whai is numarates what is ​

Answer:

14

Step-by-step explanation:

g(f(1)) simply means we first get f(1), and that result becomes the input value for g(x). and that is then the end result.

so,

f(1) = 4

and then

g(f(1)) = g(4) = 14

[tex]y=-x+5~~~~~.....(i)\\\\x-y=1~~~~~~.....(ii)\\\\\\(i)-(ii):\\\\~~~~~y-x+y=-x+5-1\\\\\implies 2y=4\\\\\implies y= \dfrac 42\\\\\implies y=2\\\\\text{Substitute y=2 in eq (ii):}\\ \\~~~~ x-2 = 1\\\\\implies x= 1+2\\\\\implies x = 3\\\\\\\text{Hence,}~~ (x,y) = (3,2)[/tex]

Answer:

x^2 +3x -4 = 0  --> (x +4)(x -1) = 0

Step-by-step explanation:

Let's find A and B such that

A*B = - 4    and  A+B = 3

Numbers are:

A = +4  and  B = -1

The factoring equation is y = (x + A)(x + B)

In this specific case: y = (x +4)(x -1)

Solutions to the quadratic equation require

y = (x +4)(x -1) = 0

and hence

x +4 = 0  --> x = -4

x - 1 = 0 -->  x= +1

Answer:

-56

Step-by-step explanation:

First lets solve the bottom of the fraction :

1/7 * -1/8 is equal to -1/56 because to multiply two fractions we multiply there numerators and denominators

Now we have to divide 1 by -1/56 which is equal to :

1 *-56= -56 becasue when dividing by a fraction you want to switch its numerator and denominator

At the end you get -56

Answer:

13

Step-by-step explanation:

The coefficient is the number placed before the given variable.

In this case:

Coefficient: 13

Variable: y

Power: 3

The coefficient is a number in front of the variable( x or y)

The coefficient would be 13

The given equation represents a hyperbola. Its main features are:

A hyperbola can be defined by its center, vertices,  foci and asymptotes. And it is represented algebraically  by the standard equation:  [tex]\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1[/tex], where:

h= x-coordinate of center

k= y-coordinate of center

a and b= semi-axis

First, you need to rewrite the given equation 9y²-x²+2x+54y+62=0 in the standard equation hyperbola:

[tex](-x^2+2x+?)+9(y^2+6y+?)=-62\\(-x^2+2x+?)+9(y^2+6y+9)=-62\\(-x^2+2x-1)+9(y^2+6y+9)=-62\\(-x^2+2x-1)+9(y^2+6y+9)=-62-1+81\\((-x^2+2x-1)+9(y^2+6y+9)=18 ) \div 18\\ \frac{(-x^2+2x-1)}{18} +\frac{9(y^2+6y+9}{18}= \frac{18}{18} \\\frac{-(x-1)^2}{18} +\frac{(y+3^2)}{2}=1\\\frac{-(x-1)^2}{(3\sqrt{2})^2} +\frac{(y+3^2)}{(\sqrt{2})^2 }=1\\\frac{\left(y+3)^2}{\left(\sqrt{2}\right)^2}-\frac{\left(x-1\right)^2}{\left(3\sqrt{2}\right)^2}=1[/tex]

Comparing the previous equation with the standard form ([tex]\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}=1[/tex]), you have:

h=1,  k=-3,  a=[tex]\sqrt{2}[/tex] and b=[tex]3\sqrt{2}[/tex] . From now, it is possible to find that the question asks:

The coordinates for center is (h,k). Thus, the center is (1,-3).

The vertices (V1 and V2) of hyperbola can be found from the coordinates of center (h,k) and the semi-axis (a).

V1= (h,k+a)= (1,-3+[tex]\sqrt{2}[/tex])

V2= (h,k-a)= (1,-3-[tex]\sqrt{2}[/tex])

The foci or the focus points can be found from the coordinates of center (h,k) and the c ([tex]\sqrt{a^2+b^2}[/tex]) which represents the distance from the center to the focus.

