PLEASE PLEASE HELP

what is the slope of the line

A. 0
B. 1
C. -2
D. no slope ​

Answer:

t = -100

Step-by-step explanation:

Simplify to isolate t:

[tex]\frac{t}{2}+10-10=-40-10[/tex]

[tex]\frac{t}{2} * \frac{2}{1} = -50 * 2[/tex]

[tex]t = -100[/tex]

Answer:

[tex]\frac{8x^{18} }{y^{2} }[/tex]

Step-by-step explanation:

Answer:

609 m²

Step-by-step explanation:

Area of unshaded:

(6 x 18) + ((13-6) x 7) = 157

Area of overall rectangle:

36 x 28 = 1008

Area of the chunck of rectangle not included:

11 x 22 = 242

Area of shaded:

1008 - 157 - 242 = 609

Answer:

26 customers

Step-by-step explanation:

First:  determine the z score from standard normal probability table with an indicative area of 0.1

Z-score from probability table  = - 1.28

mean   = 3.5 minutes

std  = 3 minutes

next determine the Z-score based on the information given in the question

Z = ( std -  mean )  / processing time

  = ( 3 - 3.5 ) / 2 = -0.25

Finally determine the number of customers

N = [tex](\frac{-1.28}{-0.25} )^2[/tex]  = 1.6384 / 0.0625 = 26.21 ≈ 26 customers  

Answer:

oh no

Step-by-step explanation:

sorry about that I guess

Answer: (b)

Step-by-step explanation:

Given

The function is given as

[tex]f(x)=\dfrac{x^2-12x+20}{x-2}[/tex]

Solving the function

[tex]f(x)=\dfrac{x^2-2x-10x+20}{x-2}\\\\f(x)=\dfrac{(x-2)(x-10)}{(x-2)}\\\\f(x)=x-10[/tex]

for [tex]x=2[/tex]

[tex]f(2)=2-10\\f(2)=-8[/tex]

The function is continuous at [tex]x=2[/tex] because [tex]\lim_{x \to 2} f(x)[/tex] exists.

If the limit exists at a point, then the function is continuous.  

Answer:

on edge its fs not b or c

Step-by-step explanation:

Answer:

54 square ft

Step-by-step explanation:

Find missing sides:

8+6 = 14

9-6 = 3

Find area of full rectangle:

9×8 = 72

Find the area of the missing part of the full rectangle

3×6 = 18

Find the area of the actual shape:

72-18 = 54

Area = 54 square ft

Answer:

d. 63%

Step-by-step explanation:

percent, will be white?

A41% b50% c59% d63%

The probability of white = P (w) = 41/65= 0.63

The  probability of yellow = P (y)= 24/65= 0.369=0.37

The probability of choosing white is 0.63 . When rounded to  nearest percent gives

0.63*100/100

=0.63*100 percent

= 63 percent

= 63%

the probability of getting   the next marble white is the same as the probability of getting a white.

Answer:

6

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Answer:

Solution given:

x²-12x+11=0

Comparing above equation with ax²+bx+c

we get

a=1

b=-12

c=11

By using Vieta's theorem

X1+X2=[tex] \frac{-b}{a} [/tex]=[tex] \frac{- -12}{1} [/tex]=12

again

X1X2=[tex] \frac{c}{a} [/tex]=[tex] \frac{11}{1} [/tex]=11

x1.x2=11

x1+x2=12

Answer:

[tex]36+27\pi\:\mathrm{in^2}[/tex]

Step-by-step explanation:

The figure consists of a square and a sector. We can add the areas of the square and sector to get the total area of the figure.

The area of a sector with measure [tex]\theta[/tex] in a circle of radius [tex]r[/tex] is equal to [tex]\frac{\theta}{360}\cdot r^2\pi[/tex]. Since there are 360 degrees in a circle and 90 degrees in each corner of a square, the measure of the sector is [tex]270^{\circ}[/tex].

Thus, its area is:

[tex]\frac{270}{360}\cdot6^2\cdot pi=\frac{3}{4}\cdot 6^2\cdot \pi=27\pi[/tex].

The area of a square with side length [tex]s[/tex] is given by [tex]s^2[/tex]. Therefore, the area of the circle is [tex]6^2=36[/tex] and the total area of the figure is [tex]\boxed{36+27\pi\:\mathrm{in^2}}[/tex]

Answer: 36+27in2

Step-by-step explanation:

Answer:

11.17

Step-by-step explanation:

arc length = 2πr(θ/360)

= 2π(8)(80/360)

= 11.1701072...

