URGENT! 15 PNTS
Points T, R, and P, define _____

A. plane B
B. line e
C. line segment PR⎯⎯⎯⎯⎯⎯⎯
D. plane M

Ethan will pay $31.99 with the discount.

How? This is the answer because:

If 39.99 is 100%, and you are trying to find 20%...

1. you need to set it up as a ratio (of course, you do not need to do this, but it is easier for me to do it this way)

2. the ratio will look like this: 39.99/100% x/20%

3. all we need to do from here is to cross multiply!

4 39.99 x

---------- = ----------

100 20

-price is on the top and percent on the bottom

-you would now do 39.99 times 20

-then divide by 100

5. once you have 20% of 39.99, you need to subtract that answer from the total

6. 39.99 - 7.998 = 31.992 (you need to round to the nearest hundredth)

Hope this helps <3

Answer:

Step-by-step explanation:

A). Let the dimensions of the rectangular pen are,

Length = l

Width = x

Since, farmer has the wire measuring 80 feet to surround the the pen.

Perimeter of the pen = 80 feet

2(l + x) = 80

l + x = 40

l = 40 - x ------(1)

Area of the rectangular pen = Length × width

                                               = lx

By substituting the value of l from equation (1),

Area (A) of the pen will be modeled by the expression,

A = (40 - x)

A = 40x - x²

B). For maximum area of the pen,

Derivative of the area = 0

[tex]\frac{d}{dx}(A)=0[/tex]

[tex]\frac{d}{dx}(A)=\frac{d}{dx}(40x-x^2)[/tex]

         = 40 - 2x

And (40 - 2x) = 0

x = 20

Therefore, width of the pen = 20 feet

And length of the pen = 40 - 20

                                     = 20 feet

Dimensions of the pen should be 20 feet by 20 feet.

Given:

The equation of the hyperbola is:

[tex]xy=-42[/tex]

To find:

The the equation which is not an asymptote of the hyperbola.

Solution:

We have,

[tex]xy=-42[/tex]

It can be written as:

[tex]y=\dfrac{-42}{x}[/tex]

Equating denominator and 0, we get

[tex]x=0[/tex]

So, the vertical asymptotic is [tex]x=0[/tex].

The degree of numerator is 0 and the degree of denominator is 1.

Since the degree of numerator is greater that the degree of denominator, therefore the horizontal asymptote is [tex]y=0[/tex] and there is no oblique asymptote.

Therefore, [tex]y=x[/tex] is not an asymptote of the given hyperbola and the correct option is C.

Answer:

8

Step-by-step explanation:

=z²-3z+4 when z is 4

=4²-3(4)+4

=16-12+4

=8

Answer:

48, 6

Step-by-step explanation:

The ratio of the pieces is 6 to 1

Add them together to get the total

6+1 = 7

Divide the total length by 7

56/7 = 8

Multiply the ratios by 8

6*8 = 48

 1*8 = 6

The peices are 48 and 6

Answer:

5

Step-by-step explanation:

(625 ^2)^(1/8)

Rewriting 625 as 5^4

(5^4 ^2)^(1/8)

We know that a^b^c = a^(b*c)

5^(4*2)^1/8

5^8 ^1/8

5^(8*1/8)

5^1

5

Answer:

[tex]5[/tex]

Step-by-step explanation:

[tex] { {(625}^{2} )}^{ \frac{1}{8} } \\ { ({25}^{2 \times 2} )}^{ \frac{1}{8} } \\ {25}^{4 \times \frac{1}{8} } \\ {5}^{2 \times 4 \times \frac{1}{8} } \\ {5}^{ \frac{8}{8} } \\ {5}^{1} \\ = 5[/tex]

Answer:

V = 1/6 π ( 5h)^3

Step-by-step explanation:

Height of napkin rings = 5h

Compute the volume V of a napkin ring

let a = 5

radius = r

express answer in terms of h

attached below is the detailed solution

Answer:

x = 50

y = 50

Step-by-step explanation:

[tex]\begin{bmatrix}x+y=100\\ 0.12x+0.38y=25\end{bmatrix}[/tex]

.12(100-y) + .38y = 25

x = 50

y = 50

Answer:

Kindly check explanation

Step-by-step explanation:

Rounding each number to 4 significant figures and expressing in standard notation :

(a) 102.53070,

Since the number starts with a non-zero, the 4 digits are counted from the left ;

102.53070 = 102.5 (4 significant figures) = 1.025 * 10^2

(b) 656.980,

Since the number starts with a non-zero, the 4 digits are counted from the left ; the value after the 4th significant value is greater than 5, it is rounded to 1 and added to the significant figure.

