In the diagram below, measure of arcABC = 230º.


What is the measure of

Answer:

100

Step-by-step explanation:

When t=0 (no time has passed), the coin is at height 100. This means the bridge must be 100 units high for this to be possible.

Answer:

100

Step-by-step explanation:

:3

Answer:

[tex]81\pi[/tex]

Step-by-step explanation:

[tex]Area = \pi * r^{2} \\\pi *9^{2} =81\pi[/tex]

Answer:

81 π

Step-by-step explanation:

formula is radius squared times pi or π so the answer would be 9x9=81 and you said to leave in terms of π so the answer is 81 π.

Answer: (-6, 0)

Step-by-step explanation:

X-intercepts of equations are any points on the equation that lie on the x-axis, or the horizontal line "y = 0".

In order to find the x-intercept of an equation, find the points that will satisfy the equation "y = 0":

y = (x + 6)(x - 3)

y = 0

(x + 6)(x - 3) = 0

With this equation, you can find which points lie on the x-axis.

When x = -6, the equation is: 0 * -9 = 0, which is correct.

When x = 3, the equation is 9 * 0 = 0, which is correct.

Make sure you're picking the correct coordinate out of the answer choices.

The x-coordinates are -6 and 3, and the y-coordinates are 0, because the points lie on the x-axis.

The correct answer is (-6, 0).

(3, 0) is also correct, but the question does not require it.

Answer:

D

Step-by-step explanation:

Answer:

[tex]r=25.7\%[/tex] will ensure that accumulated amount at the end of one year is 1.5 times more than amount invested

Step-by-step explanation:

Given: An amount was invested at r% per quarter.

To find: value of r such that accumulated amount at the end of one year is 1.5 times more than amount invested

Solution:

Let P denotes amount invested and n denotes time

As an amount (A) was invested at r% per quarter,

[tex]A=P\left ( 1+\frac{r}{400} \right )^{4n}[/tex]

According to question, accumulated amount at the end of one year is 1.5 times more than amount invested.

So,

[tex]A=1.5P+P=2.5P\\A=2.5P\\P\left ( 1+\frac{r}{400} \right )^{4n}=2.5P[/tex]

Put n = 1

[tex]P\left ( 1+\frac{r}{400} \right )^{4}=2.5P\\\left ( 1+\frac{r}{400} \right )^{4}=2.5\\1+\frac{r}{400} =(2.5)^{\frac{1}{4}}\\\frac{r}{100}=(2.5)^{\frac{1}{4}}-1\\r=100\left [ (2.5)^{\frac{1}{4}}-1 \right ]\\=25.7\%[/tex]

Answer:

11453

Step-by-step explanation:

Answer:

The first one matches with f(x)√x because a square root cannot be negative

The second one matches with f(x)=√(x-5) because the square root would be negative if it were less than five.

The third one matches with f(x)=8x because there is nothing that makes it a not possible answer

The last one matches with 7/(x-8) because there cannot be a denominator of zero.

Answer:

A. 1.8 ×[tex]10^{30}[/tex] Kg

B i. 3.0 × [tex]10^{17}[/tex] seconds

  ii. 9.6 × [tex]10^{9}[/tex] years

C. After 9.2 × [tex]10^{9}[/tex] (9.2 billion) years

Step-by-step explanation:

Given that the mass of the Sun = 2× [tex]10^{30}[/tex] Kg.

Mass of hydrogen when Sun was formed = 76% of 2× [tex]10^{30}[/tex] Kg

                            = [tex]\frac{76}{100}[/tex]  ×2× [tex]10^{30}[/tex] Kg

                           = 1.52 × [tex]10^{30}[/tex] Kg

Mass of hydrogen available for fusion = 12% of 1.52 × [tex]10^{30}[/tex] Kg

                           = [tex]\frac{12}{100}[/tex] × 1.52 × [tex]10^{30}[/tex] Kg

                           = 1.824 ×[tex]10^{30}[/tex] Kg

A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 ×[tex]10^{30}[/tex] Kg.

