You and Jake are studying together and after correctly solving a system of equations using
the substitution method the result is 3 3. Jake does not know how to interpret this
result.
In at least one complete sentence, explain to Jake the number of solutions, solution type,
and how the graph of the system of equations would look

Answers

Answer 1

Answer: its the 3rd one if im wrong im sorry

:3


Related Questions

This graph represents a quadratic function. An upward parabola on a coordinate plane vertex at (minus 2, 2) and passes through (minus 3, 5) and (minus 1, 5). What is the value of a in the function’s equation? A. -2 B. -3 C. 2 D. 3

Answers

Answer: 3

Step-by-step explanation:

Substituting into vertex form, the equation is

[tex]y=a(x+2)^2 +2[/tex]

Substituting in the coordinates (-3, 5),

[tex]5=a(-3+2)^2 +2\\\\5=a+2\\\\a=3[/tex]

Let g(x)= 18 - 3x
Find g-¹ (0). Final answer is just a number.

Answers

Answer:

g(0)⁻¹ = 6

Step-by-step explanation:

First, you must find the inverse of the function. Remember, another way of representing g(x) is with "y".  To find the inverse, you must swap the positions of the "x" and "y" variables in the equation. Then, you must rearrange the equation and isolate "y".

g(x) = 18 - 3x                                    <----- Original function

y = 18 - 3x                                        <----- Plug "y" in for g(x)

x = 18 - 3y                                        <----- Swap the positions of "x" and "y"

x + 3y = 18                                        <----- Add 3y to both sides

3y = 18 - x                                         <----- Subtract "x" from both sides

y = (18 - x) / 3                                    <----- Divide both sides by 3

y = 6 - (1/3)x                                     <----- Divide both terms by 3

Now that we have the inverse function, we need to plug x = 0 into the equation and solve for the output. In the inverse function, "y" is represented by the symbol g(x)⁻¹.

g(x)⁻¹ = 6 - (1/3)x                                  <----- Inverse function

g(0)⁻¹ = 6 - (1/3)(0)                               <----- Plug 0 in for "x"

g(0)⁻¹ = 6 - 0                                       <----- Multiply 1/3 and 0

g(0)⁻¹ = 6                                             <----- Subtract

Need help with this!!

Answers

Answer:

below

Step-by-step explanation:

1) slope = rise / run

2 coordinates are (-4, 0), (0, 2).

2 - 0 = 2

0 -- 4 = 4

2 / 4 = 1/2 so the slope is 0.5 or ½

2) it crosses the y axis at the average of the origin and 4.

4 + 0 = 4 / 2 = 2 so y intercept is 2.

3) in y= mx + b form

f(x) = ½x + 2, or, f(x) = 0.5x + 2

George cuts a rectangular piece of glass down one of the diagonals as shown below. What is the length of the diagonal that he cut to the nearest whole inch? Enter only the number. An image shows a rectangle with length = 18 inches and width = 36 inches. A red dotted line crosses the rectangle from the upper left corner to the lower right corner.

Answers

The length of the diagonal that he cut to the nearest whole inch is 36 inches

Triangle

A rectangle with length = 18 inchesWidth = 36 inches

A red dotted line crosses the rectangle from the upper left corner to the lower right corner to form a triangle.

Length of the diagonal of a rectangle;

Hypotenuse² = adjacent² + opposite²

= 18² + 36²

= 324 + 1296

hyp² = 1620

Take the square root of both sides

hyp = √1620

hyp = 35.4964786985976

Approximately,

hypotenuse = 36 inches

Therefore, the length of the diagonal that he cut to the nearest whole inch is 36 inches.

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X
-5
Probability 17
-3
-2 0
0 2
13 33 16 11
3
.10
Find the probability that x <-3

Answers

The value of the probability is 0.30

How to determine the probability?

