Suppose a ball is thrown upward to a height of h0 meters. Each time the ballâ bounces, it rebounds to a fraction r of its previous height. Let hn be the height after the nth bounce and let Sn be the total distance the ball has traveled at the moment of the nth bounce. Complete parts a. and b below. h0 = 9, r = o.4Find the first four terms of the sequence (Se).

Answer:

3.2 units

Step-by-step explanation:

Given that:

WY bisects UV at Y.

[tex]UV=x-7[/tex] and

[tex]YV = 3x-29[/tex],

To find: UV = ?

Solution:

First of all, let us draw the diagram of the given dimensions and bisector line WY of UV.

As UV is bisected i.e. divided in two equal parts at Y by the line WY

Therefore, UY = YV

UV = UY+YV

OR

UV = 2 YV

Now, let us put the given values to solve for [tex]x[/tex]:

[tex]x-7=2 \times (3x-29)\\\Rightarrow x-7=2 \times 3x-2 \times 29\\\Rightarrow x-7=6x-58\\\Rightarrow 6x-x=58-7\\\Rightarrow 5x=51\\\Rightarrow \bold{x =10.2 }[/tex]

Now, we are given that:

[tex]UV=x-7[/tex]

Putting value of [tex]x[/tex] as solved in above step to get the value of UV:

[tex]UV=10.2-7\\\Rightarrow \bold{UV=3.2}[/tex]

So, answer is UV = 3.2 units

Answer:

true

Step-by-step explanation:

Hello, please consider the following.

x is obviously different from 0 and then, dividing by x is legit.

For the first 105 miles the speed is x, so the time spent is

[tex]\dfrac{105}{x}[/tex]

For the second part, the speed is 1.4 x, so the time spent is

[tex]\dfrac{105}{1.4x}=\dfrac{75}{x}[/tex]

In total, the time she spent driving is

[tex]\dfrac{105}{x}+\dfrac{75}{x}\large \boxed{=\dfrac{180}{x}}[/tex]

Thank you.

Answer:

Yes, the table represents quadratic function.

[tex]Y = 3X^2+10[/tex]

Step-by-step explanation:

Given that table of values:

[tex]\begin{center}\begin{tabular}{ c c}X & Y \\ 1 & 13 \\ 2 & 22 \\ 3 & 37 \\ 4 & 58 \\ 5 & 85 \\\end{tabular}\end{center}[/tex]

To find:

Whether the given table represents a quadratic?

Solution:

First of all, let us plot the given values on the coordinate xy plane.

Kindly refer to the attached image for the graph of given values.

The graph seems parabolic in nature which is the graph of a quadratic equation.

Now, let us try to find the equation from the given set of values from hit and trial.

Let Quadratic equation be:

[tex]y=ax^{2} +b[/tex]

If the coefficient a = 1

[tex]f(1) = 13 = 1^2+12[/tex]

[tex]f(2) = 22 = 2^2+18[/tex]

[tex]f(3) = 37= 3^2+28[/tex]

[tex]f(4) = 58 = 4^2+42[/tex]

[tex]f(5) = 85 = 5^2+60[/tex]

value of b is not same in each case.

Now, let us try coefficient a = 3

[tex]f(1) = 13 = 3 \times 1^2+10[/tex]

[tex]f(2) = 22 = 3\times 2^2+10[/tex]

[tex]f(3) = 37= 3\times 3^2+10[/tex]

[tex]f(4) = 58 = 3\times 4^2+10[/tex]

[tex]f(5) = 85 =3\times 5^2+10[/tex]

Value of b = 10

So, we can clearly say that the given table represents a quadratic equation.

and the quadratic equation is:

[tex]Y = 3X^2+10[/tex]

Answer:

  [tex]1-\dfrac{2}{3}\sqrt{3}[/tex]

Step-by-step explanation:

Maybe you want to simplify ...

  [tex](1-\sqrt{3})\dfrac{1}{3+\sqrt{3}}[/tex]

Multiply numerator and denominator by the 'conjugate' of the denominator:

  [tex](1-\sqrt{3})\dfrac{1}{3+\sqrt{3}}\cdot\dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{(1-\sqrt{3})(3-\sqrt{3})}{9-3}=\dfrac{3-4\sqrt{3}+3}{6}\\\\\boxed{1-\dfrac{2}{3}\sqrt{3}}[/tex]

Answer:

soln,

y=37/100×x

or, 100y=37x.....is the answer

An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.

The equation of y is 37% of x is

y = 0.37x

An expression is a way of writing a statement with more than two variables or numbers with operations such as addition, subtraction, multiplication, and division.