B. Given that the Sun fuses 6 × [tex]10^{11}[/tex] Kg of hydrogen each second.

i. The Sun's initial hydrogen would last;

                                     [tex]\frac{1.8*10^{30} }{6*10^{11} }[/tex]

                                 = 3.04 × [tex]10^{17}[/tex] seconds

The Sun's hydrogen would last 3.0 × [tex]10^{17}[/tex] seconds

ii. Since there are 31536000 seconds in a year, then;

The Sun's initial hydrogen would last;

                                     [tex]\frac{3.04*10^{17} }{31536000}[/tex]

                                 = 9.640 × [tex]10^{9}[/tex] years

The Sun's hydrogen would last 9.6 × [tex]10^{9}[/tex] years.

C. Given that our solar system is now about 4.6 × [tex]10^{9}[/tex] years, then;

                               [tex]\frac{9.6*10^{9} }{4.6*10^{9} }[/tex]

                             = 2.09

So that;   2 × 4.6 × [tex]10^{9}[/tex] = 9.2 × [tex]10^{9}[/tex] years

Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × [tex]10^{9}[/tex] years.

Part(A): The total mass of hydrogen available 9.6 billion years.

Part(B): The total time is 5.10 billion years.

Part(D): The hydrogen will last [tex]5.04\times 10^9 \ years[/tex]

Our sun is the largest object in our solar system. The mass of the sun is approximately [tex]1.988\times 1030[/tex] kilograms

Part(A):

Given that,

The total mass of the Sun =[tex]2\times10^{30} kg[/tex]

Mass of hydrogen in Sun =  [tex]2\times10^{30} \times0.76\ kg[/tex]

The mass of hydrogen ever available for fusion is,

[tex]2\times10^{30} \times 0.76 \times 0.12 kg = 1.824\times 10^{29}[/tex]

Mass of hydrogen fuses each second = 600 billion kg/second.

Time hydrogen will last in seconds=[tex]1.824\times 10^{29}seconds.[/tex]

[tex]= 0.00304\times 10^{29 }= 3.04\times 10^{17}.[/tex]

Time hydrogen will last in seconds =[tex]1 year = 31,536,000 seconds.[/tex]

[tex]31,536,000 x = 3.04\times 10^{17} = 9.6 \ billion \ years.[/tex]

(B) Present age of sun = [tex]9.6-4.5 \ billion \ years[/tex]

The time when we need to worry about Sun running out of hydrogen for fusion = 5.10 billion years.

(D) The solar system is 4.6 billion years old that is [tex]4.6\times 10^9\ years[/tex]

And in part (B) we have calculated that hydrogen will last [tex]9.64\times 10^9[/tex] then,

[tex]9.64\times 10^9-4.6\times 10^9=5.04\times 10^9[/tex]

Learn more about the topic mass of the sun:

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Answer:

Amount received per brother based on number of children plus debt is given as

Roberto, S/ 55,333.33

Luis, S/ 46,000

Armando, S/ 74,666.67

Step-by-step explanation:

English Translation

A prestigious businessman decides to distribute his inheritance of S / 176,000 among his three brothers Roberto, Luis and Armando, DP to the number of his children and IP to the amount of his debts. How much corresponds to each brother?

Roberto: No of children 4, Amount of debts (S /): 2 000

Luis: No. of children 3, Amount of debts (S /): 6,000

Armando: No of children 5, Amount of debts (S /): 8,000

Solution

The man shares the inheritance according to the number of children per person and according to each brother's debts.

Assuming the debts are first settled,

The total debts = 2000 + 6000 + 8000 = S/ 16,000

We assume that each brother receives the respective debt amounts first, then the remaining cash is divided amongst the 3 brothers according to the number of their children.

Total amount available = S/ 176,000

total debt = S/ 16,000

Amount available less debts = 176,000 - 16,000 = S/ 160,000

There are 4, 3 and 5 children respectively for the 3 brothers.

Total number of children = 4+3+5 = 12.

Amount corresponding based on a per child basis =( S/ 160,000/12) = S/ 13,333.33

Meaning that each brother receives the following amount based on their children's sake

Roberto, 4 × S/ 13,333 = S/ 53,333.33

Luis, 3 × S/ 13,333.33 = S/ 40,000

Armando, 5 × S/ 13,333 = S/ 66,666.67

Total amount each brother then receives when the amount received due to debts are added

Roberto, 53,333.33 + 2,000 = S/ 55,333.33

Luis, 40,000 + 6,000 = S/ 46,000

Armando, 66,666.67 + 8,000 = S/ 74,666.67

To check, 55,333.33 + 46,000 + 74,666.67 = 176,000 (total inheritance!)

