please help!!
options are:
A. Point in quadrant 1
B. Point in negative x-axis
C. Point in quadrant 2

Answer: 44.59 or 44.6

Step-by-step explanation:

30^2+33^2= 900+ 1089

= 1989^2

square root = 44.6

hope this helps

Answer:

The number is 20.

Step-by-step explanation:

60 - 2y = 20

- 2y + 60 = 20

- 2y + 60 - 60 = 20 - 60

- 2y = - 40

- 2y ÷ - 2 = - 40 ÷ - 2

y = 20

Answer:

d. x²+2

Step-by-step explanation:

x²+2x+1-2x-2+3 = x²+2

Answer:

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

Step-by-step explanation:

Answer: Prism

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:

a 2%

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:

9

Step-by-step explanation:

Substitute 8 in place of x

f(x) = 2*8-7

f(x) = 9

Answer:

9

Step-by-step explanation:

plug in 8 where x is so it would look like

2(8)-7 = 9

do 2 *8 first which is 16

then do 16-7 which gives you 9

Answer:

X = 65 degrees

Step-by-step explanation:

180 = 70 + 45 + X

180 = 115 + X

X = 65 degrees

Answer:

Since there are 7.5 gallons in each cubic foot, multiply the cubic feet of the pool by 7.5 to arrive at the volume of the. 3.14 x 25 ft x 3 ft x 7.5 = 1766.25 gallons

Step-by-step explanation:

Answer:

I believe it would be "505 times n"

Step-by-step explanation:

If tis is not what you are looking for I am sorry, but the question was vague.

Answer:

[tex] \displaystyle 5\sqrt{2}[/tex]

Step-by-step explanation:

we have a right angle isosceles triangle

in order to figure out the length of each leg we can consider Pythagoras theorem given by

[tex] \displaystyle {a}^{2} + {b}^{2} = {c}^{2} [/tex]

remember that,

isosceles triangle has two equal legs so a=b and given that the the hypotenuse is 10

substitute:

[tex] \displaystyle {a}^{2} + {a}^{2} = {10}^{2} [/tex]

simplify addition:

[tex] \displaystyle {2a}^{2}= 100[/tex]

simplify square:

divide both sides by 2:

[tex] \displaystyle {a}^{2}= 50[/tex]

square root both sides:

[tex] \displaystyle {a}^{}= \sqrt{50}[/tex]

[tex] \displaystyle {a}^{}= 5\sqrt{2}[/tex]

hence,

the length of each leg is 5√2

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]

Answer:

y= 12x

Step-by-step explanation:

36/3=12

12× x =y

________________________

Answer:

P=34ft

Step-by-step explanation:

Solution

P=2(l+w)=2·(12+5)=34ft

Answer:

6

Step-by-step explanation:

Answer:

6

Step-by-step explanation:

Answer:

you should multiply or add

Step-by-step explanation:

Answer:

Yes the bots answer was deleted!

Anyways x = 29 because 29 + 151 = 180

Hope that helps!

Answer:

v = (1/48)π

Step-by-step explanation:

v = (4/3)πr³

v = (4/3)π(1/4)³

v = (4/3)(1/64)π

v = (1/3)(1/16)π

v = (1/48)π

Answer: Did this all by myself 1/6 is closer to 0 and 5/8 is closer to 1/2 so the answer is 1/2  0 then 1/2 then 1/3 again

Step-by-step explanation:  

1/ 6   -5 /8 =(1 × 4  6 × 4)  – (5 × 3  8 × 3)  = 4 /24  – 15 /24  =  4 – 15  24 = – 11  24

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:

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:

Step-by-step explanation:

Answer:

I think the answer is 3:5

Answer:

3.5

Step-by-step explanation:

Answer:

XY is a tangent

Step-by-step explanation:

Given

[tex]XY = 12[/tex]

[tex]WY = 20[/tex]

[tex]WZ = 8[/tex]

Required

Is XY a tangent?

XY is a tangent if:

[tex]WY^2 = XY^2 + WX^2[/tex]

Because XY should make a right angle at point X with the circle

Where

[tex]WX = 2 * WZ[/tex]

So, we have:

[tex]WY^2 = XY^2 + WX^2[/tex]

[tex]WY^2 = XY^2 + (2*WZ)^2[/tex]

[tex]WY^2 = XY^2 + (2WZ)^2[/tex]

[tex]WY^2 = XY^2 + 4WZ^2[/tex]

This gives:

[tex]20^2 = 12^2 + 4*8^2[/tex]

[tex]400 = 144 + 256[/tex]

[tex]400 = 400[/tex]

Yes, XY is a tangent

Answer:

7/3

Step-by-step explanation:

7 is in numerator's and 3 is in denominator's position.

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]