Estimate √15 to the nearest tenth. Then locate √15 on a number line.

√15~ (Round to the nearest tenth as needed.) ​

The inverse of the function f(x) = (3x)^2 is f-1(x) = 1/3√x

The function is given as:

f(x) = (3x)^2

Remove the bracket in the above equation

So, we have:

f(x) = 9x^2

Express f(x) as y

So, we have

y = 9x^2

Swap the positions of x and y

So, we have

x = 9y^2

Make y the subject of the formula

y^2 = x/9

Take the square root of both sides

y = 1/3√x

Express as an inverse function

f-1(x) = 1/3√x

Hence, the inverse of the function f(x) = (3x)^2 is f-1(x) = 1/3√x

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Answer:

Yes

Step-by-step explanation:

The scale factor of the drawing is 1 / 200 or 0.005.

Scale factor is used to scale shapes in different dimensions.

In other words, Scale factor is described as the number or the conversion factor which is used to change the size of a figure without affecting its shape.

Therefore, scale factor can be represented mathematically as follows:

Scale factor = dimensions of the new shape ÷ dimensions of the original shape.

Hence he uses the scale 10 millimetres equals to 2 meters.

We have to convert metres to millimetres to get the scale.

10millimetres = 2meters

Therefore,

1 meter  =  1000 millimetres.

2 meters = ?

cross multiply

length = 2 × 1000 = 2000 millimetres

Therefore, the scale factor of the drawing is as follows:

scale factor = 10 / 2000

scale factor = 1 / 200

scale factor = 0.005

Therefore, the scale factor of the drawing is 1 / 200 or 0.005.

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Answer:

hey from where are you...

Answer:

Step-by-step explanation:

32.1 gb is answer

77.56

-45.46

second file would be 32.10 gb

Answer:

The graph of  [tex]y=\sqrt{4x+16}+5[/tex]  is the graph of [tex]y=\sqrt{x}[/tex]  translated 4 units left, stretched horizontally by a factor of 1/4, and translated 5 units up.

Step-by-step explanation:

Transformations

[tex]\textsf{For }a > 0[/tex]

[tex]f(x+a) \implies f(x) \: \textsf{translated}\:a\:\textsf{units left}[/tex]

[tex]f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]

[tex]y=f(ax) \implies f(x) \: \textsf{stretched parallel to the x-axis (horizontally) by a factor of} \: \dfrac{1}{a}[/tex]

Given function

[tex]y=\sqrt{4x+16}+5[/tex]

Parent function

Parent functions are the simplest form of a given family of functions.

[tex]y=\sqrt{x}[/tex]

The graph of the parent function is related to the graph of the given function by a series of transformations.  To determine the series of transformations, work out the steps of how to go from the parent function to the given function.

Factor the expression under the square root sign:

[tex]y=\sqrt{4(x+4)}+5[/tex]

Transformations

Parent function:

[tex]f(x)=\sqrt{x}[/tex]

Translated 4 units left:

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

Horizontally stretched by a factor of 1/4 (compressed by a factor of 4):

[tex]\begin{aligned}f(4(x+4)) & =\sqrt{4(x+4)}\\ & = \sqrt{4x+16} \end{aligned}[/tex]

Translated 5 units up:

[tex]f(4x+16)+5=\sqrt{4x+16}+5[/tex]

Therefore, the graph of  [tex]y=\sqrt{4x+16}+5[/tex]  is the graph of [tex]y=\sqrt{x}[/tex]  translated 4 units left, stretched horizontally by a factor of 1/4, and translated 5 units up.

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The solutions of given equations are x = 0,b = -3,y = 2 and c = 5 respectively.

A formula known as an equation uses the same sign to denote the equality of two expressions.

The equation must be constrained by =,< or >.

Given the equations,

01) x/4 + 2x/3 - 4 = - 4 + 5x/2

(3x + 8x)/12 = 5x/2

x = 0

02) 3.2b + 10 = -1.7b - 4.7

2.2b + 1.7b = -4.7 - 10

b = -3

03) 6.6y + y + 6.5 = y - 6.7

5.5y = -6.7 - 6.5

y = 2

04) 2c - 7c + 8 = -17

-5c = -25

c = 5

Hence "The solutions of given equations are x = 0,b = -3,y = 2 and c = 5 respectively".