= 11.17 to nearest hundredth

Answer: The answer is B because the triangle rotated a 90 degrees counterclockwise then got a reduction.

Step-by-step explanation:

Answer: Prism

Step-by-step explanation:

Answer:

X = 65 degrees

Step-by-step explanation:

180 = 70 + 45 + X

180 = 115 + X

X = 65 degrees

Answer:

2

Step-by-step explanation:

your dog doesn't love you because it saw what you did. it knows. be cautious around your dog from now on. it knows more than you think and sees all. I'm warning you

Answer:

I think it depends how far bill wants to hike..

Step-by-step explanation:

Answer:

Hence, the [tex]y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t))\end{array}[/tex] and approximately value of [tex]y(t)[/tex] is [tex]-0.844[/tex].

Given :

 [tex]my''+cy'+ky=F(t), y(0)=0, y'(0)=0,[/tex]

Where  [tex]m=2[/tex] kilograms

           [tex]c=8[/tex] kilograms per second

           [tex]k=80[/tex] Newtons per meter

          [tex]F(t)=20\sin (6t)[/tex] Newtons

Explanation :

(1)

Solve the initial value problem. [tex]y(t)[/tex]

   [tex]my''+cy'+ky=F(t), y(0)=0, y'(0)=0,[/tex]

[tex]\Rightarrow 2y''+8y'+80y=20\sin (6t)[/tex]

[tex]\Rightarrow y''+4y'+40y=10\sin (6t)[/tex]

Auxilary equations :[tex]F(t)=0[/tex]

[tex]\Rightarrow r^2+4r+40=0[/tex]

[tex]\Rightarrow r=\frac{-4\pm\sqrt{4^2-4\times 1\times 40}}{2\times 1}[/tex]

[tex]\Rightarrow r=\frac{-4\pm\sqrt{16-160}}{2}[/tex]

[tex]\Rightarrow r=\frac{-4\pm\sqrt{-144}}{2}[/tex]

[tex]\Rightarrow r=\frac{-4\pm12i}{2}[/tex]

[tex]\Rightarrow r=-2\pm6i[/tex]

The complementary solution is [tex]y_c=e^{-2t}\left(c_1\cos 6t+c_2\sin 6t\right)[/tex]

The particular Integral, [tex]y_p=\frac{1}{f(D)}F(t)[/tex]

[tex]y_{y} &=\frac{1}{D^{2}+4 D+40} 25 \sin (6 t) \\\\ y_{y} &=\frac{25}{-6^{2}+4 D+40} \sin (6 t) \quad\left(D^{2} \text { is replaced with }-6^{2}=-36\right) \\\\y_{y} &=\frac{25}{4 D+4} \sin (6 t) \\\\y_{y} &=\frac{25}{4(D+1)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4(D+1)(D-1)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4\left(D^{2}-1\right)} \sin (6 t) \\\\y_{y} &=\frac{25(D-1)}{4(-36-1)} \sin (6 t) \\\\y_{y} &=-\frac{25}{148}(D-1) \sin (6 t) \\y_{y} &=-\frac{25}{148}\left(\frac{d}{d t} \sin (6 t)-\sin (6[/tex]

Hence the general solution is :[tex]y=y_c+y_p=e^{-2t}(c_1\cos 6t+c_2\sin 6t)-\frac{25}{148}(6\cos 6t-\sin 6t)[/tex]

Now we use given initial condition.

[tex]y(t) &=e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\\y(0) &=e^{-\alpha 0)}\left(c_{1} \cos (0)+c_{2} \sin (0)\right)-\frac{25}{148}(6 \cos (0)-\sin (0)) \\\\0 &=\left(c_{1}\right)-\frac{25}{148}(6) \\\\c_{1} &=\frac{75}{74} \\\\y(t) &=e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\[/tex]

[tex]y^{\prime}(t)=-2 e^{-2 t}\left(c_{1} \cos 6 t+c_{2} \sin 6 t\right)+e^{-2 t}\left(-6 c_{1} \sin 6 t+6 c_{2} \cos 6 t\right)-\frac{25}{148}(-36 \sin (6 t)-6 \cos (6 t)) \\\\y^{\prime}(0)=-2 e^{0}\left(c_{1} \cos 0+c_{2} \sin 0\right)+e^{0}\left(-6 c_{1} \sin 0+6 c_{2} \cos 0\right)-\frac{25}{148}(-36 \sin 0-6 \cos 0) \\\\0=-2\left(c_{1}\right)+\left(6 c_{2}\right)-\frac{25}{148}(-6) \\\\0=-2 c_{1}+6 c_{2}+\frac{75}{74} \\\\0=-2\left(\frac{75}{74}\right)+6 c_{2}+\frac{75}{74} \\\\[/tex][tex]\begin{array}{l}0=-\frac{150}{74}+6 c_{2}+\frac{75}{74} \\\\\frac{150}{74}-\frac{75}{74}=6 c_{2}\end{array}[/tex]

