Given that Q(x)=2x^2 +5x-3 find and simplify Q(a+h)-Q(a-h)

Hi there!

[tex]\large\boxed{\approx 43.33 min}}[/tex]

Recall:

d = st, where:

d = distance

s = speed

t = time

We can set up an expression where the extra ten minutes is taken into account:

50x = 65(x - 10) <--- because J left 10 minutes after, we must subtract from the time variable, or "x".

Solve for x:

50x = 65x - 650

Subtract 65x from both sides:

-15x = -650

Divide both sides by -15:

x ≈ 130/3 or 43.33 min

The answer is kindly 21 (100% correct)

Answer:

45.9

Step-by-step explanation:

Answer:

0 is the answer assuming the whole thing is a fraction where the numerator is f(a+h)-f(a) and the denominator is h.

Step-by-step explanation:

If the expression for f is really a constant, then the difference quotient will lead to an answer of 0.

If the extra for f is linear (including constant expressions), the difference quotient will be the slope of the expression.

However, let's go about it long way for fun.

If f(x)=5, then f(a)=5.

If f(x)=5, then f(a+h)=5.

If f(a)=5 and f(a+h)=5, then f(a+h)-f(a)=0.

If f(a+h)-f(a)=0, then [f(a+h)-f(a)]/h=0/h=0.

Answer:

[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Step-by-step explanation:

Solving (a): The frequency distribution

Given that:

[tex]Lowest = 138[/tex] --- i.e. the lowest class value

[tex]Class = 5[/tex] --- Number of classes

From the given dataset is:

[tex]Highest = 459[/tex]

So, the range is:

[tex]Range = Highest - Lowest[/tex]

[tex]Range = 459 - 138[/tex]

[tex]Range = 321[/tex]

Divide by the number of class (5) to get the class width

[tex]Width = 321 \div 5[/tex]

[tex]Width = 64.2[/tex]

Approximate

[tex]Width = 64[/tex]

So, we have a class width of 64 in each class;

The frequency table is as follows:

[tex]\begin{array}{cc}{Class}& {Frequency} & 138 - 202 & 14 & 203 - 267 & 5 & 268 - 332 & 3 & 333 - 397 & 1 & 398 - 462 & 3 \ \end{array}[/tex]

Solving (b) The relative frequency histogram

First, we calculate the relative frequency by dividing the frequency of each class by the total frequency

So, we have:

[tex]\begin{array}{ccc}{Class}& {Frequency} & {Relative\ Frequency} & 138 - 202 & 14 & 0.53 & 203 - 267 & 5 & 0.19 & 268 - 332 & 3 & 0.12 & 333 - 397 & 1 & 0.04 & 398 - 462 & 3 & 0.12 \ \end{array}[/tex]

See attachment for histogram

The class with the greatest is 138- 202 and the class with the least relative frequency is 333 - 397

Answer:

Third Choice - The graph of g(x) is the graph of f(x) compressed vertically by a factor of 3

Step-by-step explanation:

x^2 is the the parent function, so it opens up with a normal compression.

Any number > (greater than) 1 as a coefficient of x will lead to a vertical compression (narrower parabola), while any number < (less than) 1 as a coefficient of x will lead to a vertical stretch (wider parabola).

So, 3x^2 would have to have to be a compressed parabola.

I hope this helps!

Answer:

The graph of g(x) is the graph of f(x) stretched vertically by a factor of 3.

Step-by-step explanation:

A vertical stretch or shrink of a function, kf(x), results from multiplying the entire function by a constant, k.

In this case, g(x) equals 3 times f(x). If k > 1, then the graph will be stretched vertically (along the direction of the y-axis) by a factor of k.

So, the graph of g(x) is the graph of f(x) stretched vertically by a factor of 3.

Answer:

a) r ⋀~p

b)(r⋀p)⟶q

c) ~r ⟶ ~q

d) (~p ⋀r) ⟶q

Step-by-step explanation:

To solve this question we will make use of logic symbols in truth table.

We are told that;

p means "The user enters

a valid password,”

q means “Access is granted,”

r means “The user has paid the

subscription fee”

A) The user has paid the subscription fee, but does not enter a valid

password.”

Fist part of the statement is correct and so it will be "r". Second part of the statement is a negation and will be denoted by ~p. Since both statements are joined together in conjunction, we will use the conjuction symbol in between them which is "⋀" Thus, we have; r ⋀~p

B) Still using logic symbols, we have;

(r⋀p)⟶q

⟶ means q is true when r and p are true.

C) correct symbol is ~r ⟶ ~q

Since both statements are negation of the question. And also, if ~r is true then ~q is also true.

D) Similar to answer A to C above, applying similar conditions, we have (~p ⋀r) ⟶q

Answer:

Part A 12 ≤ 6x ≤ 36

Part B 2 ≤ x ≤ 6

Step-by-step explanation:

Answer:

it's H. 1/2 in.=1,000 ft

[tex]{hope 8 helps}}[/tex]

Answer:

False

Step-by-step explanation:

To show that this is false, all we have to do is find one example.

