi\end{aligned}[/tex]

Simpson's rule

[tex]\begin{aligned}\displaystyle \int^{2 \pi}_0 \sqrt[3]{\sin^2 (x)}\:\:\text{d}x & \approx \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(0+4(1)+2(0)+4(1)+0\right)\\\\& = \dfrac{1}{3}\cdot \dfrac{1}{2} \pi \left(8\right)\\\\& = \dfrac{4}{3} \pi\end{aligned}[/tex]

Answer:

NMK= 22

Step-by-step explanation:

From N to K equals 177.

N to M equals x

K to M equals to 107

L equals 35

35    177

x        107

it equals to 21.158

Round up to the nearest ones or whole number

D box plot correctly represents the data

A box plot or boxplot is a technique for illustratively displaying the localization, dispersion, and skewness groups of numerical data through their quartiles.

Using a five-number summary (the minimum, first quartile (Q1), median, third quartile (Q3), and "maximum"), a boxplot is a common method of visualizing data distribution.

The median value for the provided data will be the sixth data point's value, or 3.

The median of the data's lower half is now the first quartile. The third data point's value of 1 will therefore be in the lower quartile.

The median of the data's upper half is now referred to as the upper or third quartile. This means that the value of the ninth data point, which is 6, will represent the upper quartile.

Thus, box-plot D effectively displays the data from above.

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Answer:

The answer is D.

Edg 2023.

Step-by-step explanation:

In a Mathematics lab. There are some cubes and cuboids of following measurements

(i) One cube of side 4 cm

(ii) One cube of side 6 cm

(iii) 3 cuboids each of dimensions 4cm ×4 cm ×6cm

(iv) 3 cuboids each of dimensions 4cm ×6 cm ×6cm

A student wants to arrange these cubes and cuboids in the form of big cube. Is it

possible to arrange them in the form of big cube? If yes, then find the length of side of

Answer:

16 pi/ 9 inches should be the right one.

Part (a)

We'll use the distance formula.

[tex](x_1,y_1) = (-4,-7) \text{ and } (x_2, y_2) = (5,-2)\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(-4-5)^2 + (-7-(-2))^2}\\\\d = \sqrt{(-4-5)^2 + (-7+2)^2}\\\\d = \sqrt{(-9)^2 + (-5)^2}\\\\d = \sqrt{81 + 25}\\\\d = \sqrt{106}\\\\d \approx 10.2956\\\\[/tex]

The diameter is exactly [tex]\sqrt{106}[/tex] units long which approximates to roughly 10.2956 units. Round that decimal value however your teacher instructs.

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

Part (b)

We divide the diameter in half to get the radius.

Therefore, the radius is exactly [tex]\frac{\sqrt{106}}{2}[/tex] units long which is approximately 5.1478 units.

Answer: 15.53

Step-by-step explanation:

Let the height of the goalpost be y. Then,

[tex]\frac{y}{10.65}=\frac{1.75}{1.2}\\\\y \approx 15.53[/tex]

The graph of the function f(x) = -(x+1)^2 shows that the domain of the function f(x) = -(x+1)² is: -∞ < x < ∞. The range of the function is f(x) ≤ 0.

The graph of a function is the arrangement of all ordered pairs of the function. Typically, they are expressed as points in a cartesian coordinate system. The graph of f is the collection of all ordered pairings (x, f(x)) such that x lies inside the domain of f.

The graph of a function might similarly be defined as the graph of the equation y = f(x). As a result, the graph of a function is a subset of the graph of an equation.

From the given information: the graph of the function f(x) = -(x+1)² can be determined if the domain, the range, and the vertex of the function are known.

Since the parabola curve from the graph shows that the graph is facing down, then the function is negative and decreasing.

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1.) maximum value

2.) for no values of x

3,) when x > -1

4.) all real numbers

5.) all numbers less than or equal to 0

Step-by-step explanation:

step 1: multiply [tex]2^m[/tex] by [tex]3^n[/tex]

2 × 3 = 6

to times power you must add them

[tex]2^m[/tex] × [tex]3^n[/tex] = [tex]6^{m+n}[/tex]

step 2: rearrange

[tex]6^{m+n}[/tex] = 108

can you do this step by yourself

Answer:

youre answer is here I hope it's right

x=24

Answer:

1/9

Step-by-step explanation:

For each of the first die's 6 outcomes, the second die also has a possible 6 outcomes.

