Which equation is the inverse of y = x2 – 36?

Answer:

[tex]\left \{ 2,\frac{5}{4} \right \}[/tex]

Step-by-step explanation:

We know that for equation of type [tex](x-a)(x-b)=0[/tex], solutions are [tex]x=a\,,\,x=b[/tex] as both points x = a and x = b satisfy the equation (x-a)(x-b)=0

Given : equation (2x−4)(4x−5)=0

To find  : Solution set of this equation .

Solution :

On dividing this equation by 2 and 4, we get

[tex]\left ( \frac{2x-4}{2} \right )\left ( \frac{4x-5}{4} \right )=0\\\left ( x-2 \right )\left ( x-\frac{5}{4} \right )=0[/tex]

On comparing equation [tex]\left ( x-2 \right )\left ( x-\frac{5}{4} \right )=0[/tex] with [tex]\left ( x-a \right )\left ( x-b \right )=0[/tex], we get [tex]a=2\,,\,b=\frac{5}{4}[/tex]

Therefore, solution set is [tex]\left \{ 2,\frac{5}{4} \right \}[/tex]

Final answer:

The solution set to the equation (2x−4)(4x−5) = 0 is {2, 5/4}.

Explanation:

The equation (2x−4)(4x−5) = 0 represents a quadratic equation. To solve this equation, we can set each factor equal to zero and solve for x. So we have:

2x - 4 = 0, which gives x = 2

4x - 5 = 0, which gives x = 5/4

Therefore, the solution set to the equation is {2, 5/4}.

Answer:

Step-by-step explanation: the answer is b for apex

Sean used 3/4 cups of sugar to make 12 brownies. By dividing 3/4 by 12, we find that there is 1/16 cup of sugar in each brownie.

The subject of this problem is mathematics, specifically, it's about division in the context of fractions. In this scenario, Sean used 3/4 cups of sugar to make a dozen (12) brownies. To determine the amount of sugar in each brownie, we can divide the total amount of sugar by the total number of brownies.

So, 3/4 divided by 12 is 1/16. This means that there is 1/16 cup of sugar in each brownie.

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Answer:  t represents the the number of seconds after rocket is released.

Step-by-step explanation:

Given: The height of a rocket a given number of seconds after it is released is modeled by [tex]h (t) = 6t^2 + 32t + 10[/tex].

Here h (height) is the dependent variable , which depends on the number of seconds after rocket is released (independent variable).

Since the independent variable in the function is t, then t must represents the the number of seconds after rocket is released.

The variable t represents the number of seconds that have passed since the rocket was released.

Here we know that the height of a rocket a given number of seconds after it is released is modeled by:

h(t) = 6*t^2 + 32*t + 10

So this is a function that relates height with time in seconds, we know that the function models the height, so we must have that:

[h(t)] is equivalent to height.

This means that the other variable, t, must be related to time in seconds.

Then we can conclude that the variable t represents the number of seconds after the rocket has been released.

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The equation for profit is income – cost:

Profit = Income – Cost

Let us say that x is the number of sold amount

Profit = 9 x – (25,000 + 4 x)

Profit = 5 x – 25,000

Breakeven point occurs when Profit = 0, hence:

5 x = 25,000

x = 5,000

The breakeven is when 5,000 people uses the income tax app

Answer:

10,13 13, 15,11

Step-by-step explanation:

THANK YOU PREVIOUS PERSON YOU SAVED MY WHOLE MATH GRADE! I OWE YOU  YOU MADE IT SO I CAN PASS 9TH GRADE.