9 is an odd number less than 15

9 is composite

9 =3*3

Answer:

The statement is false

Step-by-step explanation:

Given

See comment for complete statement

Required

Is the statement true or false

From central limit theorem, we understand that a distribution is approximately normal if the distribution takes a sample considered to be large enough from the population.

Also, the mean and the standard deviation are known.

However, the given statement implies that the distribution will be normal depending on an underlying distribution; this is false.

the e-function stuff can be confusing sometimes, but notices that g(x) / the blue line, is just somewhat lower, rest is the same.

how much lower? look at the y-intercepts

f(0)= "about 5"

g(0)= "about -3"

with this y-intercept only option c can work

9514 1404 393

Answer:

  y = 2

Step-by-step explanation:

Subtracting the second equation from the third gives ...

  (2z +3x) -(2z -2x) = (-6) -(-6)

  5x = 0

  x = 0

Using this in the third equation, we have ...

  2z +0 = -6

  z = -3

And substituting these values into the first equation, we have ...

  5y -3(0) -4(-3) = 22

  5y = 10 . . . . . subtract 12

  y = 2

__

The solution to the system is (x, y, z) = (0, 2, -3).

Answer:

I think the choose (B)

5x/x + 3/x

Answer:

I thinkchoose no.3

5x+3

5x+3x

Answer:

The center of a circle is the point in the circle which is equidistant to all the edges of thr circle. The point a is the center, while point b is an arbitrary point in the circle. Find attachment for the diagram.

Answer:

[tex]P(\frac{A}{B'})[/tex]=0.111

Step-by-step explanation:

Given:

The probability of winning the first game is 10.1

The first game is won

The probability of winning the second game is 15

If the first is lost, the probability of winning the second game is 25

Solution:

[tex]P(B)=P(A)P(\frac{B}{A})+P(A')P(\frac{B}{A'})\\ =0.1(0.15)+(0.3)*0.25)\\P(B)=0.24 ------(1)\\P(\frac{A}{B})=\frac{P(\frac{B}{A})P(A) }{P(B)}\\ =\frac{0.15(0.1)}{0.24}\\ =0.0625 ------(2)\\P(B')=1-P(B)=0.76 ------(3)\\P(A)=P(B)P(\frac{A}{B})+P(B')P(\frac{A}{B'})\\0.1=0.24(0.0625)+0.76(p(\frac{A}{B'} ))\\P(\frac{A}{B'})=0.111[/tex]

Answer:

[tex]P(W_1/W_2')=0.1110[/tex]

Step-by-step explanation:

Probability of winning the first game be considering the given factors be, [tex]W_1=0.1[/tex]

Probability of winning the second game be considering the given factors be, [tex]W_2[/tex]= probability of winning the second game when the first game is won + probability of winning the second game when the first game is lost:

[tex]P(W_2)=P(W_1).P(W_2/W_1)+P(W_1').P(W_2/W_1')[/tex]

[tex]P(W_2)=0.1\times 0.15+0.9\times 0.25[/tex]

[tex]P(W_2)=0.24[/tex]

Hence the probability of losing the second game:

[tex]P(W_2')=1-P(W_2)[/tex]

[tex]P(W_2')=0.76[/tex]

Probability of winning the first game when the second game is won:

[tex]P(W_1/W_2)=\frac{P(W_2/W_1).P(W_1)}{P(W_2)}[/tex]

[tex]P(W_1/W_2)=\frac{0.15\times 0.1}{0.24}[/tex]

[tex]P(W_1/W_2)=0.0625[/tex]

Probability of winning the first game be considering the given factors, [tex]W_1[/tex]= probability of winning the first game when the second game is won + probability of winning the first game when the second game is lost:

[tex]P(W_1)=P(W_2).P(W_1/W_2)+P(W_2').P(W_1/W_2')[/tex]

[tex]0.1=0.24\times0.0625+0.76\times P(W_1/W_2')[/tex]

[tex]P(W_1/W_2')=0.1110[/tex]

Answer:

(a) A = 20(1.6)^t

(b) 210 rabbits

Step-by-step explanation:

Initial number of rabbits = 20

rate of growth, R = 60 % annually

(A) The general equation is  

[tex]A = P \left ( 1+\frac{R}{100} \right )^t\\\\A = 20\left ( 1+\frac{60}{100} \right )^t\\\\A = 20 (1.6)^t[/tex]

(B) Let the time, t = 5 years

So, the population after 5 years is

[tex]A = 20 (1.6)^5\\\\A = 209.7 = 210 rabbits[/tex]

Answer:

Step-by-step explanation:

i think if its -10 degrees i think the fahrenheit would be 50 degrees

Step-by-step explanation:

Let's say the rectangle's width is equal to y. We know that the length is three times the width, so the length = 3 * y. We also know that the area for a rectangle is equal to length * width, so the area, z, is equal to

(3*y) * y = z

3 * y² = z

Now, let's increase the width of the rectangle by 1. We can replace y with y+1 (as y+1 is 1 greater than y), and 3 * y with 3 * (y+1) to get

3*(y+1) * (y+1) = new area

(3y+3)*(y+1) = new area

3y²+3y +3 y + 3 = new area

3y² + 6y + 3 = new area

The difference in area is equal to the new area subtracted by the old area, or

3y²+6y+3 - 3y² = 6y +3. The variable for x is not given, so if x = (2y+1), the answer would be the second choice. However, solely using the information given, it is impossible to determine a solution outside of saying that it is not option 4, as 6y + 3 ≠ 3

Answer:

x = 3 , y = 4

Step-by-step explanation:

5x + y = 19 --------- ( 1 )

=> y = 19 - 5x

7x - 2y = 13 ------------ ( 2 )

Substitute y in ( 2 ) :

                        7x - 2( 19 - 5x ) = 13

                        7x - 38 + 10x = 13

                        17x = 13 + 38

                        17x = 51

                           x = 3

Substitute x in ( 1 ) :

                           5x + y = 19

                           5( 3 ) + y = 19

                            15 + y = 19

                              y = 19 - 15

                               y = 4

Answer:

A. 8h=m

Step-by-step explanation:

$8* h= total money m earned.

Answer:

90

Step-by-step explanation:

=> x + y = 12

=> x² + y² + 2xy = 144

=> x² + y² + 2 * 27 = 144

=> x² + y² = 144 - 54

=> x² + y² = 90

Answer:

"(5+x)/(6+2x) = 3/5" is the answer

Answer:

A

Step-by-step explanation:

the proof of the answer is shown above

Answer:

z=3x/4+1 x=4z/3-4/3

Step-by-step explanation:

Answer:

The maximum value of the objective function is 112 when x = 0 and y = 7.

Step-by-step explanation:

Given the constraints:

5x+3y≤37, 3x+5y≤35, x≥0, y≥0

Plotting the above constraints using geogebra online graphing tool, we get the solution to the constraints as:

A(0, 7), B(7.4, 0), C(5, 4) and D(0, 0)

The objective function is given as E =2x+16y, therefore:

At point A(0, 7):  E = 2(0) + 16(7) = 112

At point B(7.4, 0): E = 2(7.4) + 16(0) = 14.8

At point C(5, 4): E = 2(5) + 16(4) = 74

At point D(0, 0): E = 2(0) + 16(0) = 0

Therefore the maximum value of the objective function is at A(0, 7).

The maximum value of the objective function is 112 when x = 0 and y = 7.

Solution :

Given :

[tex]f(x) = \left\{\begin{matrix}\frac{1}{450}(x^2-x) & \text{if } 6 < x < 12 \\ 0 & \text{otherwise}\end{matrix}\right.[/tex]

1. Cumulative distribution function

[tex]$P(X \leq x) = \int_{- \infty}^x f(x) \ dx$[/tex]

              [tex]$=\int_{- \infty}^6 f(x) dx + \int_{6}^x f(x) dx $[/tex]

             [tex]$=0+\int_6^x \frac{1}{450}(x^2-x) \ dx$[/tex]

             [tex]$=\frac{1}{450} \int_6^x (x^2-x) \ dx$[/tex]

             [tex]$=\frac{1}{450}\left[\frac{x^3}{3}-\frac{x^2}{2}\right]_6^x$[/tex]

             [tex]$=\frac{1}{450}\left[ \left( \frac{x^3}{3} - \frac{x^2}{2}\left) - \left( \frac{6^3}{3} - \frac{6^2}{2} \right) \right] $[/tex]

            [tex]$=\frac{1}{450}\left[\frac{x^3}{3} - \frac{x^2}{2} - 54 \right]$[/tex]

2.  Mean [tex]$E(x) = \int_{- \infty}^{\infty} \ x \ f(x) \ dx$[/tex]

                       [tex]$=\int_{6}^{12}x . \left( \frac{1}{450} \ (x^2-x)\right)\ dx$[/tex]

                     [tex]$=\frac{1}{450} \int_6^{12} \ (x^3 - x^2) \ dx$[/tex]

                     [tex]$=\frac{1}{450} \left[\frac{x^4}{4} - \frac{x^3}{3} \right]_6^{12} \ dx$[/tex]

                     [tex]$=\frac{1}{450} \left[ \left(\frac{(12)^4}{4} - \frac{(12)^3}{3} \right) - \left(\frac{(6)^4}{4} - \frac{(6)^3}{3} \right) $[/tex]

                     [tex]$=\frac{1}{450} [4608 - 252]$[/tex]

                    = 17.2857