6 × 6 = 36

There are 36 different outcomes from

1, 1

1, 2

1, 3

...
6, 6

Many outcomes have the same sum. For example, 3, 4 and 4, 3 both add to 7.

How many outcomes add to 9?

3, 6

4, 5

5, 4

6, 3

4 different outcomes out of a possible 36 different outcomes add to 9.

p(sum of 9) = 4/36 = 1/9

gradient = rise/run

= 2/1

= 2

So,it's the first option, y=2x

Hope this helps!

Answer:

70°

Step-by-step explanation:

This is a question on angle properties.

Angle K + 35 + 75 = 180 (Sum of angles in a triangle)

Angle K + 110 = 180

Angle K = 180 - 110 = 70°

Answer:

angle k = [tex]\boxed {70}^\circ[/tex]

Step-by-step explanation:

The angles in a triangle add up to 180°.

∴ ∠k + 75° + 35° = 180°

⇒ ∠k + 110° = 180°

⇒ ∠k = 180° - 110°

⇒ ∠k = 70°

Equivalent to tan A =[tex]\frac{y}{x}[/tex]

Equivalent to tan A:

A right triangle ABC so that m∠C = 90°

The lengths are:

                           Side a is opposite ∠A,

                           Side b is opposite ∠B,

                           Side c (the hypotenuse) is opposite ∠C

Because cos A = x, therefore

[tex]x =\frac{b}{c}[/tex]  => [tex]b = cx[/tex]                  (1)

Because sin A = y, therefore

[tex]y =\frac{a}{c}[/tex] => [tex]a = cy[/tex]                 (2)

By definition,

tan A = a/b

         [tex]=\frac{cy}{cx}[/tex]

        [tex]=\frac{y}{x}[/tex]

Equivalent to tan A =[tex]\frac{y}{x}[/tex]

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It should be noted that sin(sin^-1x) = sin-1(sinx) is not an identity.

It should be noted that an identity simply means an equation that is always true no matter the values that are substituted.

In this case, should be noted that sin(sin^-1x) = sin-1(sinx) is not an identity. It's simply the inverse of the sine function.

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The weight of the package is 12 ounces

The given parameters are

Total amount = $2.76

Domestic rate = $.23 per ounce

The equation to determine the weight of the package is represented as:

Weight = Total amount/Domestic rate

Substitute the known values in the above equation

Weight = 2.76/.23

Evaluate the quotient

Weight = 12

Hence, the weight of the package is 12 ounces

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By the use of the completing the square method, the maximum height of the path of the water 3 ±√29/4 + 9 m.

What is called distance?

h = -4x² +24x -29

Now we have;

0 = -4x² +24x -29

-4x² +24x -29= 0

x² - 6x = -29/4

(x - 3)² = -29/4 + 3²

(x - 3)² =  -29/4 + 9

x = 3 ±√29/4 + 9 m

Therefore, the maximum height of the path of the water  x = 3 ±√29/4 + 9 m

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According to the total interest gained, the sum deposited by Madan Bahadur was Rs 4,000.

For the principal [tex]P[/tex] and the rate of interest [tex]r\%[/tex] per annum, the total interest after [tex]t[/tex] years is given by the formula: [tex]I=\dfrac{Prt}{100}[/tex].

Given that after 5 years Madan Bahadur got a net interest of Rs 1900 after charging 5% of the total interest as income tax where the rate of interest was 10% p.a.

Suppose the sum deposited was [tex]P[/tex] and the total interest was [tex]I[/tex].

So, the net interest would be [tex]I_n=I-I\times \frac{5}{100}=\frac{19I}{20}[/tex].

Now, given that the net interest [tex]I_n=1900[/tex]. So, we must have [tex]\frac{19I}{20}=1900\\\Longrightarrow I=\frac{1900\times 20}{19}\\\therefore I=2000[/tex]    (1)

Thus, the total interest after 5 years is Rs 2,000.

Also, using the above formula for the total interest, the total interest of the sum [tex]P[/tex] at the rate [tex]r=10\%[/tex] p.a. after [tex]t=5[/tex] years would be [tex]I=\frac{Prt}{100}=\frac{P\times 10\times 5}{100}=\frac{P}{2}[/tex].

So, we must have from (1),

[tex]\frac{P}{2}=2000\\\Longrightarrow P=2000\times 2\\\therefore P=4000[/tex]

Therefore, according to the total interest gained, the sum deposited by Madan Bahadur was Rs 4,000.

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