Please answer the one this exercises if you can send photos sent it pls and you have to use quadratic function for these Thanks​

Answer: x is -1/2,0 and y is 0,2

Step-by-step explanation:

Answer:

(4,0) , (0,2)

Step-by-step explanation:

Similar to the previous question.

x-intercept is where the line touches the x-axis , and y = 0.

when y = 0,

8x + 16(0) = 32

8x = 32

x = [tex]\frac{32}{8}[/tex]

x = 4

Therefore the coordinates of the x-intercept is (4,0)

y-intercept is where the line touches the y-axis, and x = 0.

when x = 0,

8(0) + 16y = 32

16y = 32

y = [tex]\frac{32}{16}[/tex]

y = 2

Therefore the coordinates of the y-intercept is (0,2)

The steps to solve a quadratic formula include:

In algebra, a quadratic equation is a equation that can be rearranged in standard form as where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0.

A quadratic equation is an equation that could be written as ax² + bx + c = 0

when a 0. The factoring method is illustrated:

To solve a quadratic equation by factoring, put all terms on one side of the equal sign, leaving zero on the other side.

Set each factor equal to zero.

Solve each of these equations.

Check by inserting your answer in the original equation.

Example 1

Solve x² – 6 x = 16.

Following the steps,

x² – 6 x = 16 becomes x² – 8x + 2x – 16 = 0

Factor.

( x – 8)( x + 2) = 0

Therefore, x = 8 and -2.

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Answer:

30°

30 is the value of x that’s makes AB//CD.

Step-by-step explanation:

the angles of measures 2x + 40 and 3x + 10 are two Alternate interior angles

If these two angle were congruent then the lines AB and CD

would be parallel .

2x + 40 = 3x + 10

⇔ 40 - 10 = 3x - 2x

⇔ 30 = x

⇔ x = 30

Between 2 seconds and 2.5 seconds, the rock hits the ground after it is dropped.

What are distance and time?

As you can see at 2s the height of the rock is 0.4m and at 2.5s it is -10.6m, which tells us that in between these values the height would have been 0m, as the movement of the rock is uniform in direction and hence, the rock will hit the ground between 2s and 2.5s.

The rock hits the ground between 2 seconds and 2.5 seconds after it is dropped.

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The domain of the given function [tex]f(x) = \sqrt{(4x + 9)} + 2[/tex].

So long as x ≥ -9/4, the function f(x) will be defined.

Given: "f(x) = Startroot 4 x + 9 Endroot + 2" should be written as

[tex]f(x) = \sqrt{(4x + 9)} + 2[/tex].

Note that [tex]$\sqrt{(4x + 9)}[/tex] exists a variation of the basic function [tex]y = \sqrt{x}[/tex], whose domain exists [0, ∞ ).

The domain of [tex]$f(x) = \sqrt{(4x + 9) }+ 2[/tex] exists seen by taking the "argument" 4x + 9 of [tex]\sqrt{(4x + 9)}[/tex]and setting it equivalent to zero:

4x + 9 ≥ 0

simplifying the equation, we get

4x ≥ -9

x ≥ -9/4

This exists the domain of the given function [tex]f(x) = \sqrt{(4x + 9)} + 2[/tex].

So long as x ≥ -9/4, the function f(x) will be defined.

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Answer:

B

Step-by-step explanation:

I tink old up

no its BBBBB

Answer: 46.7 in/3

Step-by-step explanation:

10 meters away is the bus from the bus stop after 10 seconds.

A series of integers in which the difference between succeeding words is always the same is referred to as an arithmetic sequence or arithmetic progression. For instance, the difference in the arithmetic series 1, 5, 9, 13, 17,... is always 4. The common difference is what we refer to as.

A bus moves away from bus stop in a straight line. After 1 second it was 10 meters away, after 2 seconds it was 20 meters and after 3 seconds it was 30 meters away.

In this way bus forms an arithmetic sequence

10, 20, 30, 40........up to 10 terms.

nth term of an arithmetic sequence is represented by

[tex]T_{n}[/tex] = a+ (n - 1)d

Here a = 10 and d = common difference = 10

Therefore 10th term = 10 + (10-1)10

                                  = 10 + 90

                                  = 100

Therefore after 10 seconds bus will be 10 meters away from the bus stop.