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Answer:

f(x)= x + 3

Step-by-step explanation:

There are multiple but here are 2

13+13+11+11 = 42

10+10+14+14 = 42

Answer:

6325

Step-by-step explanation:

[tex]6000 \\ \: \: 300 \\ \: \: \: \: 20 \\ \: \: \: + 5[/tex]

___________

6325

Answer:

Your answer would be [tex]6325[/tex]

Step-by-step explanation:

[tex]=6325[/tex]

[tex]6000 +300+20+5[/tex]

[tex]=6325[/tex]

hopefully this helps! TwT

The certificate will be worth of  $11534.9 on Ruth's 19th birthday.

Compound Interest is calculated using the formula.

[tex]Amount=P*(1+\frac{r}{n} )^{nt}[/tex]

where , P=principal

r = rate of interest

n= number of times interest is compounded

t = no of years

In the given question ;

Principal = $1500

n=1      ...(as it is compounded annually)

t= 18 years ...(as on 19th birthday Ruth will complete 18 years)

r=12%=0.12

Substituting the values of P,n,t,r in the formula we get,

[tex]Amount=P*(1+\frac{r}{n} )^{nt}[/tex]

[tex]=1500*(1+\frac{0.12}{1} )^{1*18}[/tex]

On solving further we get

=[tex]1500*(1.12)^{18}[/tex]

=1500*7.6899

=11534.85

Therefore , On Ruth's 19th birthday the certificate will be of worth $11534.9  .

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Answer:

(15 ÷ n) + (n ÷ 15)

Step-by-step explanation:

The sum of 15 divided by a number and that number divided by 15.

15 divided by a number

= (15 ÷ n)

15 is being divided by an unknown number (put n as a variable)

that number divided by 15

= (n ÷ 15)

An unknown number (put n as a variable) is being divided by 15.

The sum

Add (15 ÷ n) and (n ÷ 15)

(15 ÷ n) + (n ÷ 15)

Hope this helped and have a lovely rest of your day! :)

Answer:

670.05

Step-by-step explanation:

Rounding to the nearest dollar, Serina will have $515 at the end of five years. Therefore, correct option is A.

To calculate the amount Serina will have at the end of five years, we can use the formula for compound interest:

A = [tex]P(1 + r)^{(nt)[/tex]

Where:

A = the future value of the investment/loan, including interest

P = the principal amount (initial deposit)

r = the annual interest rate (in decimal form)

n = the number of times that interest is compounded per year

t = the number of years the money is invested for

In this case, Serina deposits $500 with an annual interest rate of 5%, compounded semiannually (n = 2), and keeps the money in her account for three years (t = 3).

Plugging in the values, we get:

A = 500(1 + 0.05)⁶

A = 500(1.005)⁶

A ≈ 500(1.030)

A ≈ $515

Therefore, correct option is A.

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Using the normal distribution, we have that:

A.

B. Due to the higher coefficient of variation, Company A has the larger dispersion.

The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

For Company A, the mean and the standard deviation are given as follows:

[tex]\mu = 500, \sigma = 40[/tex]

The Empirical Rule is applicable for a normal variable, meaning that 95% of the measures are within 2 standard deviations of the mean, hence:

The middle 95% of the wires from Company A tested between 420 and 580 psi.

The middle 50% is the measures between the 25th percentile(X when Z = -0.675) and and the 75th percentile(X when Z = 0.675), hence:

25th percentile:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

-0.675 = (X - 500)/40

X - 500 = -0.675(40)

X = 473.

75th percentile:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

0.675 = (X - 500)/40

X - 500 = 0.675(40)

X = 527.

The middle 50% of the wires of Company A tested between 473 psi and 527 psi.

The company with the larger dispersion has the highest coefficient of variation, given by the standard deviation divided by the mean.

Hence:

Hence Company A has the larger dispersion.