Using the table of values, we have:

P(x <= -3) = P(x = -5) + P(x = -3)


From the table of values, we have:

P(x = -5) = 0.17

P(x = -3) = 0.13

Substitute the known values in the above equation

P(x <= -3) = 0.17 + 0.13

Evaluate the sum

P(x <= -3) = 0.30

Hence, the value of the probability is 0.30

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Given two independent random samples with the following results:

n1=13
x‾1=141
s1=13   

n2=9
x‾2=161
s2=12

Use this data to find the 98% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3: Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 3 of 3: Construct the 98% confidence interval. Round your answers to the nearest whole number.

please explain

Answers

The point estimate of difference of the sample his will be -20.

How to illustrate the information?

Based on the information given, the. following can be depicted:

n1 = 13

x1 = 141

s1 = 13

n2 = 9

x2 = 161

s2 = 12

The point estimate of difference will be:

= 141 - 161

= -20

The margin of error to be used in constructing the confidence interval will be calculated by multiplying the standard error which is 5.467 and the critical value. This will be:

= 5.467 × 2.528

= 13.822

The margin of error is 13.822.

The confidence interval will now be:

= (-20 + 13.822) and (-20 - 13.822)

= -6.178 and -33.822

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Let g be a one-to-one function and suppose f is the inverse function of g
if g(5)=11 and g(3)=5, find f(5)

Answers

Answer:

So g(6) comes out to be 2.

Step-by-step explanation:

So f(6)=5 means when f acts on 6, the result is 5....so the inverse would take 5 back to 6...or g(5)=6

So f(2)=6 means that f reassigns 2 to 6....so the inverse would take 6 back to 2....or g(6)=2

hope this helps :)

Answer: 3

Step-by-step explanation:

If f(x) = y, then invf(y) = x
So, f(5) = invg(5) = 3

Please help me solve the problem in the image. I found that a=-65 and b=17 and when I plug it in the equation it’s wrong

Answers

Answer:

-181 - 79x

Step-by-step explanation:

I am not sure your comment fits the given problem.

I get the following approach and solution :

T(1 + 4x) = 1×a + 4x×b = 2 + 2x

a = 2 + 2x - 4xb

T(4 + 15x) = 4×a + 15x×b = -3 + 3x

4×(2 + 2x - 4xb) + 15xb = -3 + 3x

8 + 8x - 16xb + 15xb = -3 + 3x

11 + 5x - xb = 0

11 + 5x = xb

b = (11 + 5x)/x

a = 2 + 2x - 4x(11 + 5x)/x = 2 + 2x - 44 - 20x = -42 - 18x

T(3 - 5x) = 3×a - 5xb = 3(-42 - 18x) - 5x(11 + 5x)/x =

= -126 - 54x - 55 - 25x = -181 - 79x

control :

T(1 + 4x) = a + 4xb = -42 - 18x + 44 + 20x = 2 + 2x

T(4 + 15x) = 4a + 15xb = -168 - 72x + 165 + 75x = -3 + 3x

please help me figure this out ​

Answers

Answer:

-25

Step-by-step explanation:

[tex] \dfrac{(20 - 5^2)(16 + 2^2)}{-2^3 + (3 \times 2^2)} = [/tex]

First, do all exponents.

[tex] = \dfrac{(20 - 25)(16 + 4)}{-8 + (3 \times 4)} [/tex]

Now do each operation in parentheses.

[tex] = \dfrac{(-5)(20)}{-8 + 12} [/tex]

Multiply in the numerator. Add in the denominator.

[tex] = \dfrac{-100}{4} [/tex]

Divide the numerator by the denominator.

[tex] = -25 [/tex]

Answer:

-25

Step-by-step explanation:

PEMDAS

The PEMDAS rule is an acronym representing the order of operations in math:

ParenthesesExponentsMultiplication and Division (from left to right)Addition and Subtraction (from left to right)

Given expression:

[tex]\sf \dfrac{(20-5^2)(16+2^2)}{-2^3+(3 \times 2^2)}[/tex]

As the given expression is a fraction, carry out the operations in the numerator and denominator first before finally dividing them.