Hope this Helps!!!

Queremos ver como se reparte una dada suma entre 3 hermanos, siendo que tenemos unas dadas restricciones, donde debemos trabajar con relaciones directamente proporcionales e inversamente proporcionales.

Veremos que:

Roberto recibe: $112,640

Luis recibe: $28,160

Armando recibe: $35,200

Sabemos que lo que se reparte es directamente proporcional al número de hijos de cada hermano, e inversamente proporcional a las deudas de cada hijo.

Entonces, definamos las variables:

R = lo que recibe Roberto.

L = Lo que recibe Luis

A = lo que recibe Armando.

Tendremos que:

R + L + A = $176,000

Entonces como lo que recibe cada hermano es directamente proporcional al número de hijos (x) e inversamente proporcional a la deuda (z) lo que cada hermano recibe será:

Entonces podemos escribir:

R + L + A = $176,000

k*4/2,000 +  k*3/6,000 + k*5/8,000 = $176,000

k*(4/2,000 + 3/6,000 + 5/8,000) = $176,000

k*(0.003125) =  $176,000

k = $176,000/(0.003125) = $56,320,000

Ahora que conocemos el valor de k, podemos calcular lo que cada hermano recibe:

R = $56,320,000*(4/2,000) = $112,640

L = $56,320,000*(3/6,000) = $28,160

A = $56,320,000*(5/8,000) = $35,200

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Answer:

x = sqrt(5)/2 - 1/2 or x = -1/2 - sqrt(5)/2

Step-by-step explanation:

Solve for x over the real numbers:

x - 2 + x/x + 1 - 1/x = -1

x - 2 + x/x + 1 - 1/x = x - 1/x:

x - 1/x = -1

Bring x - 1/x together using the common denominator x:

(x^2 - 1)/x = -1

Multiply both sides by x:

x^2 - 1 = -x

Add x to both sides:

x^2 + x - 1 = 0

Add 1 to both sides:

x^2 + x = 1

Add 1/4 to both sides:

x^2 + x + 1/4 = 5/4

Write the left hand side as a square:

(x + 1/2)^2 = 5/4

Take the square root of both sides:

x + 1/2 = sqrt(5)/2 or x + 1/2 = -sqrt(5)/2

Subtract 1/2 from both sides:

x = sqrt(5)/2 - 1/2 or x + 1/2 = -sqrt(5)/2

Subtract 1/2 from both sides:

Answer: x = sqrt(5)/2 - 1/2 or x = -1/2 - sqrt(5)/2

Answer:

x = 6                        x = -1

Step-by-step explanation:

When we have absolute value equations, we get two solutions, one positive and one negative

2x - 5 =7             2x -5= -7

Add 5 to each side

2x-5+5 = 7+5         2x -5+5 = -7+5

2x =12                     2x = -2

Divide each side by 2

2x/2 =12/2               2x/2 = -2/2

x = 6                        x = -1

Answer:

x= 0, 2, -6

Hope this helps!

Answer:

4 ft

Step-by-step explanation:

288=16 * 18

12+4=16

14+4=18

The dimensions increased by 4 feet.

The area of the rectangle is the product of the length and width of a given rectangle.

The area of the rectangle = length × Width

Given;

Dimensions of rectangle = 12 + x and 14 + x

The area of the rectangle= (12 + x) (14 + x) = 288

x² + 26x + 168 = 288

x² + 26x - 120 = 0

(x + 30) (x - 4) = 0

x=-30, x =4

Hence, The dimensions increased by 4 feet.

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#SPJ2

Answer:

Step-by-step explanation:

The original equation is [tex]121y''+110y'-24y=0[/tex]. We propose that the solution of this equations is of the form [tex] y = Ae^{rt}[/tex]. Then, by replacing the derivatives we get the following

[tex]121r^2Ae^{rt}+110rAe^{rt}-24Ae^{rt}=0= Ae^{rt}(121r^2+110r-24)[/tex]

Since we want a non trival solution, it must happen that A is different from zero. Also, the exponential function is always positive, then it must happen that

[tex]121r^2+110r-24=0[/tex]

Recall that the roots of a polynomial of the form [tex]ax^2+bx+c[/tex] are given by the formula

[tex] x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}[/tex]

In our case a = 121, b = 110 and c = -24. Using the formula we get the solutions

[tex]r_1 = -\frac{12}{11}[/tex]

