What is the area of the region in the first quadrant bounded on the left by the graph of x=y4+y2 and on the right by the graph of x=5y ?
a)2.983
b)8.508
c)5005.880
d)5293.155

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: Prism

Step-by-step explanation:

Answer:

About 292.72

Step-by-step explanation:

We can imagine a circle circumscribed around the triangle, then solve it as we normally do using the apothem. By dividing 360 by the number of sides (3) we know that the central angle constructed by the apothem and radiuses would be 120. Divided by 2 (because we are using a right triangle to solve it) is 60. Now with this you can either use trig ratios (sin, cos, or tan) or use special right triangles (30-60-90 or 45-45-90). I used special rights. With this I know my apothem is [tex]\frac{13\sqrt{3} }{3}[/tex]. Perimeter is 26 times 3 which is 78. The formula for area of regular polygons is 1/2(perimeter times apothem). When we plug our values in the answer is about 292.72. You are lucky that I just had my test on this, I hope it helps.

Here is my work since this is very visual for me:

Answer:

y= 12x

Step-by-step explanation:

36/3=12

12× x =y

________________________

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:

AC² = 10.8²+14.4² = 324

AC = √324= 18

area = AC.DC÷2

area=18×24÷2

area=216

Answer:

216 centimeters squared

Step-by-step explanation:

working is above

Answer:

1/20

Step-by-step explanation:

According to G0ogle 5 percent equals 1/20 in lowest terms.

Answer:

5% equals 5/100 which is 1/20 in lowest terms.

Step-by-step explanation:

5% is basically equivalent to 5/100. 5/100 in lowest terms is 1/20 since you divide the numerator and denominator by 5.

I hope this helps, have a nice day.

Answer:

V = 310.5 cm³

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Geometry

Volume of a Prism Formula: V = lwh

Step-by-step explanation:

Step 1: Define

Identify variables

l = 13.8 cm

w = 4.5 cm

h = 5 cm

Step 2: Solve for V

Answer:

u=−x2−x+1

Step-by-step explanation:

Let's solve for u.

2x−u−1x^2−3x+3=2

Step 1: Add x^2 to both sides.

−x2−u−x+3+x2=2+x2

−u−x+3=x2+2

Step 2: Add x to both sides.

−u−x+3+x=x2+2+x

−u+3=x2+x+2

Step 3: Add -3 to both sides.

−u+3+−3=x2+x+2+−3

−u=x2+x−1

Step 4: Divide both sides by -1.

−u/−1=x2+x−1

−1/u=−x2−x+1

Answer:

u=−x2−x+1

Answer:

look at the picture i have sent

Answer:

The cities are 35 cm apart in map.

The scale on a map is 5 cm : 8 km.

Step-by-step explanation:

mrk me brainliest please

Answer:

discount=MP-SP

=Rs420-Rs405

=Rs15

Step-by-step explanation:

Answer: 7500

Step-by-step explanation:

multiply 150 by 50

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:

the answer is 6.

Step-by-step explanation:

Answer:

6.

Step-by-step explanation:

3+3=6 = 2×3=6

you can do draw 3 circles 2 times and add it all together.

Answer:

Angles in a triangle sum to 180.

So

x= 180 - 75 - 43

x=62°

Answer:

x=62°

Step-by-step explanation:

x= 180 - 75 - 43

Answer:

C) 96i – 280j

Step-by-step explanation:

Multiplying vector by constant:

When a vector is multiplied by a constant, each component of the vector is multiplied by this constant.

In this question:

u = -12i + 35j

-8u = -8(-12i + 35j) = (8*12i - 8*35j) = 96i - 280j.

The answer is C) 96i – 280j

Answer:

b

Step-by-step explanation:

3/3 = 1

Answer:

C

Step-by-step explanation:

Answer:

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

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:

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:

$3.74

Step-by-step explanation:

we already know the cost of the banana's and it gives us the total price including apples. So $5.24 - $1.50 = $3.74, and the 3.74 is the cost of apples

Answer:

The answer is less than that is the answer

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:

if you just wanted the equations fixed: y= -3x + 8 & y= 1/3x - 11/3

but if you wanted x then x = 7/2

Step-by-step explanation:

3x + y = 8 *subtract 3x from both sides*

y = -3x + 8

x - 3y = 11 *add 3y to both sides*

x = 3y + 11 subtract 11 from both sides*

x - 11 = 3y * divide both sides by 3*

1/3x - 11/3 = y --> y = 1/3x - 11/3

and to solve for x i just used ma thway but if you did want to solve it you'd need to put the equations equal to each other --> -3x + 8 = 1/3x - 11/3

Answer:

[tex](6\sqrt{3},\,\frac{5\pi}{6})[/tex]

Step-by-step explanation:

The radius  r  can be found from the relationship

 [tex]r^2=x^2+y^2\\r^2=(-9)^2+(3\sqrt{3})^2\\r^2=81+27=108\\r=\sqrt{108}\\r=6\sqrt{3}[/tex]

The point is in Quadrant II (-, +), so use the inverse cosine function to find the angle.

[tex]\cos{\theta}=\frac{x}{r}=\frac{-9}{6\sqrt{3}}\\\cos{\theta}=-\frac{9}{6\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\\cos{\theta}=-\frac{9\sqrt{3}}{6\cdot3}\\\cos{\theta}=-\frac{\sqrt{3}}{2}\\\\\cos^{-1}\frac{-\sqrt{3}}{2}}=\frac{5\pi}{6}[/tex]

See the attached image.

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: Hello the scatter plot related to your question is missing attached below is the scatter plot

answer :  The model fails to capture the relationship between the variables accurately, and there may exist nonlinear relationship ( C )

Step-by-step explanation:

The conclusion that can be drawn based upon the scatter chart is that The model fails to capture the relationship between the variables accurately, and there may exist nonlinear relationship

A scatter plot helps in observing the relationship within different numeric variables but the scatter plot attached fails in the showing the actual relationship