656.980 = 657.0 (4 significant figures) = 6.57 * 10^2

(c) 0.008543210,

Since number starts at 0 ; the first significant figure is the first non - zero digit ;

0.008543210 = 0.008543 (4 significant figures) = 8.543 * 10^-3

(d) 0.000257870,

Since number starts at 0 ; the first significant figure is the first non - zero digit ;

0.000257870 = 0.0002579 (4 significant figures) = 2.579 * 10^-4

(e) -0.0357202,

Since number starts at 0 ; the first significant figure is the first non - zero digit ;

-0.0357202 = - 0.03572 (4 significant figures) = - 3.572* 10^-2

Answer:

4 e

Step-by-step explanation:

dz6dxrx xrrx6 xz33x4xr4x xrx

Answer:

proportion of gamers who prefer console does not differ from 29%

Step-by-step explanation:

Given :

n = 341 ; x = 95 ; Phat = x / n = 95/341 = 0.279

H0 : p = 0.29

H1 : p ≠ 0.29

The test statistic :

T = (phat - p) ÷ √[(p(1 - p)) / n]

T = (0.279 - 0.29) ÷ √[(0.29(1 - 0.29)) / 341]

T = (-0.011) ÷ √[(0.29 * 0.71) / 341]

T = -0.011 ÷ 0.0245725

T = - 0.4476532

Using the Pvalue calculator from test statistic score :

df = 341 - 1 = 340

Pvalue(-0.447, 340) ; two tailed = 0.654

At α = 0.01

Pvalue > α ; We fail to reject the null and conclude that there is no significant evidence that proportion of gamers who prefer console differs from 29%

Answer:

it's. is now the MA plz I miss you

si pudiera escoger entre vivir eternamente y vivir dos veces

yo escogeria vivir dos veces porque vivir una vida eterna sin ti a mi lado seria el mayor sufrimiento, ahora vivir dos veces me dejaria tranquilo porque despues del final de mi vida podria volver a encontrarme contigo y vivir todos los momentos bellos una vez mas y eso seria un sueño volviendose realidad

Answer:

270 degrees

Step-by-step explanation:

If you plug in 270 in place of the x's, the function is true!

This is correct for Plate/Edmentum users!! Hope I could help :)

Answer:

0.01375 = 1.375% probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2.

Step-by-step explanation:

We have the mean during the interval, which means that the Poisson distribution is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

In which

x is the number of sucesses

e = 2.71828 is the Euler number

[tex]\mu[/tex] is the mean in the given interval.

Lost-time accidents occur in a company at a mean rate of 0.8 per day.

This means that [tex]\mu = 0.8n[/tex], in which n is the number of days.

10 days:

This means that [tex]n = 10, \mu = 0.8(10) = 8[/tex]

What is the probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2?

This is:

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]

In which

[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]

[tex]P(X = 0) = \frac{e^{-8}*8^{0}}{(0)!} = 0.00034[/tex]

[tex]P(X = 1) = \frac{e^{-8}*8^{1}}{(1)!} = 0.00268[/tex]

[tex]P(X = 2) = \frac{e^{-8}*8^{2}}{(2)!} = 0.01073[/tex]

So

[tex]P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00034 + 0.00268 + 0.01073 = 0.01375[/tex]

0.01375 = 1.375% probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2.

Answer: The mean price per share is $22.91

The required weighted mean price per share is $46.09.

Given that,
In June, an investor purchased 300 shares of Oracle (an information technology company) stock at $53 per share. In August, she purchased an additional 400 shares at $42 per share. In November, she purchased an additional 400 shares at $45.
To determine the weighted mean price per share.