Following PEMDAS, carry out the calculations inside the parentheses first, then carry out the rest of the calculations following the order of operations:

Parentheses

Calculate the exponents inside the parentheses:

[tex]\implies \sf \dfrac{(20-25)(16+4)}{-2^3+(3 \times 4)}[/tex]

Multiply:

[tex]\implies \sf \dfrac{(20-25)(16+4)}{-2^3+(12)}[/tex]

Add and subtract:

[tex]\implies \sf \dfrac{(-5)(20)}{-2^3+(12)}[/tex]

Exponents

Calculate the exponent:

[tex]\implies \sf \dfrac{(-5)(20)}{-8+(12)}[/tex]

Multiply and Divide

Multiply:

[tex]\implies \sf \dfrac{-100}{-8+(12)}[/tex]

Add and Subtract

Add:

[tex]\implies \sf \dfrac{-100}{4}[/tex]

Finally, divide the numerator by the denominator:

[tex]\implies \sf -25[/tex]

HELP ME WITH THIS PLEASEEEE

Answers

Answer:

the perpendicular bisector

Explanation:

The perpendicular bisector will intersect the segment at its midpoint.

Ryan obtains a loan for home renovations from a bank that charges simple interest at an annual rate of 9.65%. His loan is for $17,100 for 54 days. Assume 1/365 each day is of a year. Answer each part below.
Do not round any intermediate computations, and round your final answers to the nearest cent.
(a) Find the interest that will be owed after 54 days. $ (b) Assuming Ryan doesn't make any payments, find the amount owed after 54 days.

Answers

well, with the assumption that a year has 365 days, that means one day is really just 1/365th of a year, so then 54 days will be 54/365 of a year.

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\\ P=\textit{original amount deposited}\dotfill & \$17100\\ r=rate\to 9.65\%\to \frac{9.65}{100}\dotfill &0.0965\\ t=years\dotfill &\frac{54}{365} \end{cases} \\\\\\ I = (17100)(0.0965)(\frac{54}{365})\implies \stackrel{\textit{interest owed}}{I\approx 244.13}~\hfill \underset{amount~owed}{\stackrel{17100~~ + ~~244.13}{\approx 17344.13}}[/tex]

Line passes through the point (8,4) and a slope of 5/4. Write equation in slope-intercept

Answers

Answer:

Step-by-step explanation:

y - 4 = 5/4(x - 8)

y - 4 = 5/4x - 10

y = 5/4x - 6

A lateral thoracic spine on a 33-cm patient is usually taken using 100 mA (large focal spot), 0.5 seconds, 86 kVp, 40-inch SID, 12:1 grid ratio. If the mA is increased to 400 to stop motion blur from tremors, which of the following technique changes will produce a radiographic density and contrast most similar to the original?​

Answers

A 0.13 sec and a 86 kVp technical changes are the changes that would have to produce the radiographic density and the contrast that are more similar to the original.

What is the radiographic density?

The radio graphic density can de defined to be the total amount or the overall darkening that can be found in a particular radiograph. This is usually known to have a certain type of density range that lies between  0.3 to 2.0 density.

When there is a density that is less than 0.3, the problem is basically because there is the issue of the density which would be found at the base and the fact that the film that is being used has fog in it.

It is known that based on the values that we have here the density and contrast would have to be of the given kvp of 86 and 0.13 seconds in order to be said to be similar to the original.

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please help!!!

Using long division, what is the quotient of this expression?
3x42x³-x-4
x²+2
OA. 3x²
3x - 4
OB. 3x² + 2x + x² + 2
O C.
3x² 2x
O D.
-
2x
3x² + 2x
- 6 +
-
- 5+
-
3x + 8
x² + 2
-
3x+6
x²+2
5x8
x² + 2
Rese

Answers

Check the picture below.

notice, the dividend and divisor must be in descending order, and when one of the variables is "missing", is really not missing, it simply has a coefficient of 0.