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Let [tex]S[/tex] denote the sum. We can first resolve the sum in [tex]m[/tex] by factorizing and decomposing into partial fractions.

[tex]\displaystyle S = \sum_{n=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{n+1}}{mn^2 + mn + m^2n} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}n \sum_{m=1}^\infty \frac1{m(m+n+1)} \\\\ ~~~~ = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n(n+1)} \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right)[/tex]

Rewrite the [tex]m[/tex]-summand as a definite integral. Interchange the integral and sum, and evaluate the resulting geometric sums.

[tex]\displaystyle \sum_{m=1}^\infty \left(\frac1m - \frac1{m+n+1}\right) = \sum_{m=1}^\infty \int_0^1 \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{m=1}^\infty \left(x^{m-1} - x^{m+n}\right) \, dx \\\\ ~~~~~~~~ = \int_0^1 \frac{1 - x^{n+1}}{1 - x} \, dx \\\\ ~~~~~~~~ = \int_0^1 \sum_{\ell=0}^n x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \int_0^1 x^\ell \, dx \\\\ ~~~~~~~~ = \sum_{\ell=0}^n \frac1{\ell+1} \\\\ ~~~~~~~~ = \sum_{\ell=1}^{n+1} \frac1\ell = H_{n+1}[/tex]

where

[tex]H_n = \displaystyle \sum_{\ell=1}^n \frac1\ell = 1 + \frac12 + \frac 13 + \cdots + \frac1n[/tex]

is the [tex]n[/tex]-th harmonic number. The generating function will be useful:

[tex]\displaystyle \sum_{n=1}^\infty H_n x^n = -\frac{\ln(1-x)}{1-x}[/tex]

To evaluate the remaining sum to get [tex]S[/tex], let

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

and observe that [tex]S=\lim\limilts_{x\to-1^+} f(x)[/tex], which I'll abbreviate to [tex]f(-1)[/tex]. Differentiating twice, we have

[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{H_{n+1}}n x^n[/tex]

[tex]\displaystyle f''(x) = \sum_{n=1}^\infty H_{n+1} x^n[/tex]

[tex]\displaystyle \implies f''(x) = -\frac{\ln(1-x)}{x^2(1-x)} - \frac1x[/tex]

By the fundamental theorem of calculus, noting that [tex]f(0)=f'(0)=0[/tex], we have

[tex]\displaystyle \int_{-1}^0 f'(x) \, dx = f(0) - f(-1) \implies f(-1) = -\int_{-1}^0 f'(x) \, dx[/tex]

[tex]\displaystyle \int_x^0 f''(x) \, dx = f'(0) - f'(x) \implies f'(x) = -\int_x^0 f''(t) \, dt[/tex]

[tex]\displaystyle \implies S = f(-1) = \int_{-1}^0 \int_x^0 \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dt \, dx[/tex]

Change the order of the integration, and substitute [tex]t=-u[/tex].

[tex]S = \displaystyle \int_{-1}^0 \int_{-1}^t \left(\frac{\ln(1-t)}{t^2(1-t)} + \frac1t\right) \, dx \, dt \\\\ ~~~ = - \int_{-1}^0 \left(\frac{(1+t) \ln(1-t)}{t^2(1-t)} + \frac1t + 1\right) \, dt \\\\ ~~~ = -1 - \int_{-1}^0 \left(\left(\frac2{1-t} + \frac2t + \frac1{t^2}\right) \ln(1-t) + \frac1t\right) \, dt \\\\ ~~~ = -1 - \int_0^1 \left(\left(\frac2{1+u} - \frac2u + \frac1{u^2}\right) \ln(1+u) - \frac1u\right) \, du[/tex]

For the remaining integrals, substitute and use power series.