The average of the values is the ratio of the total sum of values to the number of values.

The mean of the values is the ratio of the total sum of values to the number of values.

Here,
Required  weight mean = 300 * 53 + 400*42 + 400 * 42 / [300 + 400 + 400]
Required  weight mean = 50700/ [1100]
Required  weight mean = $46.09 per share.

Thus, the required weighted mean price per share is $46.09.

Learn more about mean here:

https://brainly.com/question/15397049

#SPJ2

Answer:

answer D

Step-by-step explanation:

V=L*W*H=1 ==> L=1,W=1,H=1

A:

L-> L+2=1+2=3

W -> W+2 = 1+2=3

H -> H+2=1+2=3

V=3*3*3=27 not the doubled of the volume's cube

A is false

B:

H -> H+2=1+2=3

V=1*1*3=3 not the doubled of the volume's cube

B is false

C:

H -> 2*H=2*1=2

L -> 2*L=2*1=2

W -> 2*W = 2*1=2

V=2*2*2=8 not the doubled of the volume's cube

C is false

D:

H-> H*2=1*2=2

L=1

W=1

V=1*1*2=2 is the doubled of the volume's cube

D is true

Answer:

[tex]f(g(x))= sign(x(1-x^{2})) = sign(x-x^{3})[/tex]

Step-by-step explanation:

Answer:

55000×100/90

61,111.111

Work Shown:

1 pound = 16 ounces

4*(1 pound) = 4*(16 ounces)

4 pounds = 64 ounces

Answer:

12

Step-by-step explanation:

(1) Both equations in (a) and (b) are separable.

(a)

[tex]\dfrac xy y' = \dfrac{2y^2+1}{x+1} \implies \dfrac{\mathrm dy}{y(2y^2+1)} = \dfrac{\mathrm dx}{x(x+1)}[/tex]

Expand both sides into partial fractions.

[tex]\left(\dfrac1y - \dfrac{2y}{2y^2+1}\right)\,\mathrm dy = \left(\dfrac1x - \dfrac1{x+1}\right)\,\mathrm dx[/tex]

Integrate both sides:

[tex]\ln|y| - \dfrac12 \ln\left(2y^2+1\right) = \ln|x| - \ln|x+1| + C[/tex]

[tex]\ln\left|\dfrac y{\sqrt{2y^2+1}}\right| = \ln\left|\dfrac x{x+1}\right| + C[/tex]

[tex]\dfrac y{\sqrt{2y^2+1}} = \dfrac{Cx}{x+1}[/tex]

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

(You could solve for y explicitly, but that's just more work.)

(b)

[tex]e^{x+y}y' = 3x \implies e^y\,\mathrm dy = 3xe^{-x}\,\mathrm dx[/tex]

Integrate both sides:

[tex]e^y = -3e^{-x}(x+1) + C[/tex]

[tex]\ln(e^y) = \ln\left(C - 3e^{-x}(x+1)\right)[/tex]

[tex]\boxed{y = \ln\left(C - 3e^{-x}(x+1)\right)}[/tex]

(2)

(a)

[tex]y' + \sec(x)y = \cos(x)[/tex]

Multiply both sides by an integrating factor, sec(x) + tan(x) :

[tex](\sec(x)+\tan(x))y' + \sec(x) (\sec(x) + \tan(x)) y = \cos(x) (\sec(x) + \tan(x))[/tex]

[tex](\sec(x)+\tan(x))y' + (\sec^2(x) + \sec(x)\tan(x)) y = 1 + \sin(x)[/tex]

[tex]\bigg((\sec(x)+\tan(x))y\bigg)' = 1 + \sin(x)[/tex]

Integrate both sides and solve for y :

[tex](\sec(x)+\tan(x))y = x - \cos(x) + C[/tex]

[tex]y=\dfrac{x-\cos(x) + C}{\sec(x) + \tan(x)}[/tex]

[tex]\boxed{y=\dfrac{(x+C)\cos(x) - \cos^2(x)}{1+\sin(x)}}[/tex]

(b)

[tex]y' + y = \dfrac{e^x-e^{-x}}2[/tex]

(Note that the right side is also written as sinh(x).)

