Explain why the absolute value function y = |x| is an example of a piecewise function

Answer:

Never:

Step-by-step explanation:

because its more understandable we know this because it wouldnt be always

Step-by-step explanation:

perp. -1/2

y + 3 = -1/2(x - 4)

y + 3 = -1/2x + 2

y = -1/2x - 1

Answer:

x

∈

R

y

≤

1

Step-by-step explanation:

Answer:

[tex]y\geq -1[/tex]

Step-by-step explanation:

see graph below for reference

Answer:

54, 63

Step-by-step explanation:

6 x 9 = 54

7 x 9 = 63

Answer:

there is no pic -_-

Step-by-step explanation:

Answer:

By the way that it is worded, I believe that you are only allowed to put 1 charm on the bracelet, so there are 9 ways to pick a charm.

Hope this helps(and also if you have a more refined wording please put it in the comments)!

Answer:

By the way that it is worded, I believe that you are only allowed to put 1 charm on the bracelet, so there are 9 ways to pick a charm.

Answer A is correct. C is correct.

Step-by-step explanation:

Step-by-step explanation:

Given

GPA = NO + B1Study + B2work + B3leisure + B4sleep + u

1. No it makes no sense. There is an intercorellation between these explanatory variables. Changing study while holding the rest variables constant or fixed would not make sense. If study us changed then one or more of the other predictors should be changed so that they can still be equal to 168

2. Study is a perfect linear function of work, leisure and sleep. There is the problem of multicollinearity among these independent variables. One assumption of the MLR is no multicollinearity.

3. To reformulate model, you have to drop the variable that is having the issue. Let's say that leisure is dropped from the model. After this the assumption is satisfied.

GPA = b0 + B1Study + b2sleep + b3work + u

Answer:

D. The same point

Step-by-step explanation:

An equilateral triangle can be defined as a type of triangle that has all of its sides being equal in length.

This ultimately implies that, an equilateral triangle comprises of three (3) congruent sides and as such creates three (3) congruent angles that measures 60° each. Also, the sum of the congruent angles of an equilateral is always equal to 180°.

Hence, in an equilateral triangle, the incenter, orthocenter, circumcenter, and centroid are the same point i.e all of the three (3) centers fall on the same point or spot.

Answer: so here is what you have to do

Step-by-step explanation:Determine Whether a Number

Is a Solution to an Equation

2 Solve Linear Equations

3 Determine Whether an Equation Is a Conditional Equation, Identity, or Contradiction

Preparing for Linear Equations

Before getting started, take this readiness quiz. If you get a problem wrong, go back to the section cited and review the material.

P1. Determine the additive inverse of 5.

P2. Determine the multiplicative inverse of - 3.

P3. Use the Reduction Property to simplify 15 # 5x.

P4. Find the Least Common Denominator of 3 and 5 .

8 12

P5. Use the Distributive Property to remove the parentheses: 61z - 22

P6. What is the coefficient of - 4x?

P7. Simplify by combining like terms: 41y - 22 - y + 5 P8. Evaluate the expression -51x + 32 - 8 when x = -2

P9. Isx = 3inthedomainof 2 ?Isx = -3inthe domain? x + 3

[Section R.3, p. 21] [Section R.3, pp. 23–24]

[Section R.3, pp. 25–26] [Section R.3, pp. 27–28]

[Section R.3, pp. 29–30] [Section R.5, p. 41] [Section R.5, pp. 41–43] [Section R.5, pp. 40–41]

[Section R.5, pp. 43–44]

 In Words

In the equation 2

2 + 5 is the left side of the equation and 0 is the right side. In this equation only the left side contains the variable, y. In the equation 4 +5=-2x+10,theexpression

4 + 5 is the left side of the equation and - 2 + 10 is the right side. In this equation both sides have expressions that contain the variable, x.

1 Determine Whether a Number Is a Solution to an Equation

An equation in one variable is a statement made up of two expressions that are equal, and in the statement, at least one of the expressions contains the variable. The expres- sions are called the sides of the equation. Examples of equations in one variable are

2y + 5 = 0 4x + 5 = -2x + 10 3 = 9 z+2

In this section, we will concentrate on solving linear equations in one variable.