Answer:

1) Average velocity = 10/3 m/s

2) Instantaneous velocity = -2 m/s
   Speed = 2 m/s to the left

3) (0, 3) ∪ (6, 8]

4) Going faster: (3, 4.5) ∪ (6, 8]
   Slowing down: (0, 3) ∪ (4.5, 6)

5) Total distance = 35.67 m (nearest hundredth)

Step-by-step explanation:

The relationships between position (displacement), velocity and acceleration are:

[tex]\boxed{\boxed{\begin{array}{c}\textbf{POSITION (s)}\\\\\text{Differentiate} \downarrow\qquad\uparrow\text{Integrate}\\\\\textbf{VELOCITY (v)}\\\\\text{Differentiate}\downarrow\qquad\uparrow \text{Integrate}\\\\\textbf{ACCELERATION (a)}\end{array}}}[/tex]

Given a particle is moving with velocity v(t) = t² - 9t + 18, to find its position s(t) we can integrate v(t):

[tex]\begin{aligned}\displaystyle s(t)=\int v(t)\;\text{d}t&=\int(t^2-9t+18)\;\text{d}t\\\\&=\dfrac{t^{2+1}}{2+1}-\dfrac{9t^{1+1}}{1+1}+18t+C\\\\&=\dfrac{t^{3}}{3}-\dfrac{9t^{2}}{2}+18t+C\end{aligned}[/tex]

As s(0) = 1, then:

[tex]\begin{aligned}s(0)=\dfrac{(0)^{3}}{3}-\dfrac{9(0)^{2}}{2}+18(0)+C&=1\\0-0+0+C&=1\\C&=1\end{aligned}[/tex]

Therefore, the position function s(t) is:

[tex]\large\boxed{s(t)=\dfrac{t^3}{3}-\dfrac{9t^2}{2}+18t+1}[/tex]

Given a particle is moving with velocity v(t) = t² - 9t + 18, to find its acceleration a(t) we can differentiate v(t):

[tex]\begin{aligned}a(t)=\dfrac{\text{d}}{\text{d}t}[v(t)]&=2\cdot t^{2-1}-1\cdot9t^{1-1}+0\\&=2t-9\end{aligned}[/tex]

Therefore, the acceleration function a(t) is:

[tex]\large\boxed{a(t)=2t-9}[/tex]

[tex]\hrulefill[/tex]

To find the average velocity over the interval [0, 8], use the formula:

[tex]\textsf{Average Velocity}=\dfrac{s(t_2)-s(t_1)}{t_2-t_1}[/tex]

In this case:

Calculate the position at t₁ and t₂ by substituting t = 0 and t = 8 into s(t):

[tex]s(0)=\dfrac{(0)^3}{3}-\dfrac{9(0)^2}{2}+18(0)+1}=1[/tex]

[tex]s(8)=\dfrac{(8)^3}{3}-\dfrac{9(8)^2}{2}+18(8)+1}=\dfrac{83}{3}[/tex]

Therefore:

[tex]\textsf{Average Velocity}=\dfrac{s(8)-s(0)}{8-0}=\dfrac{\frac{83}{3}-1}{8}=\dfrac{10}{3}\; \sf m/s[/tex]

Therefore, the average velocity is 10/3 m/s.

[tex]\hrulefill[/tex]

To find the instantaneous velocity at t = 5 seconds, substitute t = 5 into v(t):

[tex]\begin{aligned}v(5)&=(5)^2-9(5)+18\\&=25-45+18\\&=-2\end{aligned}[/tex]

So, the instantaneous velocity at t = 5 seconds is -2 m/s.

Speed is a scalar quantity that measures how fast an object is moving regardless of its direction. Therefore, speed is the magnitude of velocity:

[tex]\textsf{Speed}=|v(5)|=|-2|=2\;\sf m/s[/tex]

Therefore, the speed at t = 5 is 2 m/s to the left.

[tex]\hrulefill[/tex]

The particle changes direction when v(t) = 0.

[tex]\begin{aligned}v(t)&=0\\\implies t^2-9t+18&=0\\t^2-6t-3t+18&=0\\t(t-6)-3(t-6)&=0\\(t-3)(t-6)&=0\\\\t-3&=0\implies t=3\\t-6&=0\implies t=6\end{aligned}[/tex]

Therefore, the particle changes direction at t = 3 and t = 6.

We know that the position of the particle at t = 0 is 1 meter right of zero. Therefore:

Therefore, the time intervals between 0 ≤ t ≤ 8 when the particle is moving right is:

[tex]\hrulefill[/tex]

When a(t) > 0:

[tex]\begin{aligned}a(t)& > 0\\2t-9& > 0\\2t& > 9\\t& > \dfrac{9}{2}\\t& > 4.5\; \sf s\end{aligned}[/tex]

When a(t) < 0:

[tex]\begin{aligned}a(t)& < 0\\2t-9& < 0\\2t& < 9\\t& < \dfrac{9}{2}\\t& < 4.5\; \sf s\end{aligned}[/tex]

Therefore:

(Refer to the attachment).