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Answer:

a

Step-by-step explanation:

given y varies directly as x² and inversely as z³ then the equation relating them is

y = [tex]\frac{kx^2}{z^3}[/tex] ← k is the constant of variation

to find k use the condition y = 12 when x = 4 and z = 2 , then

12 = [tex]\frac{k(4)^2}{2^3}[/tex] = [tex]\frac{16k}{8}[/tex] ( multiply both sides by 8 )

96 = 16k ( divide both sides by 16 )

6 = k

y = [tex]\frac{6x^2}{z^3}[/tex] ← equation of variation

when y = 1.728 and z = 5 , then

1.728 = [tex]\frac{6x^2}{5^3}[/tex] = [tex]\frac{6x^2}{125}[/tex] ( multiply both sides by 125 )

216 = 6x² ( divide both sides by 6 )

36 = x² ( take square root of both sides )

[tex]\sqrt{36}[/tex] = x , that is

x = 6

The equation s for each scenario in which case, the former, Prize A is represented as a linear equation while the latter, Prize B is represented as an exponential function are;

It follows from the task content that in scenario which pertains to Prize A, the initial price which represents the y-intercept is; $5,000 while the slope which represents the increase per month is; $100.

Consequently, we have; Prize A = $5,000 + $100m

While, for Prize B, in which case the function via exponential, we have;

Prize B = $2,000(1.1)^m.

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Answer:

[tex]\textsf{a)} \quad v=12t^2-6t-18[/tex]

[tex]\textsf{b)} \quad a=24t-6[/tex]

c)   1.5 s

d)  -19.25 m

e)   24.5 m

Step-by-step explanation:

Displacement

[tex]x=4t^3-3t^2-18t+1[/tex]

(where t ≥ 0 and x is in meters)

To find the equation for velocity, differentiate the equation for displacement:

[tex]\implies v=\dfrac{\text{d}x}{\text{d}t}=12t^2-6t-18[/tex]

(where t ≥ 0 and v is in meters per second)

To find the equation for acceleration, differentiate the equation for velocity:

[tex]\implies a=\dfrac{\text{d}v}{\text{d}t}=24t-6[/tex]

(where t ≥ 0 and a is in meters per second squared)

The particle comes to rest when its velocity is zero:

[tex]\begin{aligned}v & = 0\\\implies 12t^2-6t-18 & = 0\\6(2t^2-t-3) & = 0\\2t^2-t-3 & = 0\\2t^2-3t+2t-3 & = 0\\t(2t-3)+1(2t-3) & = 0\\(t+1)(2t-3) & = 0\\\implies t & =-1, \dfrac{3}{2}\end{aligned}[/tex]

As t ≥ 0, the particle comes to rest at 1.5 s.

Substitute the found value of t from part (c) into the equation for displacement to find where the particle comes to rest:

[tex]\implies 4(1.5)^3-3(1.5)^2-18(1.5)+1=-19.25\: \sf m[/tex]

We have determined that the particle is at rest at 1.5 s.

Therefore, to find how far the particle traveled in the first 2 seconds, we need to divide the journey into two parts: before and after it was at rest.  

The first leg of the journey is the first 1.5 s and the second leg of the journey is the next 0.5 s.

At the beginning of the journey, t = 0 s.

[tex]\textsf{when }t=0: \quad x=4(0)^3-3(0)^2-18(0)+1=1[/tex]

Therefore, when t = 0, x = 1

When t = 1.5 s, x = -19.25 m (from part (d)).

⇒ Total distance traveled in first 1.5 s = 1 + 19.25 = 20.25 m

When t = 2, x = -15 m.

Therefore, the particle has traveled 19.25 - 15 = 4.25 m in the last 0.5 s of its journey.

Total distance traveled:

1 + 19.25 + 4.25 = 24.5 m

Refer to the attached diagram

When the particle is at rest, it changes direction.  If we model its journey using the x-axis, for the first leg of its journey (0 - 1.5 s) it travels in the negative direction (to the left). At 1.5 s it stops, then changes direction and travels in the positive direction (to the right), arriving at -15 at 2 seconds.