[tex]r_2 = \frac{2}{11}[/tex]

So, in this case, the general solution is [tex]y = c_1 e^{\frac{-12t}{11}} + c_2 e^{\frac{2t}{11}}[/tex]

a) In the first case, we are given that y(0) = 1 and y'(0) = 0. By differentiating the general solution and replacing t by 0 we get the equations

[tex]c_1 + c_2 = 1[/tex]

[tex]c_1\frac{-12}{11} + c_2\frac{2}{11} = 0[/tex](or equivalently [tex]c_2 = 6c_1[/tex]

By replacing the second equation in the first one, we get [tex]7c_1 = 1 [/tex] which implies that [tex] c_1 = \frac{1}{7}, c_2 = \frac{6}{7}[/tex].

So [tex]y_1 = \frac{1}{7}e^{\frac{-12t}{11}} + \frac{6}{7}e^{\frac{2t}{11}}[/tex]

b) By using y(0) =0 and y'(0)=1 we get the equations

[tex] c_1+c_2 =0[/tex]

[tex]c_1\frac{-12}{11} + c_2\frac{2}{11} = 1[/tex](or equivalently [tex]-12c_1+2c_2 = 11[/tex]

By solving this system, the solution is [tex]c_1 = \frac{-11}{14}, c_2 = \frac{11}{14}[/tex]

Then [tex]y_2 = \frac{-11}{14}e^{\frac{-12t}{11}} + \frac{11}{14} e^{\frac{2t}{11}}[/tex]

c)

The Wronskian of the solutions is calculated as the determinant of the following matrix

[tex]\left| \begin{matrix}y_1 & y_2 \\ y_1' & y_2'\end{matrix}\right|= W(t) = y_1\cdot y_2'-y_1'y_2[/tex]

By plugging the values of [tex]y_1[/tex] and

We can check this by using Abel's theorem. Given a second degree differential equation of the form y''+p(x)y'+q(x)y the wronskian is given by

[tex]e^{\int -p(x) dx}[/tex]

In this case, by dividing the equation by 121 we get that p(x) = 10/11. So the wronskian is

[tex]e^{\int -\frac{10}{11} dx} = e^{\frac{-10x}{11}}[/tex]

Note that this function is always positive, and thus, never zero. So [tex]y_1, y_2[/tex] is a fundamental set of solutions.

Answer:

X+7

Step-by-step explanation:

Remove the parentheses:

4x+2-3x+5

Collect like terms:

4x-3x=x

2+5=7

Solution:

X+7

Hey there!

(4x + 2) - 3x + 5

= 4x + 2 - 3x + 5

COMBINE the LIKE TERMS

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

= 4x - 3x + 2 + 5

= 1x + 7

= x + 7

Therefore, your answer is: x + 7

Good luck on your assignment & enjoy your day!

~Amphitrite1040:)

Answer:

12 45

Step-by-step explanation:

plane leaves London at 09 35

plane arrives in Los Angeles after 11 hrs and 10 min

considering time difference:

local arrival time:

Answer: It’s 2

Step-by-step explanation:

look at picture

Answer:

The line on the graph is y = 4, where no matter what the value of x is, the value of y will always be 4. Therefore, any line parallel to this one will be y = ?. If it passes through (-4, -6), that means that the equation is y = -6.

Answer:

С)))) Y= -6

Step-by-step explanation:

just did on edg. :D

Answer:

A. it would be shifted up

Step-by-step explanation:

Y=MX+B

B is the Y-intercept.

Answer:

a. it would be shifted up

Step-by-step explanation:

the difference between the original and the new function is that the b value is changed from -6 to +8, meaning the y-intercept value has increased. this would shift the graph up by 14.

Answer:

  A.  G(x) = (x -3)^3 -(x -3)

Step-by-step explanation:

The graph before it was shifted left will be a right-shift of the equation shown. That is accomplished by replacing x with (x-1.5). Then the right-shifted equation is ...

  G(x) = ((x-1.5) -1.5)^3 -((x -1.5) -1.5)

  G(x) = (x -3)^3 -(x -3) . . . . matches choice A

Answer:

B

Step-by-step explanation:

The triangular prism must have a larger base than the cylinder

Answer: itz 605

Step-by-step explanation:

Answer:

Step-by-step explanation:

a. Null hypothesis: P = p

Alternatives hypothesis: P =/ p

Where P is the hypothesized population proportion and p is the sample proportion

b. Performing a test of proportions

Randomization: the sample was randomly selected in the study

The population size is at least 20 times as big as the sample size.