Answer:

6 units to the right, 3 units down

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

8 * 3 = 24

12 * 2 = 24

(the star is the multiplication sign)

Hope this helps :)

Answer:

it is 222

Step-by-step explanation:

1440 divided by 20 :)

Answer:

c=5/72

Step-by-step explanation:

5/9=14c-5/12

Switch sides

14c-5/12=5/9

Add 5/12 tp both sides

14c-5/12+5/12=5/9+5/12

Simplify;

14c-5/12+5/12

=14c

Simplify;

5/9+5/12

LCM of 9, 12;

36

The least common multiplier of a, b is the smallest positive number that is divisible by both a and b.

Prime factorization of 9;

3*3

Prime factorization of 12;

2*6

=2*2*3

=3*3*2*2

=36

5/9=5*4/9*4=20/36

5/12=5*3/12*3=15/36

=20/36+15/36

20+15/36

=35/36

14c=35/36

14c/14=35/36/14

14c/14=c

35/36*14

=35/504

=5/72

Answer c=5/72

Answer to the following question is as follows;

Value of C = 5 / 72

Step-by-step explanation:

Given:

[tex]\frac{5}{9} = 14c - \frac{5}{12}\\\\[/tex]

Find:

Value of C

Computation:

[tex]\frac{5}{9} = 14c - \frac{5}{12}\\\\[/tex]

By taking 5/12 to LHS Side

[tex]\frac{5}{9} + \frac{5}{12} = 14c\\\\[/tex]

By taking LCM;

[tex]\frac{(4)(5) + (3)(5)}{36} = 14c\\\\[/tex]

[tex]\frac{20 + 15}{36} = 14c\\\\[/tex]

[tex]\frac{35}{36} = 14c\\\\[/tex]

By transfer 14 to LHS side

[tex]\frac{35}{36 \times 14 } = c\\\\[/tex]

[tex]\frac{5}{36 \times 2 } = c\\\\[/tex]

[tex]\frac{5}{72} = c\\\\[/tex]

Value of C = 5 / 72

Learn more;

https://brainly.com/question/21481562?referrer=searchResults

Answer:

[tex]\theta=41.18^{\circ}[/tex]

Step-by-step explanation:

Given that,

The height of the tree, h = 14 ft

The height of the shadow, b = 16 ft

We need to find the angle of elevation of the sun from the ground. Let the angle be θ. We can use trigonometry to find it. So,

[tex]\tan\theta=\dfrac{P}{B}\\\\\tan\theta=\dfrac{14}{16}\\\\\theta=41.18^{\circ}[/tex]

So, the required angle of elevation of the sun is equal to [tex]41.18^{\circ}[/tex].

Answer:

[tex]l = \sqrt[3]{\frac{2V}{7}}[/tex]     [tex]b = \sqrt[3]{\frac{2V}{7}}[/tex]       [tex]h = \sqrt[3]{\frac{49V}{4}}[/tex]

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

[tex]V = l * b * h[/tex]

Make h the subject of the formula

[tex]h = \frac{V}{lb}[/tex]

The surface area (S) of the aquarium is:

[tex]S = lb + 2(lh + bh)[/tex]

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

[tex]C = 7 * lb + 1 * 2(lh + bh)[/tex]

[tex]C = 7lb + 2(lh + bh)[/tex]

[tex]C = 7lb + 2h(l + b)[/tex]

Substitute [tex]\frac{V}{lb}[/tex] for h in the above equation

[tex]C = 7lb + 2*\frac{V}{lb}(l + b)[/tex]

[tex]C = 7lb + \frac{2V}{lb}(l + b)[/tex]

[tex]C = 7lb + \frac{2V}{b} + \frac{2V}{l}[/tex]

Differentiate with respect to l and with respect to b

[tex]C_l=7b - \frac{2V}{l^2}[/tex] [tex]=0[/tex]

[tex]C_b=7l - \frac{2V}{b^2}[/tex] [tex]=0[/tex]

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

[tex]7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}[/tex]

[tex]7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}[/tex]

By comparison:

[tex]l =b[/tex]