Example:

2 + 3x + 4y = 7 is an expression.

2 + 3 - 4 is an expression.

2x4 + 4x = 4 is an expression.

We have,

y is 37% of x.

This can be written as,

y = 37/100 of x

y = 0.37x

Thus,

The equation of y is 37% of x is

y = 0.37x

Learn more about expressions here:

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Answer:

Values of k can be 0, 1, or 2 such that intersection of the given lines lie in the 1st quadrant.

Step-by-step explanation:

Given two lines:

[tex]x=2k[/tex] and

[tex]3x+2y=12[/tex]

To find:

Values of 'k' such that the intersection of given two lines lie in the first quadrant.

Solution:

In 1st quadrant, the values of [tex]x[/tex] and [tex]y[/tex] both are positive.

So, let us find out intersection of the two lines.

Intersection of the two lines can be found by solving the two equations for the values of [tex]x[/tex] and [tex]y[/tex].

Given that [tex]x=2k[/tex] to be in the first quadrant, the value of k must be positive.

Let us put [tex]x=2k[/tex] in the equation [tex]3x+2y=12[/tex] to find the intersection point.

[tex]3 \times 2k + 2y=12\\\Rightarrow 6k+2y=12\\\Rightarrow 2y=12-6k\\\Rightarrow \bold{y=6-3k}[/tex]

For y to be positive:

[tex]6 - 3k \geq 0\\\Rightarrow 3k \leq 6\\\Rightarrow k \leq 2[/tex]

So, values of k can be 0, 1, or 2 such that intersection of the given lines lie in the 1st quadrant.

Answer:

found this on the web. hopefully it helps

Answer:

11 years old

Step-by-step explanation:

Because the dog is 4 years older, add 7+4

Answer:oufonsrgonsrgnrsosgnsogwo0on93rhskgv oaoef

Step-by-step explanation:

iaeiuaefiaef ebf efa fcade emma jog 123e oefofoinsfnfsaefosefnhsfnalnfaogh

Answer:

  1 : 50,000,000

Step-by-step explanation:

The given scale (with units) is ...

  1 mm : 50 km

If we convert both units to meters, so we can give the scale as a pure number, then we have ...

  10^-3 m : 5×10^4 m = 1 : 5×10^7 = 1 : 50,000,000

Answer:

3x + 5

Step-by-step explanation:

We are doing variables and simplifying.

You have two types of numbers here, you have a coefficient, and you have regular numbers. Now do keep in mind that you can never add a regular number to a coefficient. So the only thing in this problem you will add is 3 and 2. Because 3x is different

3x + (3 + 2)

3x + 5

The simplified form of the expression is 3x + 5.

Option A is the correct answer.

We have,

Expression:

3x + 3 + 2

Add like terms.

3x + (3 + 2)

3x + 5

This is the simplest simplified form of the expression.

Thus,

The simplified form of the expression is 3x + 5.

Learn more about expressions here:

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Use the divergence theorem,

[tex]\displaystyle\iint_{\partial\sigma}\mathbf F(x,y,z)\cdot\mathrm d\mathbf S=\iiint_\sigma\mathrm{div}\mathbf F(x,y,z)\,\mathrm dV[/tex]

We have

[tex]\mathrm{div}\mathbf F(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(2y)}{\partial y}+\dfrac{\partial0}{\partial z}=5[/tex]

so that the flux across [tex]\sigma[/tex] is equal to 5 times the volume of the cube. The cube itself has edge length 5, so its volume is [tex]5^3=125[/tex], making the flux [tex]5^4=\boxed{625}[/tex].

Answer:

Once we got

[tex]x=-1[/tex]

[tex]y=3[/tex]

[tex]z=2[/tex]

[tex]\boxed{\text{The sum is 4}}[/tex]

Step-by-step explanation:

Given the linear system:

[tex]\begin{cases} 2x + 3y-z = 5 \\ x- 3y + 2z = -6 \\ 3x + y - 4z = -8 \end{cases}[/tex]

Let's solve it using matrices. I will use Cramer's rule

[tex]M=\left[\begin{array}{ccc}2&3&-1\\1&-3&2\\3&1&-4\end{array}\right][/tex]

Considering determinant as D.