The sample includes both successes and failures with 29 success and 171 failures.

c. To perform the hypothesis test: we have to find the standard deviation first

Sd = sqrt[ P * ( 1 - P ) / n ]

where P is the hypothesized value of population proportion, n is the sample size.

Sd = √[0.12*(1-0.12)/200]

Sd = √[0.12*(0.88/200]

Sd = √[0.12*(0.0044)]

Sd = √0.000528

Sd = 0.023

Then we can find the z score

z = (p - P) / σ where p = 29/200 = 0.145

z = (0.145-0.12)/ 0.023

z = 0.025/0.023

z = 1.09

Calculation the p value using 0.05 level of significance and a two waited test (p value calculator),

A p-value of 0.2757 which is greater than 0.05, thus we will fail to reject the null stating that there is not enough strong evidence that the left-handed rate in the state of Maine is higher than the national average.

Answer:

p-value (0.0208) is less than alpha = 0.05 reject H0.

Step-by-step explanation:

we have the following data:

sample size = n = 75

x, the number to evaluate is 45

the sample proportion would be: x / n = 45/75

p * = 0.6

Now, the null and alternative hypotheses are:

H0: P = 0.72

Ha: P no 72

two tailed test

statistic tes = z = (p * - p) / [(p * (1-p) / n)] ^ (1/2)

replacing we have:

z = (0.6 - 0.72) / [(0.72 * (1-0.72) / 75)] ^ (1/2)

z = -2.31

p-vaule = 2 * p (z <-2.31)

using z table, we get:

p-vaule = 2 * (0.0104)

p-vaule = 0.0208

Therefore, p-value (0.0208) is less than alpha = 0.05 reject H0.

Answer:   -9  ≤ f(6) - f(3)  ≤ 15

Step-by-step explanation:

In order to use the Mean Value Theorem, it must be continuous and differentiable. Both of these conditions are satisfied so we can continue.  

Find f(6) - f(3) using the following formula:

[tex]f'(c)=\dfrac{f(b)-f(a)}{b-a}[/tex]

Consider:    a = 3, b = 6

[tex]\text{Then}\ f'(c)=\dfrac{f(6)-f(3)}{6-3}\\\\\\\rightarrow \quad 3f'(c)=f(6) - f(3)[/tex]          

Given: -3 ≤  f'(x)  ≤ 5

          -9  ≤ 3f'(c) ≤ 15    Multiplied each side by 3

→   -9  ≤ f(6) - f(3)  ≤ 15  Substituted 3f'(c) with f(6) - f(3)

Answer:

|5| = 5 and |2| = 2 and because 5 > 2, our answer is |5| > |2|.

Answer:

|5|

Step-by-step explanation:

5 is greater than 2

Answer:

(0,34)

Step-by-step explanation:

For each rise of 14 in the x direction, this graph rises by -8 in the y direction. This means that, when x is 0, and the graph intersects the y axis, the y value will be 50-8-8=34. Therefore, the y intercept of this line is (0,34). Hope this helps!

Answer:

The answer is (0,34)

Answer:

[tex]x=\frac{1}{6}[/tex]

Step-by-step explanation:

[tex]\sqrt{3-5x}=\sqrt{x+2}\\Square\:both\:sides\\\left(\sqrt{3-5x}\right)^2=\left(\sqrt{x+2}\right)^2\\\mathrm{Expand\:}\left(\sqrt{3-5x}\right)^2:\quad 3-5x\\\mathrm{Expand\:}\left(\sqrt{x+2}\right)^2:\quad x+2\\3-5x=x+2\\\mathrm{Solve\:}\:3-5x=x+2:\quad x=\frac{1}{6}\\x=\frac{1}{6}\\\mathrm{Verify\:Solutions}:\quad x=\frac{1}{6}\space\mathrm{True}\\\mathrm{The\:solution\:is}\\x=\frac{1}{6}[/tex]

Answer:

A- 1/6

Step-by-step explanation:

GOT IT RIGHT ON EDGE

Answer:  The correct answer is:  " 2x² " .

________________________________

Step-by-step explanation:

________________________________

We are asked:  "What is the sum of:  "x + x² + 2" and "x² − 2 − x" ?