[tex]C_l=7b - \frac{2V}{l^2}[/tex] [tex]=0[/tex] becomes

[tex]7l - \frac{2V}{l^2}=0[/tex]

[tex]7l = \frac{2V}{l^2}[/tex]

Cross Multiply

[tex]7l^3 = 2V[/tex]

Solve for l

[tex]l^3 = \frac{2V}{7}[/tex]

[tex]l = \sqrt[3]{\frac{2V}{7}}[/tex]

Recall that: [tex]l =b[/tex]

[tex]b = \sqrt[3]{\frac{2V}{7}}[/tex]

Also recall that:

[tex]h = \frac{V}{lb}[/tex]

[tex]h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}[/tex]

[tex]h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}[/tex]

Apply law of indices

[tex]h = \sqrt[3]{\frac{49V^3}{4V^2}}[/tex]

[tex]h = \sqrt[3]{\frac{49V}{4}}[/tex]

The dimension that minimizes the cost of material of the aquarium is:

[tex]l = \sqrt[3]{\frac{2V}{7}}[/tex]     [tex]b = \sqrt[3]{\frac{2V}{7}}[/tex]       [tex]h = \sqrt[3]{\frac{49V}{4}}[/tex]

Answer:

a - Mutual

b - (x+21)+(2x+3)=90

c - 3x +24 =90

3x=90-24

3x=66

X=22

Answer:

1.5x-y=6

x-y=5+6

x-y=11

2.x + 6y =4

x+y=6-4

X+y=2

xy=2

SORRY THE WRONG

Step-by-step explanation:

FROM INDONESIAN

Answer:

Q.1 a²+10a+24

splitting the middle.

a²+6a+4a+24

=a(a+6)+4(a+6)

(a+6)(a+4)

Q.2 x²+9x+18

splitting the middle

x²+6x+3x+18

x(x+6)+3(x+6)

(x+6)(x+3)

Answer:

182.82812

Step-by-step explanation:

Answer:

$12,375

$12820.50

Step-by-step explanation:

To obtain a rounded value of $28 (to nearest dollar) the lowest and highest possible value before rounding will be : 27.50 and 28.49

Hence,

Lower bound = $27.50

Upper bound = $28.49

Therefore. Cost of making 450 chairs ;

27.50 * 450 = $12,375 (lower bound)

28.49 * 450 $12820.50 (upper bound)

Answer:

A,B

Step-by-step explanation:

I think bc I had the same question well something like this in A in B was it

Answer:

65

Step-by-step explanation:

Answers:

===========================================

Explanation:

The lowest point occurs when y = -2. The highest point occurs when y = 2.

The vertical distance between the highs and lows is 4 units (since 2-(-2) = 2+2 = 4). This cuts in half to get 4/2 = 2.

The amplitude is 2. It measures the vertical distance from the midline to either the peaks or valleys.

-----------------------------

Note that this curve has two lowest points and one highest point shown. What we'll do is shift the graph over to the right so that each of those lowest points snap to a grid point. The lowest point on the left will move to where x = -pi, while the lowest point on the right snaps to x = 3pi/2. We'll keep each point at the same y coordinate.

The horizontal distance from x = -pi to x = 3pi/2 is

(3pi/2) - (-pi) = 3pi/2 + 2pi/2 = 5pi/2

This represents the horizontal distance from the left valley point to the right valley point. This is one full period of the curve. After we arrive at the right valley point, we repeat the process of going up and coming back down in this exact same fashion. The process is repeated forever to generate the cosine curve. The curve repeats itself every 5pi/2 units along the x axis.

Side note: any phase shift of a curve will have the same period as the original curve. This is why we're allowed to shift left or right to help find the period.

You have 2 points with the same x value of 2 and 2 points with the same y value of 4.

You now need 2 points with the same x value of 4 ( you are given 1) and 2 y values of 1 ( you are given 1.

