if g(x)=-2f(x), whats the right graphing

C

1, -9 , 81 , -729 .....

ratio :-

R1 = -9/1 = -9

R2 = 81/-9 = -9

R3 = -729 / 81 =-9

so ,. R1= R2= R3

Answer:

-4

Step-by-step explanation:

7x^3 + 5x^2 - 2
= 7*-1 + 5*1 - 2

= -7 + 5 - 2

= -4

Answer:

125

Step-by-step explanation:

The sum of two interior angles in a triangle is equal to an exterior angle that is supplementary to the third interior angle.

We can write the following equation according to this information and that will help us find the value of x:

65 + 60 = x add like terms

125 = x is the answer we are looking for.

[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\textsf{equation:}[/tex]

[tex]\large\textsf{a = 60 + 65}[/tex]

[tex]\huge\textsf{solving:}[/tex]

[tex]\large\textsf{a = 60 + 65}[/tex]

[tex]\large\textsf{60 + 65 = a}[/tex]

[tex]\huge\textsf{simplify it:}[/tex]

[tex]\large\textsf{a = 125}[/tex]

[tex]\huge\textsf{therefore, your answer should be:}[/tex]

[tex]\huge\boxed{\mathsf{a =} \frak{125}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

  x = 8

Step-by-step explanation:

Diagonals of a kite cross at right angles. That gives us a relation that can be solved for x.

The measure shown is equal to the angle measure of 90°.

  14x -22 = 90

We can solve this 2-step linear equation in the usual way.

  14x = 112 . . . . . . step 1, add the opposite of the constant to get x alone

  x = 112/14 = 8 . . . step 2, divide by the coefficient of x

The value of x is 8.

Answer:

Step-by-step explanation:

The average of the five consecutive numbers ending with b in discuss when expressed in terms of a is; Choice D; a+3.

First, since it was given in the task content that the average of six positive consecutive odd integers starting with a is equal to b, it therefore follows that;

(a+a+2+a+4+a+6+a+8+a+10)/6 = b

6b=6a+30

b=a+5

Also, let the average of the consecutive intergers ending with b be denoted by; x.

(b+b-1+b-2+b-3+b-4)/5 = x

=(5b-10)/5

=b–2

The average, x=b – 2 (where b = a-5)

Ultimately, the value of the required average is; = a+5-2 = a+3.

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Answer:

It's easy to solve the problem.

let's start...

given equation:- y = 2x+2

X y

1 (2× 1 )+ 2 = 4 .

3 (3×2)+2 = 8 .

9 (9×2) + 2 = 20 .

10. (10× 2)+2 = 22 .( table completed)

just put the value of X in the expression.

Answer:

2.8m²

Step-by-step explanation:

p = F / A

A = F / p

A = 84 / 30

A = 2.8m²

Answer:

[tex]\dfrac{8}{\sin(6^\circ)} \approx 76.53[/tex]

Step-by-step explanation:

[tex]UX \sin(6^\circ) = 8 \Rightarrow \boxed{UX = \dfrac{8}{\sin(6^\circ)} \approx 76.53}[/tex]

The missing value in the logarithm are as follows:

Using logarithm rule,

logₐ b - logₐ c = logₐ (b / c)

logₐ b + logₐ c = logₐ (b × c)

Therefore,

log₃ 7 - log₃ 2 = log₃ (7 / 2)

log₉ 7 + log₉ 4 = log₉ (7 × 4) =  log₉ 28

log₆ 1 / 81 = log₆ 81⁻¹ = log₆ 3⁻⁴ =  - 4 log₆ 3

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Answer:

c)  3 units

d)  g(x) - f(x) = x² + 2x

e)  (-∞, -2] ∪ [0, ∞)

Step-by-step explanation:

To calculate the length of FC, first find the coordinates of point C.

The y-value of point C is zero since this is where the function f(x) intercepts the x-axis.  Therefore, set f(x) to zero and solve for x:

[tex]\implies 1-x^2=0[/tex]

[tex]\implies x^2=1[/tex]

[tex]\implies \sqrt{x^2}=\sqrt{1}[/tex]

[tex]\implies x= \pm 1[/tex]

As point C has a positive x-value,  C = (1, 0).

To find point F, substitute the x-value of point C into g(x):

[tex]\implies g(1)=2(1)+1=3[/tex]

⇒ F = (1, 3).

Length FC is the difference in the y-value of points C and F:

[tex]\begin{aligned} \implies \sf FC& = \sf y_F-y_C\\ & = \sf 3-0\\ & =\sf 3\:units \end{aligned}[/tex]

Given functions:

[tex]\begin{cases}f(x)=1-x^2\\ g(x)=2x+1 \end{cases}[/tex]

Therefore:

[tex]\begin{aligned}\implies g(x)-f(x) & = (2x+1) - (1-x^2)\\& = 2x+1-1+x^2\\& = x^2+2x\end{aligned}[/tex]

The values of x for which g(x) ≥ f(x) are where the line of g(x) is above the curve of f(x):

Point A is the y-intercept of both functions, therefore the x-value of point A is 0.

