to get the equation of any straight line, we simply need two points off of it, let's use the points in the picture below.
[tex](\stackrel{x_1}{-2}~,~\stackrel{y_1}{2})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{-1}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-1}-\stackrel{y1}{2}}}{\underset{run} {\underset{x_2}{4}-\underset{x_1}{(-2)}}}\implies \cfrac{-3}{4+2}\implies \cfrac{-3}{6}\implies -\cfrac{1}{2}[/tex]
[tex]\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{2}=\stackrel{m}{-\cfrac{1}{2}}(x-\stackrel{x_1}{(-2)}) \\\\\\ y-2 = -\cfrac{1}{2}(x+2)\implies y-2=-\cfrac{1}{2}x-1\implies y=-\cfrac{1}{2}x+1[/tex]
what is the x-intercept of the function graphed below
A. (2,0)
B. (0,6)
C. (0,2)
D. (6,0)
Answer:
A. (2, 0)
Step-by-step explanation:
The function crosses the x-axis at point (2,0). Therefore, that is the x-intercept.
hope this helps :)
The length breadth and height of a sectangular box are 24 cm, 6 cm and 2cm respectevely. Find the area of largest surface of box?
Answer:
[tex]408cmx^{2}[/tex]
Step-by-step explanation:
A=2(wl+hl+hw)=2·(2·6+24·6+24·2)= 408cm²
length= 6
width= 2
height= 24
to find:the area if the box.
solution:[tex]a = lwh[/tex]
[tex]a = 2(wl + hl + hw)[/tex]
[tex]a = 2 \times (2.6 + 24.6 \times + 24.2)[/tex]
[tex]a = 408 {cm}^{2} [/tex]
therefore, the area of the box is 408 square centimeters.
What is the area
Use 3.14 for pi
Round to the nearest hundredth
Answer: The answer is 120
Step-by-step explanation:
5x4x3=60 6x4x5+120 divided by 2= 60
60+60= 120
177 568 123 nearest hundred thousand
Answer:
177 600 000 is the answer
Marilyn created this box and whiskers plot to represent the number of inches of snow that fell during the winter in several different cities.
(a) What was the least amount of snowfall in any of the cities? Show your work
(b) What is the median amount of snowfall? Show your work
(c) In which quarter is the data most concentrated? Explain how you know.
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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A triangle has a perimeter of 51 cm. If the three sides of the triangle are n, 4n-4, and 4n-8, what is the length of each side?
Answer:
7,24,20
Step-by-step explanation:
The perimeter is simply the lengths of the sides, added up
n+4n-4+4n-8=9n-12=51 cm
9n=51+12
9n=63
n=7
plug in 7 for n to get:
7,4*7-4,4*7-8
or
7,24,20
Par on a golf course is 72. If a golfer shot rounds of 74, 67, and 73 in a tournament, what will she need to shoot on the final round to average par?
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
[tex] \displaystyle\rm\int \limits_{0}^{ \frac{\pi}{2} } \sqrt[3]{tanx} \ln(tanx)dx[/tex]
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]
You know how yo make 7 different types of cookies.you have time to make any 5 of them how many different combinations of cookies type can you make?
Solve 30% of blank =90
Answer:
300
Step-by-step explanation:
Let y = unknown number
30% of y = 90
⇒ 0.3 × y = 90
⇒ y = 90 ÷ 0.3
⇒ y = 300
A small refrigerator is a cube with a side length of 13 inches.
Use the formula S = 6s2 to find the surface area of the cube.
The surface area of the cube is
__________ square inches.
6 x 13^2=1,014
A=1,014 in
The surface area of the cube is 1,014 square in
Do not include "g(f(1)) =" in your answer
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
PLEASE HELP. I'M DESPERATE! 100 POINTS AND BRAINLY IF CORRECT!
The formula for the volume of a right circular cylinder is
V = πr² h. If r = 2b and h = 5b +3 then what is the
volume of the cylinder in terms of b?