Comparing it to a system of equations, the expression is represented as follows:

18g + 5.

A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

For this problem, we consider g as the variable. Then:

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The answer is A = m².

We know the general formula for finding the area of a square is :

Area = side²

Now, we are given the side is equal to m.

Hence, the expression which can be used to find the area is :

A = m²

Answer:

See attached image.

Step-by-step explanation:

If x=y, the (x,y) = (x,x) or (y,y).  All points have the same x and y values.

The inclination to the nearest tenth of a degree exists 0.24146

The given scenario includes a right-angled triangle where the length of the ramp exists hypotenuse and the rise of ramp exists the perpendicular.

Given: H = 4.6 m and P = 1.1 m

We have to use the trigonometric ratios to find the angle. The ratio that has to be used should involve both perpendicular and hypotenuse

Let x be the angle then

sin x = P/H

sin x = 1.1/4.6

sin x = 0.23913

[tex]$$sin^{-1} x[/tex] = 0.24146

The inclination to the nearest tenth of a degree exists 0.24146

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Using the binomial distribution, there is a 0.9983 = 99.83% probability that at most eleven of the thirteen babies are girls.

The formula is:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

The parameters are:

For this problem, the values of the parameters are:

p = 0.5, n = 13

The probability that at most eleven of the thirteen babies are girls is:

[tex]P(X \leq 11) = 1 - P(X > 11)[/tex]

In which

P(X > 11) = P(X = 12) + P(X = 13)

Then:

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 12) = C_{13,12}.(0.5)^{12}.(0.5)^{1} = 0.0016[/tex]

[tex]P(X = 13) = C_{13,13}.(0.5)^{13}.(0.5)^{0} = 0.0001[/tex]

So:

P(X > 11) = P(X = 12) + P(X = 13) = 0.0016 + 0.0001 = 0.0017

[tex]P(X \leq 11) = 1 - P(X > 11) = 1 - 0.0017 = 0.9983[/tex]

0.9983 = 99.83% probability that at most eleven of the thirteen babies are girls.

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Answer:

11÷7=[tex]\frac{11}{7}[/tex]

v= 11/7

the value of b is = 10.

The equation of a hyperbola is x^2/24^2 - y^2/ (10)^2= 1.

The geometric characteristics of a hyperbola or the equations for which it is the solution set characterize it as a particular kind of smooth curve that lies in a plane. Mirror reflections of each other that resemble two infinite bows make up a hyperbola's two connected components or branches.

The general formula for hyperbola = (x - h)²/a²- (y - k)²/(b)² = 1

x²/24 - y²/(b)² = 1

(x - 0)²/24 - (y -0)²/(b)² = 1

a=24,h=0 and k=0

Now equation of the directrix

x=a²/c...(1)

and we know x=576/26...(2)

Therefore from 1 and 2 we get

24²/c=576/26.

isolate the c so we get,

C=26

C= center of focii

c = √(a² + b³)

c² = a² + b²

b = c² - a²

b = 10

So we get the value of b is 10.

Therefore the equation of a hyperbola is x²/24² - y²/ (10)² =  1.

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Answer:

x = 36

Step-by-step explanation:

3x and 2x are a linear pair and sum to 180° , that is

3x + 2x = 180

5x = 180 ( divide both sides by 5 )

x = 36

The solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

According to the given question.

We have a differential equation

[tex]y^{"} + 4y^{'} + 4y = 2x^{3}[/tex]

The above differerntial equation acn be written as

[tex](D^{2} +4D+ 4)= 2x^{3}[/tex]

Now, the auxillary equation for the above differential equation is given by

[tex]m^{2} + 4m + 4 = 0[/tex]

[tex]\implies m^{2} + 2m + 2m + 4 = 0[/tex]

⇒ m (m + 2) + 2(m + 2) = 0

⇒ m(m + 2)(m + 2) = 0

Therefore,

[tex]C.F = (C_{1} + C_{2}x)e^{-2x}[/tex]