Multiply both sides by e ˣ :

[tex]e^x y' + e^x y = \dfrac{e^{2x}-1}2[/tex]

[tex]\left(e^xy\right)' = \dfrac{e^{2x}-1}2[/tex]

Integrate both sides and solve for y :

[tex]e^xy = \dfrac{e^{2x}-2x}4 + C[/tex]

[tex]\boxed{y=\dfrac{e^x-2xe^{-x}}4 + Ce^{-x}}[/tex]

(c) I've covered this in an earlier question of yours.

(d)

[tex]y'=\dfrac y{x+y}[/tex]

Multiply through the right side by x/x :

[tex]y' = \dfrac{\dfrac yx}{1+\dfrac yx}[/tex]

Substitute y(x) = x v(x), so that y' = xv' + v, and the DE becomes separable:

[tex]xv' + v = \dfrac{v}{1+v}[/tex]

[tex]xv' = -\dfrac{v^2}{1+v}[/tex]

[tex]\dfrac{1+v}{v^2}\,\mathrm dv = -\dfrac{\mathrm dx}x[/tex]

[tex]-\dfrac1v + \ln|v| = -\ln|x| + C[/tex]

[tex]\ln\left|\dfrac yx\right| -\dfrac xy = C - \ln|x|[/tex]

[tex]\ln|y| - \ln|x|  -\dfrac xy = C - \ln|x|[/tex]

[tex]\boxed{\ln|y| -\dfrac xy = C}[/tex]

Answer:

Scatter plot charts are good for relationships and distributions, but pie charts should be used only for simple compositions — never for comparisons or distributions.

Answer:

b

Step-by-step explanation:

You only need to add the real numbers and the ys.

Answer:

12 + 2y

Step-by-step explanation:

9+y+y+3

Combine like terms

9+3   + y+y

12 + 2y

Answer:

x = 1/8(k-c)

Step-by-step explanation:

8x + c = k

Subtract c from each side

8x +c-c = k-c

8x = k-c

Divide each side by 8

8x/8 = (k-c)/8

x = 1/8(k-c)

Answer:

x-1/8(k-c)

Step-by-step explanation:

Answer:

The expected value of the lottery is $80

Step-by-step explanation:

To get the expected value, we have to multiply each outcome by its probability

Then we proceed to add up all of these to get the expected value of the lottery

we have this as ;;

125(0.2) + 100(0.3) + 50(0.5)

= 25 + 30 + 25 = $80

Answer:

1 / 97290

Step-by-step explanation:

The number of ways of selecting 3 specific route capitals from 47 states can be obtained thus :

Probability = required outcome / Total possible outcomes

Total possible outcomes = 47P3

Recall :

nPr = n! / (n-r)!

47P3 = 47! / (47-3)! = 47! / 44! = 97290

Hence, probability of selecting route if 3 specific capitals is = 1 / 97290

Answer:

the missing side is 21!!!!!!!!

9514 1404 393

Answer:

  (b)  x = 1

Step-by-step explanation:

A graph shows the solution to f(x) = g(x) is x = 1.

__

We want to solve ...

  g(x) -f(x) = 0

  x^3 +2x^2 -x -2 -(-11/3x +11/3) = 0

  x^2(x +2) -1(x +2) +11/3(x -1) = 0 . . . . . factor first terms by grouping

  (x^2 -1)(x +2) +11/3(x -1) = 0 . . . . .  . the difference of squares can be factored

  (x -1)(x +1)(x +2) +(x -1)(11/3) = 0 . . . . we see (x-1) is a common factor

  (x -1)(x^2 +3x +2 +11/3) = 0

The zero product rule tells us this will be true when x-1 = 0, or x = 1.

__

The discriminant of the quadratic factor is ...

  b^2 -4ac = 3^2 -4(1)(17/3) = 9 -68/3 = -41/3

This is less than zero, so any other solutions are complex.

Answer:

Orthocenter would be in the middle of the shape.

Step-by-step explanation:

B.