If velocity and acceleration have the same sign, it means the object is speeding up.

If velocity and acceleration have opposite signs, it means the object is slowing down.

Therefore, the time intervals when the particle is going faster and slowing down are:

[tex]\hrulefill[/tex]

To find the total distance the particle has traveled between 0 and 8 seconds, we need to consider the distance traveled between the intervals when it changes direction.

To do this, find the position of the particle at t = 0, t = 3, t = 6 and t = 8.

[tex]s(0)=\dfrac{(0)^3}{3}-\dfrac{9(0)^2}{2}+18(0)+1=1[/tex]

[tex]s(3)=\dfrac{(3)^3}{3}-\dfrac{9(3)^2}{2}+18(3)+1=23.5[/tex]

[tex]s(6)=\dfrac{(6)^3}{3}-\dfrac{9(6)^2}{2}+18(6)+1=19[/tex]

[tex]s(8)=\dfrac{(8)^3}{3}-\dfrac{9(8)^2}{2}+18(8)+1=\dfrac{83}{3}\approx27.67[/tex]

Therefore, in the interval 0 ≤ t < 3, the particle travels:

[tex]|s(3)-s(0)|=|23.5-1|=22.5\; \sf meters\;(to\;the\;right)[/tex]

In the interval  3 < t < 6, it travels:

[tex]|s(6)-s(3)|=|19-23.5|=4.5\; \sf meters\;(to\;the\;left)[/tex]

In the interval 6 < t ≤ 8, it travels:

[tex]|s(8)-s(6)|=|27.67-19|=8.67\; \sf meters\;(to\;the\;right)[/tex]

So the total distance the particle has traveled between 0 and 8 seconds is:

[tex]\textsf{Total distance}=22.5+4.5+8.67=35.67\; \sf meters[/tex]

Answer:

infinite number of values in domain and range

Step-by-step explanation:

need points for finals

For a function of the form f(x) = m, the domain is all real numbers and the range is the single real number m. Thus, there are no values that are in both the domain and the range.

In a function of the form f(x) = m, where m is a real number greater than 0, the domain encompasses all real numbers since there are no restrictions on any values of x that would make the function undefined. Since the function gives the same result (m) regardless of what x is, f(x) remains constant. Thus, the range only contains one real number, which is m.

The correct option from the ones given, therefore, is: There are no values that are in the domain and the range. This is because the domain contains all real numbers, while the range contains only one number (m), and these sets do not overlap.

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Answer:

26 (I got 100%)

Step-by-step explanation:

39 - 13

Phil payment - Francesca payment = 26

Answer:

The equation that models the total fee for using this cab company

[tex]y=\$2.50+\$0.40\times x[/tex]

Step-by-step explanation:

Initial rate charged by company = $2.50

Amount charged for an each mile = $0.40

Let the miles cover during a ride = x

Total cost of ride can given as = y

The equation that models the total fee for using this cab company

[tex]y=\$2.50+\$0.40\times x[/tex]

The graphical interpretation equation in an image.

Answer:

-162

Step-by-step explanation:

Integration by substitution is a technique in calculus where a change of variables is made to simplify the integral by expressing it in terms of a new variable, making the integration easier to manage.

To evaluate the following integral, use the method of substitution:

[tex]\displaystyle \int_{-1}^0 (2x^4 + 8x)^3 (4x^3 + 4) \;\text{d}x[/tex]

Let u = 2x⁴ + 8x.