Here is the solution process:

                     3 x² -  2 x  -   6                      

x² + 0x + 2 | 3 x⁴ -  2 x³ +  0  x² -     x -   4

                    3  x⁴ + 0 x³ +  6  x²                

                             - 2 x³  -  6  x² -      x -   4

                             - 2 x³  + 0  x² -   4 x        

                                        -  6  x² +  3 x  -  4

                                        -  6  x²  + 0 x  -  12

                                                        3 x  + 8

Long division:

Standard: [tex]\sf{3x^{2} -2x-6+\dfrac{3x+8}{x^{2} +2} \ \ \to \ \ \ Option \ "A" }[/tex]Quotient: 3x² - 2x - 6Rest: 8 + 3x

Therefore, the correct option is "A".

Whole page of geometry stuff for 50 points only do 2,3 and 4 ( serious answers only or 1 star and report )

Answers

See below for the distance between the points and the lines

How to determine the distance between the lines and the points?

Question 2

The line and the points are given as:

x = y

P = (4, -2)

Rewrite the equation as:

y = x

The slope of the above equation is

m = 1

The slope of a line perpendicular to it is

m = -1

A linear equation is represented as:

y = mx + b

Substitute m = -1

y = -x + b

Substitute (4, -2) in y = -x + b

-2 = -4 + b

Solve for b

b = 2

Substitute b = 2 in y = -x + b

y = -x + 2

So, we have:

x = y and y = -x + 2

Substitute x for y

x = -x + 2

Solve for x

x = 1

Substitute x = 1 in y = x

y = 1

So, we have the following points

(1, 1) and (4, -2)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(1 - 4)² + (1 + 2)²

Evaluate

d = 3√2

Hence, the distance between x = y and P = (4, -2) is 3√2 units

Question 3

The line and the points are given as:

y = 2x + 1

Q = (2, 10)

The slope of the above equation is

m = 2

The slope of a line perpendicular to it is

m = -1/2

A linear equation is represented as:

y = mx + b

Substitute m = -1/2

y = -1/2x + b

Substitute (2, 10) in y = -1/2x + b

10 = -1/2 * 2 + b

Solve for b

b = 11

Substitute b = 11 in y = -1/2x + b

y = -1/2x + 11

So, we have:

y = 2x + 1 and y = -1/2x + 11

Substitute 2x + 1 for y

2x + 1 = -1/2x + 11

Solve for x

x = 4

Substitute x = 4 in y = 2x + 1

y = 9

So, we have the following points

(4, 9) and (2, 10)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(4 - 2)² + (9 - 10)²

Evaluate

d = √5

Hence, the distance between the line and the point is √5 units

Question 4

The line and the points are given as:

y = -x + 3

R = (-5, 0)

The slope of the above equation is

m = -1

The slope of a line perpendicular to it is

m = 1

A linear equation is represented as:

y = mx + b

Substitute m = 1

y = x + b

Substitute (-5, 0) in y = x + b

0 = 5 + b

Solve for b

b = -5

Substitute b = 5 in y = x + b

y = x + 5

So, we have:

y = x + 5 and y = -x + 3

Substitute x + 5 for y

x + 5 = -x + 3

Solve for x

x = -1

Substitute x = -1 in y = x + 3

y = 2

So, we have the following points

(-1, 2) and (-5, 0)

The distance between the above points is

d = √(x2 - x1)² + (y2 - y1)²

So, we have:

d = √(-1 + 5)² + (2 - 0)²

Evaluate

d = 2√5

Hence, the distance between the line and the point is 2√5 units

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If $6,000 principal plus $132.90 of simple interest was withdrawn on August 14, 2011, from an investment earning 5.5% interest, on what day was the money invested?