[tex]\begin{array}{l}c_{2}=\frac{25}{148}\\\\\text { Substitute } c_{1} \text { and } c_{2} \text { in } y(t) \text { . Then }\\\\y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t))\end{array}[/tex]

(2)

[tex]y(t)=e^{-2 t}\left(\frac{75}{74} \cos 6 t+\frac{75}{148} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\left(\frac{75}{74} e^{-2 t} \cos 6 t+\frac{75}{148} e^{-2 t} \sin 6 t\right)-\frac{25}{148}(6 \cos (6 t)-\sin (6 t)) \\\\y(t)=\frac{75}{74}\left(e^{-2 t}-1\right) \cos 6 t+\frac{25}{148}\left(3 e^{-2 t}+1\right) \sin 6 t \\\\|y(t)| \leq \frac{75}{74} e^{-2 t}-1|\cos 6 t|+\frac{25}{148}\left|3 e^{-2 t}+1\right||\sin 6 t| \\\\[/tex]

[tex]|y(t)| \leq \frac{75}{74}\left|e^{-2 t}-1\right|+\frac{25}{148}\left|3 e^{-2 t}+1\right| \\\\\lim _{t \rightarrow \infty} y(t) \leq \lim _{t \rightarrow \infty}\left\{\frac{75}{74}\left|e^{-2 t}-1\right|+\frac{25}{148}\left|3 e^{-2 t}+1\right|\right\} \\\\\lim _{t \rightarrow \infty} y(t)=\left\{\frac{75}{74}\left|\lim _{t \rightarrow \infty}\left(e^{-2 t}-1\right)\right|+\frac{25}{148}\left|\lim _{t \rightarrow \infty}\left(3 e^{-2 t}+1\right)\right|\right\} \\[/tex]

[tex]\lim _{t \rightarrow \infty} y(t)=\left\{\frac{75}{74}(-1)+\frac{25}{148}(1)\right\}=-\frac{75}{74}+\frac{25}{148}=-\frac{-150+25}{148}=-\frac{125}{148} \approx-0.844[/tex]

Answer:

P=34ft

Step-by-step explanation:

Solution

P=2(l+w)=2·(12+5)=34ft

The last one 8,000 + 600 + 9

Answer: If a parallelogram has diagonals that are perpendicular, it is a rhombus. … This definition can also be stated as: A square is a quadrilateral that is also a rectangle and a rhombus.

Step-by-step explanation:

Answer:

1,2,4,8,16,32

Step-by-step explanation:

power of two sequence

Answer:

2/9

Step-by-step explanation:

convert mixed number to improper fraction

8/3÷3 1/4

dividing is equivalent to multiplying by the reciprocal

8/3×1/3×1/4

reduce the numbers with the greatest common factor 4

2/3×1/3

multiply the fraction

2/3×1/3=2/9

2/9 is your answer

Answer: the answer is 4x^2

Step-by-step explanation:

hope it help

Answer:

A. Right 6, Down 5

Step-by-step explanation:

I don't know how to explain this

Answer:

423.32cm³

Step-by-step explanation:

Volume of a cylinder = [tex]V=\pi r^2h[/tex]

where r = radius and h = height

Given height = 11 cm

Given radius = 3.5 cm

* plug in these values into the formula *

[tex]V=\pi 3.5^211\\3.5^2=12.25 \\12.25\pi =38.48\\34.48*11=423.32\\V=423.32[/tex]

So we can conclude that the answer would be D

Using the sine rule,

[tex] \frac{a}{sin(a)} = \frac{b}{sin(b)} = \frac{c}{sin(c)} [/tex]

Here we are going to use the values of A and C,

[tex] \frac{12}{sin(a)} = \frac{14}{sin(90)} \\ \frac{12}{sin(a)} = \frac{14}{1} \\ sin(a) = 12 \div 14 \\ sin(a) = 0.8571[/tex]

So sin(A) = 12/14 = 6/7 = 0.8571, but since the question says in its simplest radical form, I think the closest answer to it should be

[tex] \frac{ \sqrt{3} }{2} [/tex]