Differentiate u with respect to x using the power rule for differentiation by multiplying each term by its exponent and then subtracting 1 from the exponent:

[tex]\dfrac{\text{d}u}{\text{d}x}=4 \cdot 2x^{4-1}+1\cdot 8 x^{1-1}[/tex]

[tex]\dfrac{\text{d}u}{\text{d}x}=8x^{3}+8[/tex]

Rewrite it so that dx is on its own:

[tex]\text{d}u=8x^{3}+8\;\text{d}x[/tex]

[tex]\text{d}x=\dfrac{1}{8x^{3}+8}\;\text{d}u[/tex]

Change the limits of integration from x to u:

[tex]\begin{aligned}x=-1 \implies u&=2(-1)^4 + 8(-1)\\&=2(1)-8\\&=2-8\\&=-6\end{aligned}[/tex]

[tex]\begin{aligned}x=0 \implies u&=2(0)^4 + 8(0)\\&=2(0)-0\\&=0-0\\&=0\end{aligned}[/tex]

Rewrite the integral in terms of u:

[tex]\begin{aligned}\displaystyle \int_{-1}^0 (2x^4 + 8x)^3 (4x^3 + 4) \;\text{d}x&=\int_{-6}^0 u^3 (4x^3 + 4) \cdot \dfrac{1}{8x^{3}+8}\;\text{d}u\\\\&=\int_{-6}^0 u^3 (4x^3 + 4) \cdot \dfrac{1}{2(4x^{3}+4)}\;\text{d}u\\\\&=\int_{-6}^0\dfrac{u^3}{2} \;\text{d}u\end{aligned}[/tex]

Integrate with respect to u using the power rule for integration by adding 1 to the exponent of each term and then dividing by the new exponent:

[tex]\begin{aligned}\displaystyle \int_{-6}^0\dfrac{u^3}{2} &=\left[\dfrac{u^{3+1}}{2 \cdot (3+1)}\right]^0_{-6}\\\\&=\left[\dfrac{u^{4}}{8}\right]^0_{-6}\\\\&=\dfrac{(0)^4}{8}-\dfrac{(-6)^4}{8}\\\\&=0-\dfrac{1296}{8}\\\\&=-162\end{aligned}[/tex]

Therefore, the value of the given integral is -162.

we know that

Vertical angles are a pair of opposite and congruent angles formed by intersecting lines

In this problem

m∠JNM=m∠KNL -------> by vertical angles

so

[tex](4x+6)\°=(7x-21)\°[/tex]

Solve for x

[tex]7x-4x=6+21\\3x=27\\x=9\°[/tex]

Find the value of m∠JNM

m∠JNM=[tex](4x+6)\°[/tex]

substitute the value of x

m∠JNM=[tex](4*9+6)\°[/tex]

m∠JNM=[tex]42\°[/tex]

therefore

the answer is

m∠JNM=[tex]42\°[/tex]

Answer:

D

Step-by-step explanation:

The only way I can think of to solve this problem is to assume normal distribution.

Since the total weight excess 5511, hence the weight per passenger is at least:

x = 5511 / 27 = 204.1 pounds

Solve for the z score:

z = (x – u) / s

z = (204.1 – 189.7) / 41.4

z = 0.35

From the standard probability tables, the P value using right tailed test is:

P = 0.3632

Given fx (x) = λe^λx

Fx (x) = 1 – e^-λx      x…0

To find distribution of Min (X1,….Xn)

By applying the equation

fx1 (x) = [n! / (n – j)! (j – 1)!][F(x)]^j-1[1-F(x)]^n-j f(x)

For minimum j = 1

[Min (X1,…Xn)] = [n!/(n-1)!0!][F(x)]^0[1-(1-e^-λx)]^n-1λe^-λx

= ne^[(n-1) λx] λe^(λx)

= nλe^(-λx[1+n-1])

= nλe^(-nλx)

The distribution of min(x1, . . . , xn) for independent exponential random variables with common parameter λ is given by the exponential distribution with parameter λ*n.

The distribution of min(x1, . . . , xn) for independent exponential random variables with common parameter λ is given by the exponential distribution with parameter λ*n.

To determine this, we can use the fact that the minimum of a set of random variables is less than or equal to a given value if and only if each of the individual random variables is less than or equal to that value. So, we can find the cumulative distribution function (CDF) of the minimum by raising the CDF of each individual exponential random variable to the power of the number of variables in the minimum. The resulting distribution is an exponential distribution with parameter λ*n.

For anyone who still needs it, the answer is 1:3.1264.