The missing point would need to be (4,1)

The width would be x2-x1 = 4-2 = 2 units

The length would be y2-y1 = 4-1 = 3 units

Area = 3x 2 = 6 square units

where is the question I dont see it

Answer:

Step-by-step explanation:

From the given information:

[tex]a_n = a_{n-1} + a_{n-2}; \ \ \ n \ge 2 \\ \\ a_o = 1 \\ \\ a_1 =1 \ \ \ \ \ since \ \ a_o = a_1 = 1[/tex]

A)

[tex]a_n - a_{n-1} - a_{n-2} = 0 \\ \\ \implies \sum \limits ^{\infty}_{n=2}(a_n -a_{n-1}-a_{n-2} ) x^n = 0 \\ \\ \implies \sum \limits ^{\infty}_{n=2} a_nx^n - \sum \limits ^{\infty}_{n=2} a_{n-1}x^n - \sum \limits ^{\infty}_{n=2}a_{n-2} x^n = 0 \\ \\ \implies (a(x) -a_o-a_1x) - (x(a(x) -a_o)) -x^2a(x) = 0 \\ \\ \implies a(x) (1 -x-x^2) -a_o-a_1x+a_ox = 0 \\ \\ \implies a(x)(1-x-x^2)-1-x+x=0 \\ \\ \implies a(x) (1-x-x^2) = 1[/tex]

[tex]\mathbf{Generating \ Function: a(x) = \dfrac{1}{1-x-x^2}=f(x)}[/tex]

B)

[tex]If \ \ 1 -x-x^2 = (1 - \alpha x) ( 1- \beta x) \\ \\ \implies 1 -x - ^2 = 1 + \alpha \beta x^2 - ( \alpha + \beta )x \\ \\ \text{It implies that:} \\ \\ \alpha \beta = -1 \\ \\ \alpha + \beta = 1 \\ \\ \implies \alpha = ( 1-\beta) \\ \\ ( 1- \beta) \beta = -1 \\ \\ \implies \beta - \beta^2 = -1 \implies \beta - \beta^2 -1 = 0\\ \\ \beta = \dfrac{-(-1) \pm \sqrt{(-1)^2 -4(1)(-1)}}{2(1)}[/tex]

[tex]\beta = \dfrac{1\pm \sqrt{5}}{2} \\ \\ \beta = \dfrac{1 + \sqrt{5}}{2} \ \ and \ \ \alpha = \dfrac{1 - \sqrt{5}}{2}[/tex]

C)

[tex]\dfrac{1}{1-x-x^2}= \dfrac{A}{1-\alpha x}+ \dfrac{\beta}{1-\beta x} \\ \\ = \dfrac{A(1-\beta x) + B(1-\alpha x)}{(1-\alpha x) (1 - \beta x)} \\ \\ = \dfrac{(A+B)-(A\beta+B\alpha)x}{(1-\alpha x) (1-\beta x)}[/tex]

[tex]\text{It means:} \\ \\ A+B=1 \\ \\ B = (1-A) \\ \\ A\beta+ B \alpha =0 \\ \\ A\beta ( 1 -A) \alpha = 0 \\ \\ A( \beta - \alpha ) = -\alpha \\ \\ A = \dfrac{\alpha}{\alpha - \beta } \\ \\ \\ \\ B = 1 - \dfrac{\alpha }{\alpha - \beta} \implies \dfrac{\alpha - \beta - \alpha }{\alpha - \beta } \\ \\ =\dfrac{-\beta }{\alpha - \beta} \\ \\ \mathbf{B = \dfrac{\beta }{\beta - \alpha }}[/tex]

D)

[tex]\text{The formula for} a_n: \\ \\ a(x) = \dfrac{\alpha }{\alpha - \beta }\sum \limits ^{\infty}_{n=0} \alpha ^n x^n - \dfrac{\beta}{\beta - \alpha }\sum \limits ^{\infty}_{n=0} \beta x^n \\ \\ \implies \sum \limits ^{\infty}_{n =0} \dfrac{\alpha ^{n+1}- \beta ^{n+1}}{\alpha - \beta}x^n \\ \\ a_n = \dfrac{\alpha ^{n+1}- \beta ^{n+1}}{\alpha - \beta } \\ \\ \\ a_n = \dfrac{1}{\sqrt{5}} \Big (\Big( \dfrac{\sqrt{5}+1}{2}\Big)^{n+1}- \Big ( \dfrac{1-\sqrt{5}}{2}\Big) ^{n+1}\Big)[/tex]