Answers

[tex]~~~~~~ \textit{Simple Interest Earned} \\\\ I = Prt\qquad \begin{cases} I=\textit{interest earned}\dotfill & \$132.90\\ P=\textit{original amount deposited}\dotfill & \$6000\\ r=rate\to 5.5\%\to \frac{5.5}{100}\dotfill &0.055\\ t=years \end{cases} \\\\\\ 132.90 = (6000)(0.055)(t)\implies \cfrac{132.90}{(6000)(0.055)}=t\implies \cfrac{443}{1100}=t \\\\\\ \stackrel{\textit{converting that to days}}{\cfrac{443}{1100}\cdot 365} ~~ \approx ~~ 147~days[/tex]

now, if we move back from August 14th by 147 days backwards, that'd put us on March 20th.

Which pair of angles are vertical angles? AngleWRU and AngleSRT AngleWRS and AngleVRT AngleVRU and AngleTRS AngleVRT and AngleSRT

Answers

Step-by-step explanation:

important of festival in nepali languages for class seven

Answer:

b

Step-by-step explanation:

A polynomial f (x) has the
given zeros of 6, -1, and -3.
Part A: Using the
Factor Theorem, determine the
polynomial f (x) in expanded form. Show all necessary
calculations.
*
Part B: Divide the polynomial f (x) by (x2 - x - 2) to
create a rational function g(x) in simplest factored form.
Determine g(x) and find its slant asymptote.

Part C: List all locations and types of discontinuities of
the function g(x).

Answers

a) The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function g(x) in factored form is g(x) = [(x - 6) · (x + 3)] / (x - 2). there is no slant asymptotes.

c) There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

How to analyze polynomial and rational functions

a) In the first part of this question we need to determine the equation of a polynomial in expanded form, derived from its factor form defined below:

f(x) = Π (x - rₐ), for a ∈ {1, 2, 3, 4, ..., n}         (1)

Where rₐ is the a-th root of the polynomial.

If we know that r₁ = 6, r₂ = - 1 and r₃ = - 3, then the polynomial in factor form is:

f(x) = (x - 6) · (x + 1) · (x + 3)

f(x) = (x - 6) · (x² + 4 · x + 4)

f(x) = (x - 6) · x² + (x - 6) · (4 · x) + (x - 6) · 4

f(x) = x³ - 6 · x² + 4 · x² - 24 · x + 4 · x - 24

f(x) = x³ + 10 · x² - 20 · x - 24

The polynomial f(x) in expanded form is f(x) = x³ + 10 · x² - 20 · x - 24.

b) The rational function is introduced below:

g(x) = (x³ + 10 · x² - 20 · x - 24) / (x² - x - 2)

g(x) = [(x - 6) · (x + 1) · (x + 3)] / [(x - 2) · (x + 1)]

g(x) = [(x - 6) · (x + 3)] / (x - 2)

The slope of the slant asymptote is:

m = lim [g(x) / x] for x → ± ∞

m = [(x - 6) · (x + 3)] / [x · (x - 2)]

m = 1

And the intercept of the slant asymptote is:

n = lim [g(x) - m · x] for x → ± ∞

n = Non-existent

Hence, there is no slant asymptotes.

c) There is vertical asymptote at a x-point if the denominator is equal to zero. There is one evitable discontinuity at x = - 1, and one definitive discontinuity at x = 2, where there is a vertical asymptote.

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A rectangular table is six times as long as it is wide. If the area is 150 ft2, find the length and the width of the table.
The width of the table is
The length of the table is?

Answers

Answer:

Length = 30ft, Width = 5ft

Step-by-step explanation:

Let x be the width.

Area of Rectangle = Length * Width

Given from the information in the question,

Length = 6x ft

Width = x ft

Substitute the values into the formula:

6x * x = 150

[tex]6x^{2}[/tex] = 150

[tex]x^{2} =\frac{150}{6}[/tex]

[tex]x^{2} =25[/tex]

[tex]x=\sqrt{25}[/tex]

x = 5 ft.