A 10b2 + 67b
B 20πb³ + 12πb²
C 20π²6³ +12π²b²
D 50пь3 + 20пb2 + 90nb
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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24) Find center vertices foci asymptotes
The given equation represents a hyperbola. Its main features are:
center - (1,-3)vertices - V1= (1,-3+[tex]\sqrt{2}[/tex]) / V2= (1,-3-[tex]\sqrt{2}[/tex])foci - F1= (1 , -3+ [tex]2\sqrt{5}[/tex] )/ F2= (1 , -3 - [tex]2\sqrt{5}[/tex] )asymptotes = [tex]\pm \frac{1}{3}\left(x-1\right)-3[/tex]HyperbolaA 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:
Find the centerThe coordinates for center is (h,k). Thus, the center is (1,-3).
Find the verticesThe 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])
Find the fociThe 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] )
Find the asymptotesThe 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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Liam wants to put some money in a simple interest account.
It pays 6% interest annually for 2 years.
Liam would like to earn $750 in interest.
How much money does he need to put in?
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,
[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.
Sean went bike riding yesterday. Use the graph below to determine how fast he was riding in miles per hour.
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
Karen thought of a number between 10 and 20. She multiplied it by 4, then divide the result by 2. Between which two numbers does the final number lie?
Vincent's Restaurant bought 6 pounds of onions. The restaurant bought 5 1/3 times as much potatoes as onions. How many pounds of potatoes did the restaurant buy?
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.
Which relationship is NOT a function?
HELPPP!!!!!!!!!! HELP!!!!!!
Answer:
I think that it is the one on the bottom left
Step-by-step explanation:
I am so sorry if it is wrong he he
Identify the coefficient in the monomial 13y^3.
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
What can be concluded if <1≈ <7?
4
→ P
3
2
8
15
→ 9
7
6
A. q||t
B. p||q
C. t|q
D. t|p
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
A rectangular picture has a length of 40 inches and a width of 9 inches.
What is the length of the diagonal of the picture in inches?
Enter your answer in the box.
inches
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
You operate a dog-walking service. You have 50 customers per week when you charge walk. For each $1 decrease in per your fee for walking a dog, you get 5
more customers per week. Cand get 5. ever earn. you $7.50 in a week? Explain.
What quadratic equation in standard form use to model this situation?
How can the discriminate of the equation help you solve the problem?
Please help! I’m really confused :(
Answer:
No.
Step-by-step explanation:
The maximum that can be earned is $720, which results from walking 60 dogs at $12 per walk.
Please help!
calculate [tex]$\frac{1}{\frac{1}{7}\cdot\frac{1}{-8}}$[/tex]
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
You are currently paying $1,800 per year for insurance. The insurance company informs you that after five years of no
claims and moving into a different risk pool your rates will go down 13%. How much will you be paying next year?
a) $234
b) $1,566
c) $1,777
d) $1,800
An ipod is regularly $140 but it is on sale for 30% off the price. What is the discount? What is the sale price?
Answer:
discount: 30% sale price: $98
Step-by-step explanation:
Simply multiply 140 by 70%(0.7) to get 98
Question
Write the following function in terms of its cofunction.
tan(78)
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.
Pls help pls helppp i will giving breanlist
Let f (x) = 7+3 and g(x) = Vx+5. What is the domain of (fºg)(x)?
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, ∞)
I forgot how to convert the inequalities so if someone could please help me that would be great!
WHEN GRAPHING
Step 1. Convert to y = mx + b
Equation 1: y = 3x/2 - 16/2Equation 2: y = -5 - 2xStep 2. Graph like normal
Find zeros & plotFind y-intercept & plotStep 3. Shade as indicated by the inequality symbol
> or ≥ = above line< or ≤ = below lineStep 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!
when solving a quadratic equation by factoring you must set y equal to what value? y=x2 + 3x - 4
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