Given:- L (2, 3), M (1, 2), and N make up the triangle known as LMN (4, 4). The image L'M'N' is created by first transforming the triangle according to the rule

To Locate:

Solution:-

Rule for Points = (x + 2) on the x-axis.

Thus, the new x-points axis's are as follows:

L' = 2 + 5 = 7

M' = 1 + 5 = 6

N' = 4 + 5 = 9 .

and, Rule for y - axis Points = (y + 2) .

Therefore, New y - axis points of new ∆ are :-

L' = 3 + 2 = 5

M' = 2 + 2 = 4

N' = 4 + 2 = 6 .

Hence, New coordinates of L', M', and N' :-

L' = (7 , 5)

M' = (6, 4)

N' = (9, 6)

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show that the point of trisection for the line connecting the points (-5,12) and (-1,12) is the midpoint of the line connecting the...

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Answer:

[tex]x=1[/tex]

Step-by-step explanation:

We have been given an equation [tex]\sqrt{x+3}+4=6[/tex]. We are asked to find the solution of our given equation.

[tex]\sqrt{x+3}+4-4=6-4[/tex]

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

Now, we will square both sides of our given equation.

[tex]x+3=2^2[/tex]

[tex]x+3=4[/tex]

[tex]x+3-3=4-3[/tex]

[tex]x=1[/tex]

To see whether [tex]x=1[/tex] is an extraneous solution or not, we will substitute [tex]x=1[/tex] in our given equation as:

[tex]\sqrt{1+3}+4=6[/tex]

[tex]\sqrt{4}+4=6[/tex]

[tex]2+4=6[/tex]

[tex]6=6[/tex]

Since both sides of our given equation are equal, therefore, [tex]x=1[/tex] is a solution for our given equation.

2 plates can be glazed and 0.50 glaze will be left over.

The unitary technique involves first determining the value of a single unit, followed by the value of the necessary number of units.

for example, Let's say Ram spends 36 Rs. for a dozen (12) bananas.

12 bananas will set you back 36 Rs. 1 banana costs 36 x 12 = 3 Rupees.

As a result, one banana costs three rupees. Let's say we need to calculate the price of 15 bananas.

This may be done as follows: 15 bananas cost 3 rupees each; 15 units cost 45 rupees.

Given:

Total bowls= 5

fraction of pint to glaze bowl= 7/8

So, pint to glaze 5 bowls

= 5 bowls x  7/8 of a pint

=  4.375 pints

and, left over after glazing the bowls.

= 6 pints - 4.375pints

= 1.625

Now, we know to glaze plate pint used= 9/16

So, number of plates will be  = 1.625 x 16/9

                                                 = 2.88 plates

Thus, we can only glaze 2 plates.

Also,  glaze left over

= 1.625 - 1.125

= 0.50 glaze left over.

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first one is 80%

second one is 50%

1st. is 80%

2nd. is 50%

we know that

The Law of Syllogism says that if the following two statements are true:

(1) If p -------> then q .

(2) If q-------> then r .

Then we can derive a third true statement:

(3) If p--------> then r .

In this problem

(1) If a triangle is an isosceles triangle, then it has two sides of equal length

(2) If a triangle has two sides of equal length, then it has two angles of equal measure

Let

p-------> the statement " an isosceles triangle"

q--------> the statement " has two sides of equal length"

r--------->  the statement "has two angles of equal measure"

Then (1) and (2) can be written

1) If p , then q .

2) If q , then r .

So, by the Law of Syllogism, we can deduce

3) If p , then r

or

If a triangle is an isosceles triangle, has two angles of equal measure

therefore

the answer is

The argument is valid by the law of syllogism.

A skyscraper is a very tall building. The biconditional equivalent to this definition is 'A building is a skyscraper if and only if it is very tall.'

The biconditional equivalent to the definition 'A skyscraper is a very tall building' is:

A building is a skyscraper if and only if it is very tall.

This biconditional statement means that for a building to be classified as a skyscraper, it must satisfy the condition of being very tall. Conversely, if a building is very tall, it can be classified as a skyscraper. This definition establishes a two-way relationship between being a skyscraper and being very tall.

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