To find the x-value of point E, equate the two functions and solve for x:

[tex]\begin{aligned}g(x) & = f(x)\\\implies 2x+1 & = 1-x^2\\x^2+2x & = 0\\x(x+2) & = 0\\\implies x & = 0, -2\end{aligned}[/tex]

As the x-value of point E is negative ⇒ x = -2.

Therefore, the values of x for which g(x) ≥ f(x) are:

Answer:

a)

A = (0, 1)

B = (-1, 0)

C = (1, 0)

D = (-0.5, 0)

b) E = (-2, -3)

c) FC = 3 units

d) x² + 2x

e) x ≤ -2 and x ≥ 0

Explanation:

This question displays one equation of a linear function g(x) = 2x + 1 and a parabolic function f(x) = 1 - x².

a)

A point is where the linear function cuts the y axis.

y = 1 - (0)²

y = 1

A = (0, 1)

B and C point is where the parabolic function cuts the x axis.

1 - x² = 0

-x² = -1

x² = 1

x = ±√1

x = -1, 1

B = (-1, 0), C = (1, 0)

D point is where the linear function cuts x axis.

2x + 1 = 0

2x = -1

x = -1/2 or -0.5

D = (-0.5, 0)

b)

E point is where both equations intersect each other.

y = y

2x + 1 = 1 - x²

x² + 2x = 0

x(x + 2) = 0

x = 0, x = -2

y = 1, y = -3

E = (-2, -3)

c)

C : (1, 0)

To find F point

y = 2(1) + 1

y = 3

F : (1, 3)

[tex]\sf Distance \ between \ two \ points = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}[/tex]

[tex]\sf d = \sqrt{(1 - 1)^2 + (3 - 0)^2}[/tex]

[tex]\sf d = \sqrt{0 + 3^2}[/tex]

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

FC length = 3 units

d)

g(x) - f(x)

(2x + 1) - (1 - x²)

2x + 1 - 1 + x²

x² + 2x

e)

g(x) ≥ f(x)

2x + 1 ≥ 1 - x²

x² + 2x ≥ 0

x(x + 2) ≥ 0

[tex]\boxed{If \ x \ \geq \ \pm \ a \ then \ -a \ \leq x \ \ and \ x \ \geq \ a }[/tex]

x ≤ -2 and x ≥ 0

Answer:

1. f(x) is reflected across the x-axis

2. f(x) is translated 1 unit up

3. f(x) is vertically scaled by a factor of 2

4. f(x) is reflected across the x-axis AND is vertically scaled by a factor of 2

5. f(x) is vertically scaled by a factor of 3 AND is translated 1 unit down

6. f(x) is vertically scaled by a factor of 1/6 AND is translated 1 unit up

Solving question:

(1)  [tex]g(x) = -f(x)[/tex]

This graph has been reflected in the x axis. Equation:  [tex]\sf g(x) = -\dfrac{2}{x}[/tex]

(2) [tex]g(x) = f(x) + 1[/tex]

Graph has been translated 1 units up vertically. Equation: [tex]\sf g(x) = \dfrac{2}{x} +1[/tex]

(3) [tex]g(x) = 2f(x)[/tex]

This graph has been stretched vertically by a factor of 2. Equation: [tex]\sf g(x) = \dfrac{4}{x}[/tex]

(4) [tex]g(x) = -2f(x)[/tex]

This graph has been reflected in the x axis and stretched vertically by a factor of 2. Equation: [tex]\sf g(x) = -\dfrac{4}{x}[/tex]

(5)  [tex]g(x) = 3f(x) - 1[/tex]

This graph has been stretched vertically by a factor of 3 and translated 1 units down. Equation: [tex]\sf g(x) = \dfrac{6}{x} -1[/tex]

(6)   [tex]g(x) = \frac{1}{6} f(x) + 1[/tex]

This graph has been stretched vertically by a factor of 1/6 and translated 1 units up. Equation: [tex]\sf g(x) = \dfrac{1}{3x} +1[/tex]

The span of 3 vectors can have dimension at most 3, so 9 is certainly not correct.