Therefore,

Length = 6 * 5 = 30ft

Width = 5 ft.

Fraction how do you make 0.475 a simple fraction

Answers

Answer: 19/40

Step-by-step explanation:

First Write 0.475 as 0.4751Multiply both numerator and denominator by 10 for every number after the decimal point0.475 × 10001 × 1000 = 4751000. Reducing the fraction gives The answer

i believe the correct answer should be19/40


Twenty years ago, 51% of parents of children in high school felt it was a
serious problem that high school students were not being taught
enough math and science. A recent survey found that 232 of 700
parents of children in high school felt it was a serious problem that high
school students were not being taught enough math and science. Do
parents feel differently today than they did twenty years ago? Use the
α = 0.1 level of significance

Answers

No the parents do not feel differently as they used to feel twenty years ago

Given :-

Po: p=0.46

Pa: p NE 0.46

alpha=0.01

Test statistic is one sample proportion [tex]z=(phat-p)/\sqrt{{ {0.46)(0.54)/800}}[/tex]

Critical value z> |2.576|

z=0.025/0.0176

=1.418

This is a p-value of 0.078

Fail to reject Po  and insufficient evidence to support the claim that parents feel differently.

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A student purchased 7 binders for a total of $8.61. Write an equation that can be used to find the cost of each binder, n, in dollars.

Answers

The equation that can be used to find the cost of each binder n in dollars is 861=7n.

Given that the cost of 7 binders is $8.61.

We are required to form an equation that represents the total cost of each binder n in dollars.

Equation is like a relationship between all the variables that are expressed in equal to form.It may be linear equation or may be more types.

Suppose the cost of 1 binder is n dollar.

We know that the total cost is basically the product of price of 1 unit and number of quantities of units.

Total cost=Price of 1 unit* number of units

8.61=n*7

8.61=7n

Hence the equation that can be used to find the cost of each binder n in dollars is 861=7n.

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What is the ordered pair of X' after point X (3, 4) is rotated 180°?
OX' (3,-4)
OX' (-3,-4)
OX' (-4, 3)
OX' (-4,-3)

Answers

Rotating 180° would be changing (x to -x and y to -y) therefore, X’ (-3, -4)

PLEASE HELP ME! I WILL AWARD BRAINLIEST TO WHOEVER ANSWERS THE QUESTION BEST!

Answers

Considering it's discriminant, it is found that:

A. The classmate is wrong, as the discriminant is of zero, hence the equation has one solution.

B. The quadratic equation has 1 x-intercept.

What is the discriminant of a quadratic equation and how does it influence the solutions?

A quadratic equation is modeled by:

y = ax^2 + bx + c

The discriminant is:

[tex]\Delta = b^2 - 4ac[/tex]

The solutions are as follows:

If [tex]\mathbf{\Delta > 0}[/tex], it has 2 real solutions.If [tex]\mathbf{\Delta = 0}[/tex], it has 1 real solutions.If [tex]\mathbf{\Delta < 0}[/tex], it has 2 complex solutions.

In this problem, the equation is:

y = 9x² - 6x + 1.

The coefficients are a = 9, b = -6 and c = 1, hence the discriminant is:

[tex]\Delta =(-6)^2 - 4(9)(1) = 36 - 36 = 0[/tex]

Since the discriminant is zero, the classmate is wrong, as it means that the equation has one solution = one x-intercept.

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Help whats the answer and an explanation to it

Answers

Answer:

C

Step-by-step explanation:

the answer is c

Prove Sin(90-A)=cosA

Answers

Step-by-step explanation:

sin(90-A) = sin90cosA-sinAcos90

= cosA*1-0*sinA

= cos90

hence proved

Step-by-step explanation:

sin(90-A)=cosA

sin(90-A)=sin(90-A)

90-A=90-A

-A+A=90-90

=0

What is the answer (X^2)(X)(4)

Answers

Answer:

Simplified: 4X^3

Step-by-step explanation:

Simplify the expression.

