Find the value of x.

Answer:

64 Bottles

Step-by-step explanation:

that is the procedure above

Answer:

14=-2x-4

18=-2x

x=-9

Hope This Helps!!!

Answer:

By using a pair of compasses and a ruler you can draw all angles

im struggling with the same one

Answer:

The price elasticity of demand is -10

Step-by-step explanation:

Given

[tex]p_1,p_2 = 50,40[/tex]

[tex]q_1,q_2 = 400,500[/tex]

Solving (a): The coefficient of price elasticity of demand (k)

This is calculated as:

[tex]k = \frac{\triangle q}{\triangle p}[/tex]

So, we have:

[tex]k = \frac{500 - 400}{40 - 50}[/tex]

[tex]k = \frac{100}{-10}[/tex]

[tex]k = -10[/tex]

Because |k| > 0, then we can conclude that the company made a wise decision.

Answer:

cos(∝) = 1/√3

cos(β) = 1/√3

cos(γ) = 1/√3

∝ = 55°

β = 55°

γ = 55°

Step-by-step explanation:

Given the data in the question;

vector is z = < c,c,c >

the direction cosines and direction angles of the vector = ?

Cosines are the angle made with the respect to the axes.

cos(∝) = z < 1,0,0 > / |z|

so

cos(∝) = < c,c,c > < 1,0,0 > / √[c² + c² + c²] = ( c + 0 + 0 ) / √[ 3c² ]

cos(∝) = c / √[ 3c² ] = c / c√3 = 1/√3

∝ = cos⁻¹( 1/√3 ) = 54.7356° ≈ 55°

cos(β) = < c,c,c > < 0,1,0 > / √[c² + c² + c²] = ( 0 + c + 0 ) / √[ 3c² ]

cos(β) = c / √[ 3c² ] = c / c√3 = 1/√3

β = cos⁻¹( 1/√3 ) = 54.7356° ≈ 55°

cos(γ) = < c,c,c > < 0,0,1 > / √[c² + c² + c²] = ( 0 + 0 + c ) / √[ 3c² ]

cos(γ) = c / √[ 3c² ] = c / c√3 = 1/√3

γ = cos⁻¹( 1/√3 ) = 54.7356° ≈ 55°

Therefore;

cos(∝) = 1/√3

cos(β) = 1/√3

cos(γ) = 1/√3

∝ = 55°

β = 55°

γ = 55°

In cylindrical coordinates, W is the set of points

W = {(r, θ, z) : 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π and -2 ≤ z ≤ 5}

(A) Then the integral of f(x, y, z) over W is

[tex]\displaystyle\iiint_W(x^2+y^2+z^2)\,\mathrm dV = \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta[/tex]

(B)

[tex]\displaystyle \int_0^{2\pi}\int_0^2\int_{-2}^5r(r^2+z^2)\,\mathrm dz\,\mathrm dr\,\mathrm d\theta = 2\pi \int_0^2\int_{-2}^5(r^3+rz^2)\,\mathrm dz\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(zr^3+\frac13rz^3\right)\bigg|_{z=-2}^{z=5}\,\mathrm dr \\\\\\= 2\pi \int_0^2\left(\frac{133}3r+7r^3\right)\,\mathrm dr \\\\\\= 2\pi \left(\frac{133}6r^2+\frac74r^4\right)\bigg|_{r=0}^{r=2} \\\\\\= 2\pi \left(\frac{110}3\right) = \boxed{\frac{220\pi}3}[/tex]

9514 1404 393

Answer:

  36.6 km

Step-by-step explanation:

We assume the initial bearing of the boy is 35°. Then he will make a 90° turn to a heading of 125°. A diagram shows the distance of interest is the hypotenuse of a right triangle in which 35° is the angle opposite the side of length 21 km.

The relevant trig relation is ...

  Sin = Opposite/Hypotenuse

  sin(35°) = (21 km)/XZ

  XZ = (21 km)/sin(35°) ≈ 36.61 km

The distance from X to Z is about 36.61 km.

_____

The attached diagram has the angles measured in the usual way for a Cartesian plane: CCW from the +x axis. This will correspond to bearing measures if we relabel the axes so that +x is North, and +y is East.

Answer:

Step-by-step explanation:

Cos 2A = 2Cos² A - 1

             [tex]= 2*(\frac{3\sqrt{2}}{5})^{2}-1\\\\=2*(\frac{3^{2}*(\sqrt{2})^{2}}{5^{2}})-1\\\\=2*\frac{9*2}{25} - 1\\\\=\frac{36}{25}-1\\\\=\frac{36}{25}-\frac{25}{25}\\\\=\frac{11}{25}[/tex]

Answer:

[tex]P = 8 + 2\sqrt{26}[/tex]

Step-by-step explanation:

Given

[tex]W = (-2, 4)[/tex]

[tex]X = (2, 4)[/tex]

[tex]Y = (1, -1)[/tex]

[tex]Z = (-3,-1)[/tex]

Required

The perimeter

First, calculate the distance between each point using:

[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 -y_2)^2[/tex]

So, we have:

[tex]WX = \sqrt{(-2- 2)^2 + (4-4)^2 } =4[/tex]

[tex]XY = \sqrt{(2- 1)^2 + (4--1)^2 } =\sqrt{26}[/tex]

[tex]YZ = \sqrt{(1- -3)^2 + (-1--1)^2 } =4[/tex]

[tex]ZW = \sqrt{(-3--2)^2 + (-1-4)^2 } =\sqrt{26}[/tex]

So, the perimeter (P) is:

[tex]P = 4 + \sqrt{26} + 4 + \sqrt{26}[/tex]

[tex]P = 8 + 2\sqrt{26}[/tex]

Answer:

its D.

