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DochEvi [55]
3 years ago
6

Rhett was solving the algebraic proof below. What reason is missing from his proof? 2. 2(3x - 4) + 5x = -41 Given. (6x - 8) + 5x

= -41 Distributive Property. 6x + 5x – 8= -41 ?. 11x - 8 = -41 Write an equivalent equation. 11x = -33 Addition or Property of Equality. x = -3 Division Property of Equality.​
Mathematics
1 answer:
serg [7]3 years ago
7 0

Answer:

Step-by-step explanation:

From the above question, we are solving for an Algebraic Equation

2(3x - 4) + 5x = -41 = Given

(6x - 8) + 5x = -41 Distributive Property. 6x + 5x – 8= -41

11x - 8 = -41 Write an equivalent equation

11x = -33 Addition of Property of Equality

x = -3 Division Property of Equality

The reason that is missing from the proof is:

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If you are offered one slice from a round pizza (in other words, a sector of a circle) and the slice must have a perimeter of 24
VARVARA [1.3K]

Answer:

A 12 in diameter will reward you with the largest slice of pizza.

Step-by-step explanation:

Let r be the radius and \theta be the angle of a circle.

According with the graph, the area of the sector is given by

A=\frac{1}{2}r^2\theta

The arc lenght of a circle with radius r and angle \theta is r \theta

The perimeter of the pizza slice is composed of two straight pieces, each of length r inches, and an arc of the circle which you know has length s = rθ inches. Thus the perimeter has length

2r+r\theta=24 \:in

We need to express the area as a function of one variable, to do this we use the above equation and we solve for \theta

2r+r\theta=24\\r\theta=24-2r\\\theta=\frac{24-2r}{r}

Next, we substitute this equation into the area equation

A=\frac{1}{2}r^2(\frac{24-2r}{r})\\A=\frac{1}{2}r(24-2r)\\A=12r-r^2

The domain of the area is

0

To find the diameter of pizza that will reward you with the largest slice you need to find the derivative of the area and set it equal to zero to find the critical points.

\frac{d}{dr} A=\frac{d}{dr}(12r-r^2)\\A'(r)=\frac{d}{dr}(12r)-\frac{d}{dr}(r^2)\\A'(r)=12-2r

12-2r=0\\-2r=-12\\\frac{-2r}{-2}=\frac{-12}{-2}\\r=6

To check if r=6 is a maximum we use the Second Derivative test

if f'(c)=0 and f''(c), then f(x) has a local maximum at x = c.

The second derivative is

\frac{d}{dr} A'(r)=\frac{d}{dr} (12-2r)\\A''(r)=-2

Because A''(r)=-2 the largest slice is when r = 6 in.

The diameter of the pizza is given by

D=2r=2\cdot 6=12 \:in

A 12 in diameter will reward you with the largest slice of pizza.

3 0
3 years ago
Noah drew 26 hearts, 21 stars, and 38 circles. what is the ratio of stars to hearts to circles?
mamaluj [8]

Answer:

21:26:38

Step-by-step explanation:

Ratio in mathematics is the relationship between two or more amounts, showing how many times they can be contained in one another.

Example: 1:2 means there are two times the first amount in the second.

For the case above;

Given;

Number of each shape noah drew;

hearts = 26

Stars = 21

circles = 38

The ratio of stars to hearts to circles

N(stars) : N(hearts) : N(circles)

21:26:38

Since they are not divisible by a common factor the ratio remains;

21:26:38

5 0
3 years ago
1 The 14 members of the baseball team are going to stop at an ice cream shop after the game. The ice cream shop has a special tr
sergeinik [125]

Answer:

###########################################################

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Which of the following sets are subspaces of R3 ?
Ratling [72]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

for point A:

\to A={(x,y,z)|3x+8y-5z=2} \\\\\to  for(x_1, y_1, z_1),(x_2, y_2, z_2) \varepsilon A\\\\ a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                        =3(aX_l +bX_2) + 8(ay_1 + by_2) — 5(az_1+bz_2)\\\\=a(3X_l+8y_1- 5z_1)+b (3X_2+8y_2—5z_2)\\\\=2(a+b)

The set A is not part of the subspace R^3

for point B:

\to B={(x,y,z)|-4x-9y+7z=0}\\\\\to for(x_1,y_1,z_1),(x_2, y_2, z_2) \varepsilon  B \\\\\to a(x_1, y_1, z_1)+b(x_2, y_2, z_2) = (ax_1+bx_2,ay_1+by_2,az_1+bz_2)

                                             =-4(aX_l +bX_2) -9(ay_1 + by_2) +7(az_1+bz_2)\\\\=a(-4X_l-9y_1+7z_1)+b (-4X_2-9y_2+7z_2)\\\\=0

\to a(x_1,y_1,z_1)+b(x_2, y_2, z_2) \varepsilon  B

The set B is part of the subspace R^3

for point C: \to C={(x,y,z)|x

In this, the scalar multiplication can't behold

\to for (-2,-1,2) \varepsilon  C

\to -1(-2,-1,2)= (2,1,-1) ∉ C

this inequality is not hold

The set C is not a part of the subspace R^3

for point D:

\to D={(-4,y,z)|\ y,\ z \ arbitrary \ numbers)

The scalar multiplication s is not to hold

\to for (-4, 1,2)\varepsilon  D\\\\\to  -1(-4,1,2) = (4,-1,-2) ∉ D

this is an inequality, which is not hold

The set D is not part of the subspace R^3

For point E:

\to E= {(x,0,0)}|x \ is \ arbitrary) \\\\\to for (x_1,0 ,0) ,(x_{2},0 ,0) \varepsilon E \\\\\to  a(x_1,0,0) +b(x_{2},0,0)= (ax_1+bx_2,0,0)\\

The  x_1, x_2 is the arbitrary, in which ax_1+bx_2is arbitrary  

\to a(x_1,0,0)+b(x_2,0,0) \varepsilon  E

The set E is the part of the subspace R^3

For point F:

\to F= {(-2x,-3x,-8x)}|x \ is \ arbitrary) \\\\\to for (-2x_1,-3x_1,-8x_1),(-2x_2,-3x_2,-8x_2)\varepsilon  F \\\\\to  a(-2x_1,-3x_1,-8x_1) +b(-2x_1,-3x_1,-8x_1)= (-2(ax_1+bx_2),-3(ax_1+bx_2),-8(ax_1+bx_2))

The x_1, x_2 arbitrary so, they have ax_1+bx_2 as the arbitrary \to a(-2x_1,-3x_1,-8x_1)+b(-2x_2,-3x_2,-8x_2) \varepsilon F

The set F is the subspace of R^3

5 0
3 years ago
A baseball field is being constructed. The builders noticed that the batters would have the sun in their eyes when batting from
Leona [35]

Answer:

A' =(-y,x)

Step-by-step explanation:

Given

Represent point A as thus;

A = (x,y)

Required

Determine the new position of A, when rotated counterclockwise

The new position will be denoted as A'

When a point is rotated counterclockwise, we start by switching the positions of x and y as follows:

Pos = (y,x)

Then, y is negated to give A'

A' =(-y,x)

5 0
3 years ago
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