[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}{1+u} \, du = \int_0^1 \ln(1+u) d(\ln(1+u)) = \frac{\ln^2(2)}2[/tex]

[tex]\displaystyle \int_0^1 \frac{\ln(1+u)}u \, du = - \int_0^1 \frac1u \sum_{k=1}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}k \int_0^1 u^{k-1} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=1}^\infty \frac{(-1)^k}{k^2} = \frac{\pi^2}{12}[/tex]

[tex]\displaystyle \int_0^1 \frac{\ln(1+u) - u}{u^2} \, du = - \int_0^1 \frac1{u^2} \left(\sum_{k=1}^\infty \frac{(-u)^k}k + u\right) \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\int_0^1 \frac1{u^2} \sum_{k=2}^\infty \frac{(-u)^k}k \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = - \sum_{k=2}^\infty \frac{(-1)^k}k \int_0^1 u^{k-2} \, du \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = -\sum_{k=2}^\infty \frac{(-1)^k}{k(k-1)} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \sum_{k=1}^\infty \frac{(-1)^k}{k(k+1)} = 1 - 2\ln(2)[/tex]

Tying everything together, we end up with

[tex]S = -1 - \left(2 \cdot \dfrac{\ln^2(2)}2 - 2 \cdot \dfrac{\pi^2}{12} + (1-2\ln(2))\right) \\\\ ~~~ = \boxed{\frac{\pi^2}6 - 2 + 2\ln(2) - \ln^2(2)}[/tex]

Answer:

46

Step-by-step explanation:

4x-4=180

4x=184

x=46

Answer:

x=46

Step-by-step explanation:

used by scientific calculator.

Answer:

a = 115°

Step-by-step explanation:

Two angles are supplementary angles because they make a straight line.

The sum of two angles is equal to 180° then to find the measure of the missing angle we simply subtract 65 from 180:

180 - 65 = 115

The sequence or pattern [[tex]x_{n}[/tex]] [tex]x_{n}[/tex] Converges to Zero, meaning that the power, limit, and final solution all get closer to zero.

A grouping of numbers in a specific order is known as a sequence. On the other hand, a series is described as the accumulation of a sequence's constituent parts. The length of the series is the number of elements (potentially infinite).Unlike a set, a sequence may contain the same things more than once at different locations, and unlike a set, the sequence's order is crucial.

According to given information;

First put;

[tex]x_{n}[/tex]−1=−√(1+x[tex]_n_-_2[/tex]+1)

into

[tex]x_{n}[/tex]=(−1+x[tex]_n_-_1[/tex]+1)

You'll notice a general form:

[tex]x_{n}[/tex]=−1+(x[tex]_n_-_r[/tex]+1)^2^−r

Then put r = n-1, and take the limit of both sides with n tending to infinity. On the right hand side, you have

−1 +[tex]\lim_{n \to \infty}[/tex](x+1)^[tex]\frac{1}{2n-1}[/tex]

       n→∞

he power approaches 0, the limit approaches 1 and the final answer approaches 0.

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Answer:

1. He get get to school 10 minutes faster riding by hisself than ridding with Christina.

Step-by-step explanation:

Christina is 22 minutes - Luis 12 minutes = 10 minutes.

Answer:

Step-by-step explanation:

Cost of one rubber = $1

Cost of one pen = 0.25

The consecutive odd numbers that sum up to 321 are 105, 107 and 109.

The sum of three consecutive odd integers is 321.

Odd numbers are numbers that when divided by 2 produces a remainder.

let

x = first odd numbers

Therefore,

x + x + 2 + x + 4 = 321

combine the like terms

x + x + x + 2 + 4 = 321

add the like terms

x + x + x + 2 + 4 = 321

3x + 6 = 321

subtract 6 from both sides

3x + 6 = 321

3x + 6 - 6 = 321 - 6

3x  = 315

divide both sides by 3

x = 315 / 3

x = 105

Therefore, the consecutive odd numbers that sum up to 321 are 105, 107 and 109.

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a formula for the perimeter of the rectangle in terms of its length, l will be given as l = P / 2 - b.

We know that the perimeter of rectangle is given by:

Perimeter ( P ) = 2 ( l + b )

Now, we need to represent this formula in terms of length that is l.