[tex]c=\sqrt{a^2+b^2}\\ c=\sqrt{(3\sqrt{2} )^2+(\sqrt{2} )^2}\\c=\sqrt{18+2} \\c=\sqrt{20}=2\sqrt{5}[/tex]

Thus,

F1= (h,k+c)= (1 , -3+ [tex]2\sqrt{5}[/tex] )

F2= (h,k-c)= (1 , -3 - [tex]2\sqrt{5}[/tex] )

The asymptotes are the lines the hyperbola tends to at ±∞. For hyperbola, the asymptotes are defined as: [tex]y=\pm \frac{a}{b}\left(x-h\right)+k[/tex]. Then, for this question:

[tex]y=\pm \frac{\sqrt{2}}{3\sqrt{2}}\left(x-1\right)-3\\y=\pm \frac{1}{3}\left(x-1\right)-3[/tex]

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Answer:

12 miles per hour.

Step-by-step explanation:

1. If you pick 5 hours, then you will noticed that the graph shows you 60 miles.

2. Find the unit rate.

5 hours = 60  miles

1 hour = x miles

x= 60 ÷ 5

x= 12 miles per hour

Answer:

3

Step-by-step explanation:

((7x-4)^2) - (7x-4)(7x-4) + 3

(7x-4)(7x-4) is 7x-4^2

So its (7x-4)^2 - (7x-4)^2 + 3

(7x-4)^2 cancel out.

Answer is 3

Answer:

Step-by-step explanation: yes, we belive that they start young with pyromania.  They may see as something powerful

Answer:

  32 pounds

Step-by-step explanation:

Pounds of potatoes can be found by multiplying pounds of onions by 5 1/3:

  (5 1/3)(6 pounds) = 32 pounds

The restaurant bought 32 pounds of potatoes.

Step-by-step explanation:

the length and the width of a rectangle create a right-angled triangle, with the diagonal being the Hypotenuse of that triangle (the baseline, opposite of the 90° angle).

so, we can use Pythagoras

c² = a² + b²

with c being the Hypotenuse.

so, we have here

diagonal² = 40² + 9² = 1600 + 81 = 1681

diagonal = sqrt(1681) = 41 in

Answer:

It's a fraction

Step-by-step explanation:

You have to figure out the meaning of the letters and find what goes on top

Answer:

[-5, 4) ∪ (4, ∞)

Step-by-step explanation:

Given functions:

[tex]f(x)=\dfrac{1}{x-3}[/tex]

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

Composite function:

[tex]\begin{aligned}(f\:o\:g)(x)&=f[g(x)]\\ & =\dfrac{1}{\sqrt{x+5}-3} \end{aligned}[/tex]

Domain: input values (x-values)

For [tex](f\:o\:g)(x)[/tex] to be defined:

[tex]x+5\geq 0 \implies x\geq -5[/tex]

[tex]\sqrt{x+5}\neq 3 \implies x\neq 4[/tex]

Therefore, [tex]-5\leq x < 4[/tex]  and  [tex]x > 4[/tex]

⇒  [-5, 4) ∪ (4, ∞)

Answer:

74

Step-by-step explanation:

By average definition:

(74 + 67 +73 +x) / 4 = 72

Multiply by 4

(74 + 67 +73 +x)  = 288

shift the first three terms on the right side

x = 288 -74 -67 -73 =  74

Answer:

trigonometric function whose value for the complement of an angle is equal to the value of a given trigonometric function of the angle itself the sine is the cofunction of the cosine.

The amount of money does Liam need to put in the simple interest account to earn  $750 in interest, is $6250.

Simple interest is the amount charged on the principal amount with a fixed rate of interest for a time period. Simple interest calculated only on the principal amount.

The formula for the simple interest can be given as,

[tex]I=\dfrac{Prt}{100}[/tex]

Here, (I) is the interest amount earned on the principal amount of (P) with the rate of (r) in the time period of (t).