[tex]D=\begin{vmatrix}2&3&-1\\1&-3&2\\3&1&-4\\\end{vmatrix}=40[/tex]

[tex]M_x = \left[\begin{array}{ccc}5&3&-1\\-6&-3&2\\-8&1&-4\end{array}\right] \implies D_x = \begin{vmatrix}5&3&-1\\-6&-3&2\\-8&1&-4\\\end{vmatrix}=-40[/tex]

[tex]M_y = \left[\begin{array}{ccc}2&5&-1\\1&-6&2\\3&-8&-4\end{array}\right] \implies D_y = \begin{vmatrix}2&5&-1\\1&-6&2\\3&-8&-4\\\end{vmatrix}=120[/tex]

[tex]M_z = \left[\begin{array}{ccc}2&3&5\\1&-3&-6\\3&1&-8\end{array}\right] \implies D_z= \begin{vmatrix}2&3&5\\1&-3&-6\\3&1&-8\\\end{vmatrix}=80[/tex]

So, we have

[tex]$x=\frac{D_x}{D} =\frac{-40}{40}=-1 $[/tex]

[tex]$y=\frac{D_y}{D} =\frac{120}{40}=3$[/tex]

[tex]$z=\frac{D_z}{D} =\frac{80}{40}=2 $[/tex]

Answer: 3

Step-by-step explanation: you add 7+3 which is 10 - 7 =3

Answer: the real solution: x = - 2

Find the attached file for the remaining solution

Step-by-step explanation:

The equation given is:

x3 - 5x2 + 28 = 0 

Let assume that -2 is one of the root of the equation. Substitute -2 for x

(-2)^3 - 5(-2)^2 + 28

-8 - 20 + 28 = 0

Therefore, -2 is one of the root of the equation since the equation tend to zero.

If x = -2, then x+2 is one of the factors of the equation. Therefore, the real solution is x = -2

Please find the attached file for the remaining solution.

Answer:

x = 16

Step-by-step explanation:

Step 1: We know DF - DE = EF

9x - 39 - 47 = EF

Step 2: Simplify

9x - 86 = EF

Step 3: Set the equation equal to 3x + 10

9x - 86 = 3x + 10

6x = 96

x = 16

Therefore x is equal to 16

Answer:

58

Step-by-step explanation:

DE + EF = DF

47+ 3x+10 = 9x-39

Combine like terms

57 +3x = 9x-39

Subtract 3x from each side

57+3x-3x = 9x-3x-39

57 = 6x-39

Add 39 to each side

57+39 = 6x-39+39

96 = 6x

Divide by 6

96/6 = 6x/6

16 =x

We want the length of EF

EF = 3x+10

     = 3*16 +10

    = 48+10

   = 58

Looks like your function is

[tex]f(x)=\dfrac x{6x^2+1}[/tex]

Rewrite it as

[tex]f(x)=\dfrac x{1-(-6x^2)}[/tex]

Recall that for [tex]|x|<1[/tex], we have

[tex]\dfrac1{1-x}=\displaystyle\sum_{n=0}^\infty x^n[/tex]

If we replace [tex]x[/tex] with [tex]-6x^2[/tex], we get

[tex]f(x)=\displaystyle x\sum_{n=0}^\infty\frac(-6x^2)^n=\sum_{n=0}^\infty (-6)^n x^{2n+1}[/tex]

By the ratio test, the series converges if

[tex]\displaystyle\lim_{n\to\infty}\left|\frac{(-6)^{n+1} x^{2(n+1)+1}}{(-6)^n x^{2n+1}}\right|=6|x^2|\lim_{n\to\infty}1=6|x|^2<1[/tex]

Solving for [tex]x[/tex] gives the interval of convergence,

[tex]|x|^2<\dfrac16\implies|x|<\dfrac1{\sqrt6}\implies -\dfrac1{\sqrt 6}<x<\dfrac1{\sqrt 6}[/tex]

We can confirm that the interval is open by checking for convergence at the endpoints; we'd find that the resulting series diverge.

The interval of the convergence is (-1/√6 < x < 1/√6). We can confirm that the interval is open by checking for convergence at the endpoints.

The function is an expression, rule, or law that defines the relationship between one variable to another variable. Functions are ubiquitous in mathematics and are essential for formulating physical relationships.

The given function is

[tex]\rm f(x) = \dfrac{x}{6x^2 + 1} \\\\or \\\\f(x) = \dfrac{x}{1 - (-6x^2)}[/tex]

For |x| < 1, we have

[tex]\rm \dfrac{1}{1-x} = \Sigma_{n=0}^{\infty} \ x^n[/tex]

If x is replaced with -6x², then we have

[tex]\rm f(x)= \Sigma_{n=0}^{\infty} (-6x^2 )^n = \Sigma_{n=0}^{\infty} (-6)^n x^{2n+1}[/tex]

Then by the ratio test, the series converges if

[tex]\displaystyle \lim_{n \to \infty} \left| \dfrac{(-6)^{n+1}x^{2(n+1)+1}}{(-6)^{n}x^{2n+1}} \right|=6|x^{2}| \displaystyle \lim_{n \to \infty }1=6|x^{2}| < 1[/tex]

Solving for x, the interval of convergence will be

[tex]|x^2| < \dfrac{1}{6} \\\\|x| < \dfrac{1}{\sqrt6} \\\\-\dfrac{1}{\sqrt6} < x < \dfrac{1}{\sqrt6}[/tex]

We can confirm that the interval is open by checking for convergence at the endpoints.

