urgent help needed algebra 2

Answer:

First simply it as 2x²+3

(a) vertex (0,3)

(b)f(0)=3

(c) no x-intercept

Answer:

Step-by-step explanation:

(xm , ym )  = x1 + x2 / 2               and            y1 + y2 / 2

               = 9 +3 / 2                                      = 10 -10 / 2

               = 12/2                                            = 0/2

                = 6                                                  = 0

                So midpoints are (6 , 0)

Answer:

midpoint =  [tex](9\frac{1}{2} , -3\frac{1}{2} )[/tex]

Step-by-step explanation:

To find the midpoint of a line segment, you have to find the average of the x and y-values of the end-points, i.e., add the x-coordinate values and divide the answer by 2, and do the same for the y-coordinate values.

• midpoint = [tex](\frac{x_{2} + x_1}{2}, \frac{y_2 + y_1}{2} )[/tex]

                 = [tex](\frac{9 + 10}{2}, \frac{3 + (-10)}{2} )[/tex]

                 = [tex](\frac{19}{2}, \frac{-7}{2} )[/tex]

                 = [tex](9\frac{1}{2} , -3\frac{1}{2} )[/tex]

Answer:

approximately $23,655.91 (rounded to the nearest hundredths place)

Step-by-step explanation:

[tex]22,000=0.93x\\x=23,655.91[/tex]

Answer:

Step-by-step explanation:

let the list price=x

93% of x=22,000

x=22,000 ×100/93

≈23,655.91 $

Answer:

Function g(x) is function f(x) vertically stretched by a factor of 6, reflected in the x-axis, and translated 2 units up.

Step-by-step explanation:

The graph of function f(x) is the parent function.

(Parent functions are the simplest form of a given family of functions).

The graph of g(x) is related to the graph of f(x) by a series of transformations.  To determine the series of transformations, work out the steps of how to go from f(x) to g(x).

Transformations

For a > 0

[tex]\begin{aligned} y =a\:f(x) \implies & f(x) \: \textsf{stretched/compressed vertically by a factor of }\:a \\& \textsf{If }a > 1 \textsf{ it is stretched by a factor of}\: a\\& \textsf{If }0 < a < 1 \textsf{ it is compressed by a factor of}\: a \end{aligned}[/tex]

[tex]y=-f(x) \implies f(x) \: \textsf{reflected in the} \: x \textsf{-axis}[/tex]

[tex]y=f(x)+a \implies f(x) \: \textsf{translated}\:a\:\textsf{units up}[/tex]

Parent function:

[tex]f(x)=x^3[/tex]

Step 1

Multiply the parent function by 6:

[tex]\implies 6f(x)=6x^3[/tex]

Therefore, this is a vertical stretch by a factor of 6.

Step 2

Now make the function negative:

[tex]\implies -6f(x)=-6x^3[/tex]

Therefore, this is a reflection in the x-axis.

Step 3

Finally, add 2 to the function:

[tex]\implies -6f(x)+2=-6x^3+2[/tex]

Therefore, the function has been translated 2 units up.

Conclusion

Function g(x) is function f(x) vertically stretched by a factor of 6, reflected in the x-axis, and translated 2 units up.

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The ratio of the area of the red rectangle to the blue rectangle in graph A is 3 : 5

The median weekly earnings on the graphs are

Represent as a ratio

Ratio = $750 : $1250

Divide by 250

Ratio = 3 : 5

Hence, the ratio of the median weekly earnings is 3 : 5

In (a), we have:

Ratio = 3 : 5

The horizontal scale is given as:

Ratio = 1 unit : 1 grid mark

The rectangles in graph A have a width of 1 unit.