Review the graph of function j(x).


On a coordinate plane, a line starts at open circle (2, 6) and goes down through (negative 2, 2). A solid circle is at (3, 6). A curve goes from solid circle (2, 3) to open circle (3, 4). A line goes from the open circle to closed circle (6, 5).


What is Limit of j (x) as x approaches 3?


3

4

5

6

Answers

The limit of j (x) as x approaches 3 is 4.

According to the question, A line begins at an open circle (2, 6) on a coordinate plane and descends through ( -2, 2). At, a complete circle is (3, 6). From a solid circle (2, 3), a curve leads to an open circle (3, 4). From the open circle to the closed circle, a line runs (6, 5).

From the graph, it can be seen that the limit of the function j(x) as the value of x approaches 3 is 4.

A diagram or pictorial representation that organizes the depiction of data or values is known as a graph.

The relationships between two or more items are frequently represented by the points on a graph.

Learn more about limits here:

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Answer:

4

Step-by-step explanation:

egg 2023

I honestly need help with these

Answers

9. The curve passes through the point (-1, -3), which means

[tex]-3 = a(-1) + \dfrac b{-1} \implies a + b = 3[/tex]

Compute the derivative.

[tex]y = ax + \dfrac bx \implies \dfrac{dy}{dx} = a - \dfrac b{x^2}[/tex]

At the given point, the gradient is -7 so that

[tex]-7 = a - \dfrac b{(-1)^2} \implies a-b = -7[/tex]

Eliminating [tex]b[/tex], we find

[tex](a+b) + (a-b) = 3+(-7) \implies 2a = -4 \implies \boxed{a=-2}[/tex]

Solve for [tex]b[/tex].

[tex]a+b=3 \implies b=3-a \implies \boxed{b = 5}[/tex]

10. Compute the derivative.

[tex]y = \dfrac{x^3}3 - \dfrac{5x^2}2 + 6x - 1 \implies \dfrac{dy}{dx} = x^2 - 5x + 6[/tex]

Solve for [tex]x[/tex] when the gradient is 2.

[tex]x^2 - 5x + 6 = 2[/tex]

[tex]x^2 - 5x + 4 = 0[/tex]

[tex](x - 1) (x - 4) = 0[/tex]

[tex]\implies x=1 \text{ or } x=4[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]\boxed{x=1} \implies y = \dfrac{1^3}3 - \dfrac{5\cdot1^2}2 + 6\cdot1 - 1 = \boxed{y = \dfrac{17}6}[/tex]

[tex]\boxed{x = 4} \implies y = \dfrac{4^3}3 - \dfrac{5\cdot4^2}2 + 6\cdot4 - 1 \implies \boxed{y = \dfrac{13}3}[/tex]

11. a. Solve for [tex]x[/tex] where both curves meet.

[tex]\dfrac{x^3}3 - 2x^2 - 8x + 5 = x + 5[/tex]

[tex]\dfrac{x^3}3 - 2x^2 - 9x = 0[/tex]

[tex]\dfrac x3 (x^2 - 6x - 27) = 0[/tex]

[tex]\dfrac x3 (x - 9) (x + 3) = 0[/tex]

[tex]\implies x = 0 \text{ or }x = 9 \text{ or } x = -3[/tex]

Evaluate [tex]y[/tex] at each of these.

[tex]A:~~~~ \boxed{x=0} \implies y=0+5 \implies \boxed{y=5}[/tex]

[tex]B:~~~~ \boxed{x=9} \implies y=9+5 \implies \boxed{y=14}[/tex]

[tex]C:~~~~ \boxed{x=-3} \implies y=-3+5 \implies \boxed{y=2}[/tex]

11. b. Compute the derivative for the curve.