Check whether the 3 vectors are linearly independent. If they are not, then there is some choice of scalars [tex]c_1,c_2,c_3[/tex] (not all zero) such that

[tex]c_1 (2,-1,1) + c_2 (3,1,1) + c_3 (1,2,0) = (0,0,0)[/tex]

which leads to the system of linear equations,

[tex]\begin{cases} 2c_1 + 3c_2 + c_3 = 0 \\ -c_1 + c_2 + 2c_3 = 0 \\ c_1 + c_2 = 0 \end{cases}[/tex]

From the third equation, we have [tex]c_1=-c_2[/tex], and substituting this into the second equation gives

[tex]-c_1 + c_2 + 2c_3 = 2c_2 + 2c_3 = 0 \implies c_2 + c_3 = 0 \implies c_2 = -c_3[/tex]

and in turn, [tex]c_1=c_3[/tex]. Substituting these into the first equation gives

[tex]2c_1 + 3c_2 + c_3 = 2c_3 - 3c_3 + c_3 = 0 \implies 0=0[/tex]

which tells us that any value of [tex]c_3[/tex] will work. If [tex]c_3 = t[/tex], then [tex]c_1=t[/tex] and [tex]c_2 = -t[/tex]. Therefore the 3 vectors are not linearly independent, so their span cannot have dimension 3.

Repeating the calculations above while taking only 2 of the given vectors at a time, we see that they are pairwise linearly independent, so the span of each pair has dimension 2. This means the span of all 3 vectors taken at once must be 2.

The speed of the train is 150 km/hour.

The distance covered by the train in 3 4/5 hours is 570 km.

The speed of a body is calculated using the formula, speed = distance/time.

The distance covered is calculated using the formula, distance = speed*time.

The time taken by a body is calculated using the formula, time = distance/speed.

In the question, we are asked for the speed of a train, when it covers a distance of 225 km in 1 1/2 hours.

Distance = 225 km.

Time = 1 1/2 hours = 1.5 hours.

Speed = Distance/Time = 225/1.5 km/hour = 150 km/hour.

Now, we are asked to calculate the distance covered by the train in 3 4/5 hours.

Speed = 150 km/hour.

Time = 3 4/5 hours = 3.8 hours.

Distance = Speed*Time = 150*3.8 km = 570 km.

Thus, the speed of the train is 150 km/hour.

The distance covered by the train in 3 4/5 hours is 570 km.

The provided question is incorrect. The correct question is:

"A train at a uniform speed covers a distance of 225 km in 1 1/2 hours. Find the speed of the train and the distance covered in 3 4/5 hours.

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Answer:

x = 12cos(35)

Step-by-step explanation:

Well you have hyp and adj, meaning cosine. and u have the angle and need the adjacent, so side(cos(theta)).

12cos(35).

If you fix a point on the curve in the given interval, and revolve that point about the [tex]x[/tex]-axis, it will trace out a circle with radius given by the function value [tex]y[/tex] for that point [tex]x[/tex]. The perimeter of this circle is then [tex]2\pi(8\sqrt x) = 16\pi \sqrt x[/tex].

The surface in question is essentially what you get by joining infinitely many of these circles at every point in the interval [0, 9].

So, the surface area is given by the definite integral

[tex]\displaystyle \int_0^9 16\pi \sqrt x \, dx = 16\pi\times\frac23 x^{3/2}\bigg|_{x=0}^{x=9} = \frac{32\pi}3 \left(9^{3/2} - 0^{3/2}\right) = \boxed{288\pi}[/tex]

[tex] \sqrt[3]{b {}^{2} } = b {}^{ \frac{2}{3} } [/tex]

[tex]f(4) = 2(3) {}^{4} = 2 \times 81 = 162[/tex]

[tex]f( \frac{1}{2} ) = \frac{1}{3} (9) {}^{ \frac{1}{2} } + 5 = \frac{ \sqrt{9} }{ 3} + 5 = 6[/tex]

[tex]f(5) = - 4(2) {}^{ - 5 + 1} = - 4(2) {}^{ -4} = \frac{2 {}^{2} }{2 {}^{4} } = \frac{1}{2 {}^{2} } = \frac{1}{4} [/tex]

[tex]f(10) = 2e {}^{0.15 \times 10} = 2e {}^{1.5} = 8.96[/tex]

[tex]y = 10(b) {}^{x} \: \: \\ 10 = 10(b) {}^{0} \: duh \\ 2 = 10b {}^{1} \\ b = \frac{2}{10} = \frac{1}{5} [/tex]

[tex]y = a(b) {}^{x} \\ 3 = a(b) {}^{0} \\ a = 3 \\ \\75 = 3(b) {}^{2} \\ b {}^{2} = 25 \\ b = + 5 \: \: \: or \: \: \: \: b = - 5[/tex]

[tex]s = 1000(1 + \frac{0.045}{4} ) {}^{4 \times 10} = 1564.38[/tex]

[tex]s = 2500.e {}^{0.07 \times 20} = 10138[/tex]

Answer:

the correct answer is 25

Answer:

Angle x= 60°

Step-by-step explanation:

For the triangle HIL, you're going to add 30+90=120

Angles in a triangle add up to 180.

So, therefore,

180-120=60   so angle L must equal 60.

HJ is a straight line and angles in a straight line add up to 180.