So, we have:

Ratio = 3 * 1: 5 * 1

Ratio = 3 : 5

Hence, the ratio of the area of the red rectangle to the blue rectangle in graph A is 3 : 5

Recall that:

Ratio = 3 : 5

Ratio = 1 unit : 1 grid mark

From the graph, we have the following widths:

Red = 3 units

Blue = 5 units

So, we have:

Ratio = 3 * 3 : 5 * 5

Simplify

Ratio = 9 : 25

Hence, the ratio of the area of the red rectangle to the blue rectangle in graph B is 9 : 25

Recall that:

Ratio = 3 : 5

Ratio = 1 unit : 1 grid mark

From the graph, we have the following widths:

Red = 3 units

Blue = 5 units

Since the base are squares, we have:

Ratio = 3 * 3  * 3 : 5 * 5 * 5

Simplify

Ratio = 27 : 125

Hence, the ratio of the volume of the red cube to the blue cube in graph C is 27 : 125

The most misleading graph is graph B.

This is so because the blue rectangle and the red rectangle do not have the same width when plotted on the same scale

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Answer: 180 degrees

Step-by-step explanation:

Angles on a line add to 180 degrees since they form a straight angle.

The minimum percentage of recently sold homes with prices between $197,650 and $242,350 is 88.9%.

Mean is the ratio of the sum of all the data points to the number of data points.

It is given that

mean of $220,000 with a standard deviation of $7450.

The range is given , let the range is represented by x - --y

It is given that x = 197650 and y = 242350

Let the number of homes sold is k

To determine the value of k

upper level = (y-mean)/standard deviation = (242350-220000)/7450 = 3

lower level = (mean-x)/standard deviation = (220000-197650)/7450 = 3

probability = 1-(1/k²)

k= 3

= 1 - (1/3^2)

= 1 - 1/9

= 0.889 or 88.9%

So, the minimum percentage of recently sold homes with prices between $197,650 and $242,350 is 88.9%.

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Answer:

A. negative

Step-by-step explanation:

It is clear from the accompanying graph that the value of y decreases as the value of x grows. The line has a negative slope, and the right answer is C, as shown by the inverse connection between the values of x and y.

Reasoning for other incorrect options:

Answer:

One solution

Step-by-step explanation:

The equation can be rewrite as 3x - 6 = 22 - x

so we try to get all the x on one side so we add x and 6 to both sides so we get 4x = 28 then we divide by 4 both side and get x=7

1600 children and 1400 adults attended

Let the children be x and adult be y.

So, we have the following equations:

x + y = 3000

1.5x + 5y = 9400

Make x the subject in x + y = 3000

x = 3000 - y

Substitute x = 3000 - y  in 1.5x + 5y = 9400

1.5(3000 - y) + 5y = 9400

Expand

4500 - 1.5y + 5y = 9400

Evaluate the like terms

3.5y = 4900

Divide both sides by 3.5

y = 1400

Substitute y = 1400 in x = 3000 - y

x = 3000 - 1400

Evaluate

x = 1600

Hence, 1600 children and 1400 adults attended

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Answer:

etrf4f3dvef3rf3rfr2wrgwrf2rg3rf3rgerferferfef I'm

Answer:

[tex]20a^3b^3[/tex]

Step-by-step explanation:

Binomial Series

[tex](a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n[/tex]

Factorial is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example:  4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

[tex]\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3[/tex]

[tex]\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3[/tex]

[tex]\implies \dfrac{6!}{3!3!}a^{3}b^3[/tex]

[tex]\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3[/tex]

[tex]\implies \left(\dfrac{120}{6}\right)a^{3}b^3[/tex]

[tex]\implies 20a^3b^3[/tex]

distance = speed * time

the first bus speed = x

second bus = x - 12

the sum of distance = 765 miles

first distance = 6x

second distance = 6 (x -12)

6x + 6(x-12) = 765

x + x -12 = 127.5

2x = 139.5

x = 69.75 mi/h which is the faster bus

the second bus speed = 57.75 mi/h

|2x + 5| ≤ 11

x = 3

2(3) + 5

6 + 5

11 ≤ 11

The solution for the x in the inequality |2x + 5| ≤ 11 is {-8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3}

Inequality shows relation between two expression which are not equal to each others.

The given inequality is,

|2x + 5| ≤ 11

To find the solution for the x, solve the inequality,

|2x + 5| ≤ 11

-11 ≤ 2x + 5 ≤ 11

Subtract 5 from whole the expression,

-11 - 5 ≤ 2x + 5 - 5 ≤ 11 - 5

-16 ≤ 2x ≤ 6

-8 ≤ x ≤ 3

The value of x for the given inequality varies from -8 to 3.