More about the function link is given below.

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Answer:

20

Step-by-step explanation:

Given:

As per the picture:

Comparing the two equations above:

x = 20 is the correct answer for this question

Otherwise something is wrong with the question.

Answer:

False

Step-by-step explanation:

In an essay, you have to cite every source you use because

a) the teacher needs to KNOW you didn't copy

b) the teacher needs to know if you got your info from a reliable source

c) the teacher needs to see your ability to use info from a source to put into your essay

This may be different for books, but citation is always a rule of thumb for essays in school, so don't forget unless you want points to be taken off of your score.

Answer:

0.1875

Step-by-step explanation:

The well formatted expression for the series has been attached to this response.

Each term in the series is govern by the rule, Tₙ

Where

n = term position

Tₙ = [tex]\frac{(-1)^{n+1}n}{2^n}[/tex]

To get the first six terms, we substitute n = 1 through 6 into Tₙ as follows:

When n = 1, we have;

T₁ = [tex]\frac{(-1)^{1+1}(1)}{2^1} = \frac{1}{2}[/tex] = 0.50000

When n = 2, we have;

T₂ = [tex]\frac{(-1)^{2+1}(2)}{2^2} = \frac{-2}{4} = \frac{-1}{2}[/tex] = -0.50000

When n = 3, we have;

T₃ = [tex]\frac{(-1)^{3+1}(3)}{2^3} = \frac{3}{8}[/tex] = 0.37500

When n = 4, we have;

T₄ = [tex]\frac{(-1)^{4+1}(4)}{2^4} = \frac{-1}{4}[/tex] = -0.25000

When n = 5, we have;

T₅ = [tex]\frac{(-1)^{5+1}(5)}{2^5} = \frac{5}{32}[/tex] = 0.15625

When n = 6, we have;

T₆ = [tex]\frac{(-1)^{6+1}(6)}{2^6} = \frac{-6}{64} = \frac{-3}{32}[/tex] = -0.09375

Therefore, the approximate sum of the series using the sum of the first six terms is

=> T₁ + T₂ + T₃ + T₄ + T₅ + T₆

=> 0.50000 + -0.50000 + 0.37500 + -0.25000 + 0.15625 + -0.09375

=> 0.1875

Answer:

-2°F

Step-by-step explanation:

So on Sunday night, the temperature was -10°F.

And by Monday morning, the temperature has increased by 8°F.

In other words, to find the temperature on Monday morning, we just have to add 8 to -10. Therefore:

[tex](-10)+(8)=-2[/tex]

The temperature on Monday morning is -2°F

Answer:

-2 degrees F.

If you have -10, you have to add positive 10 just to get to 0.

Answer:

0.412

Step-by-step explanation:

Given the stem plot figure out which ones are below 17g.

So: 7g / 17g = 0.412 which is the proportion

Answer: $130.66

Step-by-step explanation:

Answer:

130.66

Step-by-step explanation:

i took the asignment

Answer:

D

Step-by-step explanation:

So we have the expression:

[tex](6m+3)(m-2)[/tex]

Use the distributive property and distribute:

[tex]=(6m+3)(m)+(6m+3)(-2)[/tex]

Distribute:

[tex]=(6m^2+3m)+(-12m-6)[/tex]

Combine like terms:

[tex]=6m^2+3m-12m-6\\=6m^2-9m-6[/tex]

The correct answer is D

Edit: Typo

Answer:

y = -7x + 2

Step-by-step explanation:

Use the slope-intercept form y = mx + b.

Substitute -7 for m, 0 for x and 2 for y.  Then

2 = (-7)(0) + b, so b must be 2.

The desired equation is y = -7x + 2.

Answer:

58.33

Step-by-step explanation:

35 isn't exactly 60% of an number but by rounding you will get 58.33

Answer: True.

Step-by-step explanation:

The one-way analysis of variance usually (abbreviated as one-way ANOVA). is a method that is used to compare the means of two or more samples ( by make use of the F distribution). This method only applies to numerical response data, the "Y", usually one variable, and numerical or (usually) categorical input data, the "X", always one variable, hence the reason it’s know as "one-way" beca it takes it account one variae at a time.