Liam wants to put some money in a simple interest account. He pays R=6% interest annually for t=2 years. Liam would like to earn I=$750 in interest.

Put the values in the formula,

[tex]750=\dfrac{P\times6\times2}{100}\\P=\dfrac{750\times100}{6\times2}\\P=6250[/tex]

Thus, the amount of money does Liam need to put in the simple interest account to earn  $750 in interest, is $6250.

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The amount of money does Liam need to put in the simple interest account to earn  $750 in interest, is $6250.

What is simple interest?

Simple interest is the amount charged on the principal amount with a fixed rate of interest for a time period. Simple interest calculated only on the principal amount.

The formula for the simple interest can be given as,

Answer: $6,250

Step By Step: Here, (I) is the interest amount earned on the principal amount of (P) with the rate of (r) in the time period of (t).

Liam wants to put some money in a simple interest account. He pays R=6% interest annually for t=2 years. Liam would like to earn I=$750 in interest.

Put the values in the formula,

Thus, the amount of money does Liam need to put in the simple interest account to earn  $750 in interest, is $6250.

Answer:

Step-by-step explanation:

Answer:

177 600 000 is the answer

WHEN GRAPHING

Step 1. Convert to y = mx + b

Step 2. Graph like normal

Step 3. Shade as indicated by the inequality symbol

Step 4. If  ≤ or ≥ ONLY, then also shade the line

FOR THIS PROBLEM

1. Graph each equation

2. Shade ABOVE line for each

3. Shade line first equation as well

Hope this helps and God bless!

Answer:

B. 20πb³ + 12πb²

Step-by-step explanation:

The equation for the volume is incorrect, the volume of a cylinder with radius (r), and height (h), is πr²h.

The volume of the cylinder in terms of[tex]20\pi b^3 + 12\pi b^2[/tex]. Option B is correct.

A cylinder is a three-dimensional figure with two bases that are joined with a curved surface.

The total space occupied by the three-dimensional figure is called volume.

Given that:

The Volume is V = [tex]\pi r^2 h[/tex]

Radius r =  2b

Height,h = 5b +3

Substitute the values into the formula for the volume of a cylinder

V = πr²h

V =[tex]\pi \times (2b)^2 \times (5b + 3)[/tex]

Simplify the expression:

V = [tex]4\pi b^2 \times (5b + 3)[/tex]

Use the distributive property to get the values.

V = [tex]20\pi b^3+ 12\pi b^3[/tex]

So, the volume of the cylinder in terms of[tex]20\pi b^3 + 12\pi b^2[/tex]. Option B is correct.

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Answer:

B

Step-by-step explanation:

because the angles 1 and 7 are alternate and equal , we can conclude that p and q are parralel

Answer:

x=1 y=2

Step-by-step explanation:

Isolate x for 4x -y =2: x=2+y/4

Substitute x= 2+y/4

(2+13y/4) +3y=7

simplify 2+13y/4) =7

Isolate y for 2+13y/4) =7: y=2

Forx= 2+y/4

Substitute y =2

x=2+2/4

x=1

a. The least amount as indicated in the box plot is: 10

b. Median amount of snowfall is: 20

c. Data is concentrated at the third quartile

The median of a data in a box plot is the value indicated by the vertical line that divides the box.

a. The least amount is the minimum value which is at the extreme end of the whisker to your left = 10

b. Median amount of snowfall is the value indicated by the vertical line = 20

c. The largest section is the third quarter, so the data is most concentrated within the third quarter.

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Answer:

it is 57

Step-by-step explanation:

Answer:

A. √57

Step-by-step explanation:

√47 would be off the number line because it's below the number 7

√65 would be past 8 but the red dot comes before the number 8

√72 would be past 8 as well and it would be past √65 but the dot is before number 8

This leaves our answer as √57 beause the square root of 57 is 7.549 (etc.) which is between the numbers 7 and 8, while also being slightly higher than 7.5 :)

Have a good day!

Answer:

No.