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Answer:

First find the x and y intercepts. Recall that intercepts intersect the x and y axis of a graph, therefore either the x or y value of a coordinate point must be 0 in order for it to be an intercept.

After you find the intercept, plot the them on the cartesian coordinate system and draw a straight line (it’s a linear equation) going through the two intercepts.

Side note: Not sure how credentials work on Quora yet. I meant to say I am a student

The equation of the lines can be plotted on the graph after calculating the coordinates on each line.

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have two linear equation:

x = 4y  

x + y = 7.0

To plot the linear equation first we will find the few coordinates to plot on the coordinate plane.

For the equation of line:

x = 4y

x 0 1 2 3 -1 -2 -3

y 0 4 8 12 -4 -8 -12

For the equation of line:

x + y = 7.0

x 0 1 2 3 -1 -2 -3

y 7 6 5 4 8 9 10

Thus, the equation of the lines can be plotted on the graph after calculating the coordinates on each line.

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Th equation that can be used to find the umber of minutes Teri ran is m + 2 = 28 minutes.

Addition is a mathematical operation that is used to determine the sum of two or more numbers.

The total minutes run by Julie = minutes ran by Teri + 2

28 = m + 2

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Step-by-step explanation:

(4²-x²)³/²

or,(4+X) (4-X)

Substitute [tex]x = 4 \sin(y)[/tex], so that [tex]dx = 4\cos(y)\,dy[/tex]. Part of the integrand reduces to

[tex]16 - x^2 = 16 - (4\sin(y))^2 = 16 - 16 \sin^2(y) = 16 (1 - \sin^2(y)) = 16 \cos^2(y)[/tex]

Note that we want this substitution to be reversible, so we tacitly assume [tex]-\frac\pi2\le y\le \frac\pi2[/tex]. Then [tex]\cos(y)\ge0[/tex], and

[tex](16-x^2)^{3/2} = 16^{3/2} \left(\cos^2(y)\right)^{3/2} = 64 |\cos(y)|^3 = 64 \cos^3(y)[/tex]

(since [tex]\sqrt{x^2} = |x|[/tex] for all real [tex]x[/tex])

So, the integral we want transforms to

[tex]\displaystyle \int (16 - x^2)^{3/2} \, dx = 64 \int \cos^3(y) \times 4\cos(y) \, dy = 256 \int \cos^4(y) \, dy[/tex]

Expand the integrand using the identity

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

to write

[tex]\displaystyle \int (16 - x^2)^{3/2} \, dx = 256 \int \left(\frac{1 + \cos(2y)}2\right)^2 \, dy \\\\ = 64 \int (1 + 2 \cos(2y) + \cos^2(2y)) \, dy \\\\ = 64 \int (1 + 2 \cos(2y) + \frac{1 + \cos(4y)}2\right) \, dy \\\\ = 32 \int (3 + 4 \cos(2y) + \cos(4y)) \, dy[/tex]

Now integrate to get

[tex]\displaystyle 32 \int (3 + 4 \cos(2y) + \cos(4y)) \, dy = 32 \left(3y + 2 \sin(2y) + \frac14 \sin(4y)\right) + C \\\\ = 96 y + 64 \sin(2y) + 8 \sin(4y) + C[/tex]

Recall the double angle identity,

[tex]\sin(2y) = 2 \sin(y) \cos(y)[/tex]

[tex]\implies \sin(4y) = 2 \sin(2y) \cos(2y) = 4 \sin(y) \cos(y) (\cos^2(y) - \sin^2(y))[/tex]

By the Pythagorean identity,

[tex]\cos(y) = \sqrt{1 - \sin^2(y)} = \sqrt{1 - \dfrac{x^2}{16}} = \dfrac{\sqrt{16-x^2}}4[/tex]

Finally, put the result back in terms of [tex]x[/tex].