Now,

[tex]PI = \frac{1}{D^{2} +4D+4} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4(1+\frac{D^{2}+4D }{4} )} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [1+(\frac{D^{2}+4D }{4} )]^{-1} 2x^{3}[/tex]

[tex]\implies PI = \frac{1}{4} [ 1 - (\frac{D^{2} +4D}{4} )+(\frac{D^{2} +4D}{4} )^{2} -(\frac{D^{2}+4D }{4}) ^{3} ...]2x^{3}[/tex]

[tex]\implies PI =\frac{1}{2} [ x^{3} -\frac{1}{4} (6x)-3x^{2} +3x^{2} -6][/tex]

[tex]\implies PI = \frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Therefore, the solution of the differential equation will be

y = CI + PI

[tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex]

Hence, the solution of the differential equation is [tex]y = (C_{1} + C_{2} x)e^{-2x} +\frac{1}{2} [x^{3} -\frac{3}{2} x-6][/tex] .

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Answer:

Step-by-step explanation:

Use periodic compound formula:

[tex]F=P\cfrac{(1+r/n)^{nt}-1}{r/n}[/tex]

Substitute the values and calculate:

[tex]F=200\cfrac{(1+0.032/12)^{12*2}-1}{0.032/12} =4950.12[/tex]     rounded

Step-by-step explanation:

Use periodic compound formula:

[tex] \sf \: F=P\cfrac{(1+ \frac{r}{n})^{nt}-1}{\frac{r}{n}}[/tex]

Substitute the values and calculate:

[tex]\sf \: F=200\cfrac{(1+ \frac{0.032}{12})^{12 \times 2}-1}{\frac{0.032}{12}}[/tex]

[tex]\sf \: F=200\cfrac{( \frac{ 12 + 0.032}{12})^{24}-1}{\frac{0.032}{12}}[/tex]

[tex]\sf \: F=200\cfrac{( {11.002 }{})^{24}-1}{0.002} [/tex]

[tex]\sf \: F=200 \times 24.7506[/tex]

[tex]\sf \: F=4950.12rounded[/tex]

Answer:

D. 195

Step-by-step explanation:

The value of A is A(1,-15) and the distance between AB is 8 so these are the correct answers.

According to the statement

We have given that the

AB = 8 and B lies at −7, and we have to find the position of the A on the number line.

Let a be a point on number line and b be the second point on number line.

And The distance between two numbers on number line is calculate by subtracting the smaller number from larger number.

Here

AB = 8

B = -7

We have to look at the options one by one.

For A = -1

As A>B

AB = -1-(7) = -1+7 = 6

For A = 1

As A>B so

AB = 1-(-7) = 8

So with A = 1 , AB = 8

This is one of the right answers.

For A=15

As A>B

AB = 15-(-7) = 22

For A=-15

As B>A

AB = -7 -(-15) = 8

As with A=1 and A=-15, the distance between AB is 8 so these are the correct answers.

So, The value of A is A(1,-15) and the distance between AB is 8 so these are the correct answers.

Disclaimer: This question was incomplete. Please find the full content below.

Question:

Suppose A and B are points on the number line. If AB = 8 and B lies at −7, where could A be located?

Select ALL that apply.

A −1

B 1

C 15

D −15

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Step-by-step explanation:

every single person will do 1/4 of each job.

divide into 2 group and each person will do half of one job and half of another.

The local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

For given question,

We have been given a function f(x) = 6 + 9x² - 6x³

We need to find the local maximum and local minimum of the function  f(x)

First we find the first derivative of the function.

⇒ f'(x) = 0 + 18x - 18x²

⇒ f'(x) = - 18x² + 18x

Putting the first derivative of the function equal to zero, we get

⇒ f'(x) = 0

⇒ - 18x² + 18x = 0

⇒ 18(-x² + x) = 0

⇒ x (-x + 1) = 0

⇒ x = 0    or    -x + 1 = 0

⇒ x = 0     or    x = 1

Now we find the second derivative of the function.