So, we get that:

P = 2 l + 2 b

Now subtract 2 b from both the sides:

P - 2 b = 2 l + 2 b - 2 b

P - 2 b = 2 l

2 l = P - 2 b

Divide both the sides by 2:

2 l / 2 = (P - 2 b) / 2

l = P / 2 - b

Therefore, we get that, a formula for the perimeter of the rectangle in terms of its length, l will be given as l = P / 2 - b.

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The side lengths could be used to create a scale drawing of the field is 6600 yard².

A rectangle is a type of quadrilateral, whose opposite sides are equal and parallel. It is a four-sided polygon that has four angles, equal to 90 degrees. A rectangle is a two-dimensional shape.

Given:

A lacrosse field is rectangular with a length of 110 yards and a width of 60 yards.

According to given question we have

We know that

Area of rectangle= Length × width

Area of rectangle= 110 × 60

Area of rectangle= 6600 yard².

Scale ratio can be 1:55

Therefore, the side lengths could be used to create a scale drawing of the field is 6600 yard².

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Interest rate compounded annually for the amount of $5,000 invested would have been worth $23,125.59 after 10 years is equal to 16.55% per year.

As given in the question,

Principal (P) = $5,000

Time (t) = 10 years

Amount = $23,125.59

[tex]r = n[(A/P)^{\frac{1}{nt}}-1]\\\\\implies r = 1[(23125.29/5000)^{\frac{1}{10} }-1]\\\\\implies r = 0.1655\\[/tex]

Convert r into percentage

r = 0.1655 × 100

 = 16.55% compounded annually

Therefore, interest rate compounded annually for the amount of $5,000 invested would have been worth $23,125.59 after 10 years is equal to 16.55% per year.

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true, 3:7 is a ratio and is equal to 3/7

If the interval is [0,3] then the second function whose graph is given has the greater average rate of change.

Given two functions, one is in the table and the other one is in the form of graph.

We are required to choose the function which has the greater average rate of change.

Function is basically the relationship between two or more variables that are expressed in equal to form. The values that we enter are known as part of domain and the values that we get from the function are known as part of codomain or range of the function.

If we observe the table then we will find that in the interval [0,3] there is not any change in the value of function, it is constant to be 4.

If we observe the graph then we will find that the value of function is continuously decreasing.

So, the second function has greater average rate of change over the interval [0,3].

Hence if the interval is [0,3] then the second function whose graph is given has the greater average rate of change.

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The function g(x) reflected along x-axis is |x| + 5.

A modulus function always outputs positive values, hence the outputs are greater than or equal to zero,f(x) ≥ 0.

When a graph is reflected along x-axis f(x) becomes -f(x).

∴ The function f(x) = |x| - 5 when reflected along x-axis it will become

f(x) = - (|x| - 5).

f(x) = - |x| + 5.

Or

g(x) = - |x| + 5.

Graph of g(x) is shown in the image attached.

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In the picture we have to solve the individual variables A=2πr²+2πrh. We got function with r as subject is[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]

Given that,

In the picture we have to solve the individual variables

A=2πr²+2πrh

We have to find function with r as subject.

Taking A to left side we get

2πr²+2πrh-A=0

We can see the equation is in the form of quadratic equation with variable r.

So, The factor we find by using the formula

That is [tex]\frac{-b\pm\sqrt{b^{2}-4ac } }{2a}[/tex]

Here, a=2π,b=2πh and c=-A

r=[tex]\frac{-2\pi h\pm\sqrt{(2\pi h)^{2}-4(2\pi)(-a) } }{2(2\pi)}[/tex]

r=[tex]\frac{-2\pi h\pm\sqrt{(4\pi^{2}h^{2} +8\pi a) } }{4\pi}[/tex]

r=[tex]\frac{-h}{2} \pm\frac{\sqrt{(4\pi^{2}h^{2} +8\pi a )} }{4\pi}[/tex]

r=[tex]\frac{-h}{2} \pm\sqrt{\frac{4\pi^{2}h^{2}+8\pi a }{16\pi^{2} } }[/tex]

r=[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]

Therefore, We got  function with r as subject is[tex]\frac{-h}{2} \pm\sqrt{\frac{h^{2}}{4 } +\frac{a}{2\pi} }[/tex]

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