[tex]\displaystyle \int (16 - x^2)^{3/2} \, dx \\\\ = 96 \sin^{-1}\left(\frac x4\right) + 128 \frac x4 \frac{\sqrt{16-x^2}}4 + 32 \frac x4 \frac{\sqrt{16-x^2}}4 \left(\frac{16-x^2}{16} - \frac{x^2}{16}\right) + C \\\\ = 96 \sin^{-1}\left(\frac x4\right) + 8 x \sqrt{16 - x^2} + \frac14 x \sqrt{16 - x^2} (8 - x^2) + C \\\\ = \boxed{96 \sin^{-1}\left(\frac x4\right) + \frac14 x \sqrt{16 - x^2} \left(40 - x^2\right) + C}[/tex]

Answer:

x = sqrt(105)

Step-by-step explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2 + b^2 = c^2  where a and b are the legs and c is the hypotenuse

16^2 + x^2 = 19^2

256 + x^2 = 361

x^2 = 361 -256

x^2 =105

Take the square root of each side

sqrt(x^2) = sqrt(105)

x = sqrt(105)

Answer:

h(-7) = 44

Step-by-step explanation:

h(x) = h (-7) means x= -7

h(x) = (-7)² - 5

49 - 5

44

Answer: [tex]x=0,x=-\frac{9}{2}, x=\frac{9}{2}[/tex]

Step-by-step explanation:

[tex]-3x(4x^2-81)(x^2+64)=0\\[/tex]

multiply the terms together

[tex]-12x^5-525x^3+15552x=0[/tex]

factor left side of the equation

[tex]3x(-x^2-64)(2x+9)(2x-9)=0[/tex]

set factors equal to 0

[tex]x=0,x=-\frac{9}{2}, x=\frac{9}{2}[/tex]

Answer:

-6/11 goes with -1/2

-7/9 goes with -3/4

-3/13 goes with -1/4

Step-by-step explanation:

6/11 is close to 6/12 which is the same as 1/2, so -6/11 goes with -1/2

7/9 is close to 6.75/9 which is 3/4, so -7/9 is close to -3/4

3/13 is close to 3/12 which is the same as 1/4, so -3/13 goes with -1/4

Answer:

1.  "a"  [tex]u'=6x[/tex]

2.  "d"  [tex]v'=15x^2[/tex]

3.  "b"  [tex]y'=75x^4+30x^2+6x[/tex]

Step-by-step explanation:

Part 1.

Given [tex]u=3x^2+2[/tex], find [tex]\frac{du}{dx} \text{ or } u'[/tex].

[tex]u=3x^2+2[/tex]

Apply a derivative to both sides...

[tex]u'=(3x^2+2)'[/tex]

Derivatives of a sum are the sum of derivatives...

[tex]u'=(3x^2)'+(2)'[/tex]

Scalars factor out of derivatives...

[tex]u'=3(x^2)'+(2)'[/tex]

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

[tex]u'=3(2x)+0[/tex]

Simplify...

[tex]u'=6x[/tex]

So, option "a"

Part 2.

Given [tex]v=5x^3+1[/tex], find [tex]\frac{dv}{dx} \text{ or } v'[/tex].

[tex]v=5x^3+1[/tex]

Apply a derivative to both sides...

[tex]v'=(5x^3+1)'[/tex]

Derivatives of a sum are the sum of derivatives...

[tex]v'=(5x^3)'+(1)'[/tex]

Scalars factor out of derivatives...

[tex]v'=5(x^3)'+(1)'[/tex]

Apply power rule for derivatives (decrease power by 1; mutliply old power as a factor to the coefficient); Derivative of a constant is zero...

[tex]v'=5(3x^2)+0[/tex]

Simplify...

[tex]v'=15x^2[/tex]

So, option "d"

Part 3.

Given [tex]y=(3x^2+2)(5x^3+1)[/tex]

[tex]\text{Then if } u=3x^2+2 \text{ and } v=5x^3+1, y=u*v[/tex]

To find [tex]\frac{dy}{dx} \text{ or } y'[/tex], recall the product rule:  [tex]y'=uv'+u'v[/tex]

[tex]y'=uv'+u'v[/tex]

Substituting the expressions found from above...

[tex]y'=(3x^2+2)(15x^2)+(6x)(5x^3+1)[/tex]

Apply the distributive property...