180-90=90

Angle JIL is equal to 90.

To find y you add 90+45 which equals 135.

Again, angles in a triangle add up to 180.

180-135=45

So,

y = 45

Now we are told that IJK is a right angle and that we are given that IJL is 45. 45 is half of 90 so LJK must be 45.

To find angle JLK we must add angle L and angle y.

60+45=105

Angles in a straight line add up to 180. So,

180-105=75

75 = Angle JLK

75+45=120

Angles in a triangle add up to 180 so,

180-120= 60

Angle x= 60°

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Answer:

D.

Step-by-step explanation:

We can solve by simply isolating x:

[tex]-122 < -3(-2-8x)-8x\\-122 < 6+24x-8x\\-122 < 6+16x\\-128 < 16x\\-8 < x\\OR\\x > -8[/tex]

You can check by plugging in any value greater than -8

-122<-3(-2-8(-7))-8*-7

-122<-3(-2+56)+56

-122<-3(54)+56

-122<-162+56

-122<-106

Answer:

80 in.²

Step-by-step explanation:

The total surface area of the pyramid is the sum of the area of the base and the areas of the 4 triangular sides.

Square: area = s²

Square: side = 4 in.

Triangular side: area = bh/2

Triangular side: base = 4 in.; height = 8 in.

Area of the base: s² = (4 in.)² = 16 in.²

Total area of the 4 triangular sides: 4 × bh/2 = 2bh = 2 × 4 in. × 8 in. = 64 in.²

Surface area = 16 in.² + 64 in.² = 80 in.²

The behavior of the graph with the equation y=x66x5+50x3+45x2108x108 at each of its zeros will be as follows: two of them will resemble a quadratic function, while one of them will resemble a linear function.

This is further explained below.

Generally, a diagram that illustrates the relationship between variable quantities, usually consisting of two variables, with each variable being measured along one of a pair of axes that are intersected at right angles.

In conclusion, At each of its zeroes, the graph y=x66x5+50x3+45x2108x108 will exhibit one linear behavior and two behaviors that resemble quadratic functions.

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Answer:

0

Step-by-step explanation:

because 1-0=1*2^3=8

because2^3=8

Answer:

3:8 3 sodas for 8 drinks

Step-by-step explanation:

6/16

3/8

Answer:

line MN

Step-by-step explanation:

If two planes intersect, then the intersection is a line.

Answer: line MN

Answer:

Step-by-step explanation:

x² + 2x = x(x + 2)

[tex]\sf \dfrac{x +5}{x + 2}-\dfrac{x+1}{x^2+2x}=\dfrac{x + 5}{x +2}-\dfrac{x+1}{x(x+2)}[/tex]

LCM = x(x+2)

                        [tex]\sf =\dfrac{(x+5)*x}{(x+2)*x}-\dfrac{x+1}{x(x+2)}\\\\=\dfrac{x*x + 5*x}{x^2+2x}-\dfrac{x+1}{x^2+2x}\\\\=\dfrac{x^2+5x - (x+1)}{x^2+2}\\\\=\dfrac{x^2+5x -x - 1}{x^2+2x)}\\\\=\dfrac{x^2+4x-1}{x^2+2x}[/tex]

By changing the central angle, ∠CAB, the inscribed angle, ∠CDB has the following data sets:

m∠BAC (β)                                m∠BDC (α)

42°                                               84°

40°                                               80°

45°                                               90°

35°                                               70°

52°                                               104°

A circle can be defined as a closed, two-dimensional curved geometric shape with no edges or corners. Also, a circle refers to the set of all points in a plane that are located at a fixed distance (radius) from a fixed point (central axis).

In Geometry, a circle is considered to be a conic section which is formed by a plane intersecting a double-napped cone that is perpendicular to a fixed point (central axis) because it forms an angle of 90° with the central axis.

The inscribed angle theorem states that the measure of an inscribed angle is one-half the measure of the intercepted arc in a circle. Thus, this is given by this mathematical expression:

m∠BDC = ½ × m∠BAC.

For this exercise, we would change the central angle, ∠CAB, so that the inscribed angle, ∠CDB can have the following data sets:

m∠BAC (β)                                m∠BDC (α)

42°                                               84°

40°                                               80°

45°                                               90°

35°                                               70°

52°                                               104°

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Answer:

plato

Step-by-step explanation:

m∠BAC m∠BDC

50° 25°

70° 35°

90° 45°

125° 62.5°

150° 75°

Answer:

EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE

Answer:

3.2

Step-by-step explanation:

4 x 8 / 8 + 2

32/10

=3.2

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The best sampling strategy would be a stratified sample.

Samples may be classified as:

For this problem, the 4 different brands of the recorders must be considered, hence the buyers should be divided into groups, and a proportion of each group should be sampled, hence a stratified sample should be used.

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