⇒ f"(x) = - 36x + 18

At x = 0 the value of second derivative of function f(x),

⇒ f"(0) = - 36(0) + 18

⇒ f"(0) = 0 + 18

⇒ f"(0) = 18

Here, at x=0, f"(x) > 0

This means, the function f(x) has the local minimum value at x = 0,  which is given by

⇒ f(0) = 6 + 9(0)² - 6(0)³

⇒ f(0) = 6 + 0 - 0

⇒ f(0) = 6

At x = 1 the value of second derivative of function f(x),

⇒ f"(1) = - 36(1) + 18

⇒ f"(1) = - 18

Here, at x = 1, f"(x) < 0

This means, the function f(x) has the local maximum value at x = 1,  which is given by

⇒ f(1) = 6 + 9(1)² - 6(1)³

⇒ f(1) = 6 + 9 - 6

⇒ f(1) = 9

So, the function f(x) = 6 + 9x² - 6x³ has local minimum at x = 0 and local maximum at x = 1.

Therefore, the local minimum value of the function f(x) = 6 + 9x² - 6x³ is 6 and the local maximum value of the function f(x) = 6 + 9x² - 6x³ is 9.

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Answer: Yes, the correlation coefficient is statistically significant

The correlation coefficient describes how one variable actions when it comes to every other. A high-quality correlation shows that the 2 circulate within the identical path, with a +1.zero correlation once they move in tandem. A poor correlation coefficient tells you that they as a substitute flow in opposite directions. A correlation of 0 shows no correlation in any respect.

where ρ corresponds to the population correlation.

The sample size is n = 17,

so then the number of degrees of freedom is df = n-2 = 17-2 = 15

The corresponding critical correlation value re for a significance level of a 0.05, for a two-tailed test is た= 0.482

Observe that in this case,

              the null hypothesis H0 : ρ-0 is rejected

                             if |r> re 0.482.

Based on the sample correlation provided, we have that lrl = 0.595 >=0.482,                            

which is concluded that the null hypothesis is rejected.

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In this situation, the cases are the bags or each piece of baggage.

In statistics, the word case is used to refer to the individual from which data is collected. For example, in a study about students' performance, the cases are each of the students.

Considering the administrators will need to weight and record each bag, the cases are each piece of baggage.

Note: This question is incomplete; here are the missing options:

A. Number of bags weighing more than 75 pounds.

B. Average weight of the bags.

C. Each piece of baggage.

D. The airport administrators.

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Answer:

a) x = 3

b) x = 10

c) x = 28

Explanation:

a)

[tex]\sf 6x + 3 = 21[/tex]

collect term

[tex]\sf 6x = 21-3[/tex]

simplify

[tex]\sf 6x = 18[/tex]

divide both both sides by 6

[tex]\sf x = 3[/tex]

b)

[tex]\sf 15(x - 5) = 75[/tex]

distribute

[tex]\sf 15x - 75 = 75[/tex]

collect terms

[tex]\sf 15x = 75 + 75[/tex]

combine

[tex]\sf 15x = 150[/tex]

divide both sides by 15

[tex]\sf x = 10[/tex]

c)

[tex]\sf \frac{3}{4} x +5 = 26[/tex]

collect like terms

[tex]\sf \frac{3}{4} x = 26-5[/tex]

combine

[tex]\sf \frac{3}{4} x = 21[/tex]

cross multiply

[tex]\sf x = \frac{ 21 (4)}{3}[/tex]

simplify

[tex]\sf x =28[/tex]

Answer:

c^2 = a^2 + b^226^2 = 10^2 + b^2 676 = 100 + b^2 576 = b^2 shortest side length = 24 units

Step-by-step explanation:

Answer:

4. 4.125 inches by 3.5 inches.

5. 8.4 hours.

Step-by-step explanation:

4. multiply 16.5 and 14 by .25 because each foot in the house is .25 of an inch on the paper.

5. 252 miles / 9 hours = 28 mph. Add 2 so you get 30 mph. 252 miles / 30 mph = 8.4 hours.

Answer:

50.24 ft^2.

Step-by-step explanation:

Area = pi r^2

Here the radius r = 4

Therefore the area of the circle

= pi * 4^2

= 16 pi

= 16 *3.14

= 50.24 ft^2.