[tex]y'=(45x^4+30x^2)+(30x^4+6x)[/tex]

Use the associative and commutative property of addition to combine like terms, and rewrite in descending order:

[tex]y'=75x^4+30x^2+6x[/tex]

So, option "b"

Answer:

D.

Step-by-step explanation:

It is the only one that contains all the x-values of the points. (Domain is the set of x-values).

Answer:

x = 25

Step-by-step explanation:

Solve by isolating x:
(x-7)2 = 36
x-7 = 18
x = 25

Answer:

$0.90

Step-by-step explanation:

If each piece of chocolate is $1, then 8 pieces is $8. Wei Xun bought 8 pieces of chocolate and 10 lollipops, which came out to $17. The chocolate was $8 total, leaving $9 for the 10 lollipops. 9/10 = 0.9. So, each lollipop was $0.90, or 90 cents.

Answer:

C=55m + 30

Step-by-step explanation:

You would multiply 55 by the number of months so you'd get 55m, and since 30 is a one time fee you would just add 30 to your monthly payment.

The appropriate expression to identify the amount of formalin is: x = 3200 ÷ 100 × 44%

To calculate the amount of formalin in this substance we must perform the following mathematical operation:

According to the above, the mathematical expression would be:

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Answer:

a = 1

b = 2

c = 1

Factored form: (y + 1)^2 or (y + 1)(y + 1)

Step-by-step explanation:

The flux is 9.

Flux is the presence of a force field in a specified physical medium, or the flow of energy through a surface.

Given:

2x+2y+z= 1

F = 2i+2j + 2k

Now,

r = xi + yj + z( 1-2x-2y) K

dr/dx= i - 2k

dr/dy = j-2k

dr/dx* dr/dy

= ( i - 2k) * (j-2k)

= 2i + 2j + k

F(x)= 2i+2j + 2k

F(x). da = 4 +4 +2 = 10 dxdy

Hence, flux

= [tex]\int\limits^1_0 {\int\limits^{1-2y}_0 {10 } \, dx dy } \,[/tex]

=  [tex]\int\limits^1_0[/tex]  10(1-2y) dx

= [tex]\int\limits^1_0[/tex] 10-2y

= 10(1) - (1)²

=9

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The plane has intercepts (1/2, 0, 0), (0, 1/2, 0), and (0, 0, 1). Parameterize [tex]S[/tex] by the vector function

[tex]\vec s(u,v) = \dfrac{(1-u)(1-v)}2 \, \vec\imath + \dfrac{u(1-v)}2 \, \vec\jmath + v \,\vec k[/tex]

with [tex]0\le u\le1[/tex] and [tex]0\le v\le1[/tex]. (More explicitly, we have the parameterization

[tex]\vec s(u,v) = (1-v)((1-u) p_1 + u p_2) + v p_3[/tex]

where [tex]p_i[/tex] denote the given points.)

The normal vector to [tex]S[/tex] is

[tex]\vec n = \dfrac{\partial\vec s}{\partial u} \times \dfrac{\partial\vec s}{\partial v} = \dfrac{1-v}2\,\vec\imath + \dfrac{1-v}2\,\vec\jmath + \dfrac{1-v}4\,\vec k[/tex]

Then the flux of [tex]\vec F = 2\,\vec\imath+2\,\vec\jmath+2\,\vec k[/tex] across [tex]S[/tex] is given by the surface integral,

[tex]\displaystyle \iint_S \vec F \cdot d\vec\sigma = \iint_S \vec F \cdot \vec n \, dA[/tex]

[tex]\displaystyle = \int_0^1 \int_0^1 \left(2\,\vec\imath+2\,\vec\jmath+2\,\vec k) \cdot \left(\frac{1-v}2\,\vec\imath + \frac{1-v}2\,\vec\jmath + \frac{1-v}4\,\vec k\right) \, du \, dv[/tex]

[tex]\displaystyle = \frac52 \int_0^1 \int_0^1 (1-v) \, du \, dv[/tex]

[tex]\displaystyle  = \frac52 \int_0^1 (1-v) \, dv = \boxed{\frac54}[/tex]