[tex]y = \dfrac{x^3}3 - 2x^2 - 8x + 5 \implies \dfrac{dy}{dx} = x^2 - 4x - 8[/tex]

Evaluate the derivative at the [tex]x[/tex]-coordinates of A, B, and C.

[tex]A: ~~~~ x=0 \implies \dfrac{dy}{dx} = 0^2-4\cdot0-8 \implies \boxed{\dfrac{dy}{dx} = -8}[/tex]

[tex]B:~~~~ x=9 \implies \dfrac{dy}{dx} = 9^2-4\cdot9-8 \implies \boxed{\dfrac{dy}{dx} = 37}[/tex]

[tex]C:~~~~ x=-3 \implies \dfrac{dy}{dx} = (-3)^2-4\cdot(-3)-8 \implies \boxed{\dfrac{dy}{dx} = 13}[/tex]

12. a. Compute the derivative.

[tex]y = 4x^3 + 3x^2 - 6x - 1 \implies \boxed{\dfrac{dy}{dx} = 12x^2 + 6x - 6}[/tex]

12. b. By completing the square, we have

[tex]12x^2 + 6x - 6 = 12 \left(x^2 + \dfrac x2\right) - 6 \\\\ ~~~~~~~~ = 12 \left(x^2 + \dfrac x2 + \dfrac1{4^2}\right) - 6 - \dfrac{12}{4^2} \\\\ ~~~~~~~~ = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4[/tex]

so that

[tex]\dfrac{dy}{dx} = 12 \left(x + \dfrac14\right)^2 - \dfrac{27}4 \ge 0 \\\\ ~~~~ \implies 12 \left(x + \dfrac14\right)^2 \ge \dfrac{27}4 \\\\ ~~~~ \implies \left(x + \dfrac14\right)^2 \ge \dfrac{27}{48} = \dfrac9{16} \\\\ ~~~~ \implies \left|x + \dfrac14\right| \ge \sqrt{\dfrac9{16}} = \dfrac34 \\\\ ~~~~ \implies x+\dfrac14 \ge \dfrac34 \text{ or } -\left(x+\dfrac14\right) \ge \dfrac34 \\\\ ~~~~ \implies \boxed{x \ge \dfrac12 \text{ or } x \le -1}[/tex]

13. a. Compute the derivative.

[tex]y = x^3 + x^2 - 16x - 16 \implies \boxed{\dfrac{dy}{dx} = 3x^2 - 2x - 16}[/tex]

13. b. Complete the square.

[tex]3x^2 - 2x - 16 = 3 \left(x^2 - \dfrac{2x}3\right) - 16 \\\\ ~~~~~~~~ = 3 \left(x^2 - \dfrac{2x}3 + \dfrac1{3^2}\right) - 16 - \dfrac13 \\\\ ~~~~~~~~ = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3[/tex]

Then

[tex]\dfrac{dy}{dx} = 3 \left(x - \dfrac13\right)^2 - \dfrac{49}3 \le 0 \\\\ ~~~~ \implies 3 \left(x - \dfrac13\right)^2 \le \dfrac{49}3 \\\\ ~~~~ \implies \left(x - \dfrac13\right)^2 \le \dfrac{49}9 \\\\ ~~~~ \implies \left|x - \dfrac13\right| \le \sqrt{\dfrac{49}9} = \dfrac73 \\\\ ~~~~ \implies x - \dfrac13 \le \dfrac73 \text{ or } -\left(x-\dfrac13\right) \le \dfrac73 \\\\ ~~~~ \implies \boxed{x \le 2 \text{ or } x \ge \dfrac83}[/tex]

What is the smallest odd number of using 9,3,6,8,1,9​​​​

Answers

Answer: well one is

Bc its the smallest besides zero, but zero is neither odd or even

Step-by-step explanation:

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