Since we are to find the "sum" ;

  →  We are to "add" these 2 (two) expressions together:

     →     (x + x² + 2) +  (x² − 2 − x) ;

Note:  Let us rewrite the above, by adding the number "1" as a coefficient to:  the values "x" ; and "x² " ;  since there is an "implied coefficient of "1" ;

            → {since:  "any value" ; multiplied by "1"; results in that exact same value.}.

            →     (1x + 1x² + 2) +  (1x² − 2 − 1x) ;

Rewrite as:

             →     1x + 1x² + 2) +  (1x² − 2 − 1x) ;

Now, let us add the "coefficient" , "1" ; just before the expression:

             "(1x² − 2 − 1x)" ;  

         {since "any value", multiplied by "1" , equals that same value.}.

And rewrite the expression; as follows:

            →     (1x + 1x² + 2) +  1(1x² − 2 − 1x) ;

Now, let us consider the following part of the expression:

                     →  " +1(1x² − 2 − 1x) " ;

________________________________

Note the distributive property of multiplication:

   →  " a(b+c) = ab + ac " ;

and likewise:

   →  " a(b+c+d) = ab + ac + ad " .

________________________________

So; we have:

→  " +1(1x² − 2 − 1x) " ;

  = (+1 * 1x²)  +  (+1 *-2) + (+1*-1x) ;

  =       + 1x²    +    (-2)     +  (-1x) ;

  =       +1x²    −    2   −  1x  ;

  ↔    ( + 1x²  −  1x  −  2)

Now, bring down the "left-hand side of the expression:

1x + 1x² + 2 ;

and add the rest of the expression:

     →  1x  +  1x²  +  2  +   1x²  −  1x  −  2 ;

________________________________

Now, simplify by combining the "like terms" ; as follows:

   +1x² + 1x² = 2x²  ;

   +1x −  1x = 0 ;  

   + 2 −  2  = 0 ;

________________________________

The answer is: " 2x² " .

________________________________

Hope this is helpful to you!

   Best wishes!

________________________________

Answer:

The correct answer is:  " 2x² " .

________________________________

Step-by-step explanation:

________________________________

We are asked:  "What is the sum of:  "x + x² + 2" and "x² − 2 − x" ?

Since we are to find the "sum" ;

 →  We are to "add" these 2 (two) expressions together:

    →     (x + x² + 2) +  (x² − 2 − x) ;

Note:  Let us rewrite the above, by adding the number "1" as a coefficient to:  the values "x" ; and "x² " ;  since there is an "implied coefficient of "1" ;

           → {since:  "any value" ; multiplied by "1"; results in that exact same value.}.

           →     (1x + 1x² + 2) +  (1x² − 2 − 1x) ;

Rewrite as:

            →     1x + 1x² + 2) +  (1x² − 2 − 1x) ;

Now, let us add the "coefficient" , "1" ; just before the expression:

            "(1x² − 2 − 1x)" ;  

        {since "any value", multiplied by "1" , equals that same value.}.

And rewrite the expression; as follows:

           →     (1x + 1x² + 2) +  1(1x² − 2 − 1x) ;

Now, let us consider the following part of the expression:

                    →  " +1(1x² − 2 − 1x) " ;

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Note the distributive property of multiplication:

  →  " a(b+c) = ab + ac " ;

and likewise:

  →  " a(b+c+d) = ab + ac + ad " .

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So; we have:

→  " +1(1x² − 2 − 1x) " ;

 = (+1 * 1x²)  +  (+1 *-2) + (+1*-1x) ;

 =       + 1x²    +    (-2)     +  (-1x) ;

 =       +1x²    −    2   −  1x  ;

 ↔    ( + 1x²  −  1x  −  2)

Now, bring down the "left-hand side of the expression:

1x + 1x² + 2 ;

and add the rest of the expression:

    →  1x  +  1x²  +  2  +   1x²  −  1x  −  2 ;

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Now, simplify by combining the "like terms" ; as follows:

  +1x² + 1x² = 2x²  ;

  +1x −  1x = 0 ;  

  + 2 −  2  = 0 ;

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The answer is: " 2x² " .

Step-by-step explanation:

Answer:

see below

Step-by-step explanation:

[tex]\frac{x-c}{d} > y[/tex]

x - c > dy

x > dy + c