Step-by-step explanation:

The maximum that can be earned is $720, which results from walking 60 dogs at $12 per walk.

Replace [tex]x\mapsto \tan^{-1}(x)[/tex] :

[tex]\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \int_0^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx[/tex]

Split the integral at x = 1, and consider the latter one over [1, ∞) in which we replace [tex]x\mapsto\frac1x[/tex] :

[tex]\displaystyle \int_1^\infty \frac{\sqrt[3]{x} \ln(x)}{1+x^2} \, dx = \int_0^1 \frac{\ln\left(\frac1x\right)}{\sqrt[3]{x} \left(1+\frac1{x^2}\right)} \frac{dx}{x^2} = - \int_0^1 \frac{\ln(x)}{\sqrt[3]{x} (1+x^2)} \, dx[/tex]

Then the original integral is equivalent to

[tex]\displaystyle \int_0^1 \frac{\ln(x)}{1+x^2} \left(\sqrt[3]{x} - \frac1{\sqrt[3]{x}}\right) \, dx[/tex]

Recall that for |x| < 1,

[tex]\displaystyle \sum_{n=0}^\infty x^n = \frac1{1-x}[/tex]

so that we can expand the integrand, then interchange the sum and integral to get

[tex]\displaystyle \sum_{n=0}^\infty (-1)^n \int_0^1 \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \ln(x) \, dx[/tex]

Integrate by parts, with

[tex]u = \ln(x) \implies du = \dfrac{dx}x[/tex]

[tex]du = \left(x^{2n+\frac13} - x^{2n-\frac13}\right) \, dx \implies u = \dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac23}}{2n+\frac23}[/tex]

[tex]\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \int_0^1 \left(\dfrac{x^{2n+\frac43}}{2n+\frac43} - \dfrac{x^{2n+\frac13}}{2n-\frac13}\right) \, dx \\\\ = \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{\left(2n+\frac43\right)^2} - \frac1{\left(2n+\frac23\right)^2}\right) \\\\ = \frac94 \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)[/tex]

Recall the Fourier series we used in an earlier question [27217075]; if [tex]f(x)=\left(x-\frac12\right)^2[/tex] where 0 ≤ x ≤ 1 is a periodic function, then

[tex]\displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \sum_{n=1}^\infty \frac{\cos(2\pi n x)}{n^2}[/tex]

[tex]\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(2\pi(3n+1)x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(2\pi(3n+2)x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(2\pi(3n)x)}{(3n)^2}\right)[/tex]

[tex]\implies \displaystyle f(x) = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{\cos(6\pi n x + 2\pi x)}{(3n+1)^2} + \sum_{n=0}^\infty \frac{\cos(6\pi n x + 4\pi x)}{(3n+2)^2} + \sum_{n=1}^\infty \frac{\cos(6\pi n x)}{(3n)^2}\right)[/tex]

Evaluate f and its Fourier expansion at x = 1/2 :

[tex]\displaystyle 0 = \frac1{12} + \frac1{\pi^2} \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{(3n+1)^2} + \sum_{n=0}^\infty \frac{(-1)^n}{(3n+2)^2} + \sum_{n=1}^\infty \frac{(-1)^n}{(3n)^2}\right)[/tex]

[tex]\implies \displaystyle -\frac{\pi^2}{12} - \frac19 \underbrace{\sum_{n=1}^\infty \frac{(-1)^n}{n^2}}_{-\frac{\pi^2}{12}} = - \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right)[/tex]

[tex]\implies \displaystyle \sum_{n=0}^\infty (-1)^{n+1} \left(\frac1{(3n+2)^2} - \frac1{(3n+1)^2}\right) = \frac{2\pi^2}{27}[/tex]

So, we conclude that

[tex]\displaystyle \int_0^{\frac\pi2} \sqrt[3]{\tan(x)} \ln(\tan(x)) \, dx = \frac94 \times \frac{2\pi^2}{27} = \boxed{\frac{\pi^2}6}[/tex]