Step-by-step explanation:

took test

Answer:

0.5217

Step-by-step explanation:

P(more than 5 customer arrive):

P(X>=6)=1-P(X<=5)=  1-∑x=0x e-λ*λx/x!= 0.5217

Answer:

y=4√3 units

Step-by-step explanation:

Hi there!

We are given ΔABC, which is a right triangle (m<C=90°), m<A=60°, AB=8, and BC=y

We need to find the value of y (BC)

The side AB is the hypotenuse of the  (the side opposite from the right angle).

BC is a leg, which is a side that makes up the right angle.

Now, if we have a right triangle that has one of the acute angles as 60°, the side OPPOSITE from that 60° angle (in this case, BC) is equal to [tex]\frac{a\sqrt{3}}{2}[/tex], where a is the length of the hypotenuse

Since we have the hypotenuse given as 8, the length of BC (y) is [tex]\frac{8\sqrt{3}}{2}[/tex], or 4√3

so y=4√3 units

Hope this helps!

Answer:b

Step-by-step explanation:

Answer:

just be smart trust me u dont need us to give u the answer ur super smart

Step-by-step explanation:

make me brainliest to help people be encourage

Answer:

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

Step-by-step explanation:

Unit circle

Answer:

[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]

Step-by-step explanation:

Given

[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]

[tex]P(1) = 2[/tex]

Required

The solution

We have:

[tex]\frac{dP}{dt} = \sqrt{Pt[/tex]

[tex]\frac{dP}{dt} = (Pt)^\frac{1}{2}[/tex]

Split

[tex]\frac{dP}{dt} = P^\frac{1}{2} * t^\frac{1}{2}[/tex]

Divide both sides by [tex]P^\frac{1}{2}[/tex]

[tex]\frac{dP}{ P^\frac{1}{2}*dt} = t^\frac{1}{2}[/tex]

Multiply both sides by dt

[tex]\frac{dP}{ P^\frac{1}{2}} = t^\frac{1}{2} \cdot dt[/tex]

Integrate

[tex]\int \frac{dP}{ P^\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

Rewrite as:

[tex]\int dP \cdot P^\frac{-1}{2} = \int t^\frac{1}{2} \cdot dt[/tex]

Integrate the left hand side

[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{-1}{2}+1} = \int t^\frac{1}{2} \cdot dt[/tex]

[tex]\frac{P^{\frac{-1}{2}+1}}{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

[tex]2P^{\frac{1}{2}} = \int t^\frac{1}{2} \cdot dt[/tex]

Integrate the right hand side

[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{1}{2} +1 }}{\frac{1}{2} +1 } + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2} } + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex] ---- (1)

To solve for c, we first make c the subject

[tex]c = 2P^{\frac{1}{2}} - \frac{2}{3}t^\frac{3}{2}[/tex]

[tex]P(1) = 2[/tex] means

[tex]t = 1; P =2[/tex]

So:

[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1^\frac{3}{2}[/tex]

[tex]c = 2*2^{\frac{1}{2}} - \frac{2}{3}*1[/tex]

[tex]c = 2\sqrt 2 - \frac{2}{3}[/tex]

So, we have:

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + c[/tex]

[tex]2P^{\frac{1}{2}} = \frac{2}{3}t^\frac{3}{2} + 2\sqrt 2 - \frac{2}{3}[/tex]

Divide through by 2

[tex]P^{\frac{1}{2}} = \frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3}[/tex]

Square both sides

[tex]P = (\frac{1}{3}t^\frac{3}{2} + \sqrt 2 - \frac{1}{3})^2[/tex]

Answer:

x=9

Step-by-step explanation:

Answer:

[tex]X=6.67ft/s[/tex]

Step-by-step explanation:

From the question we are told that:

Height of pole [tex]H_p=15[/tex]

Height  of man [tex]h_m=6ft[/tex]

Speed of Man [tex]\triangle a =4ft/s[/tex]

Distance from pole [tex]d=35ft[/tex]

Let

Distance from pole to man=a

Distance from man to shadow =b

Therefore

 [tex]\frac{a+b}{15}=\frac{b}{6}[/tex]

 [tex]6a+6b=15y[/tex]

 [tex]2a=3b[/tex]

Generally the equation for change in velocity is mathematically given by

 [tex]2(\triangle a)=3(\triangle b )[/tex]

 [tex]2*4=3(\triangle b)[/tex]

 [tex]\triangle a=\frac{8}{3}[/tex]

Since

The speed of the shadow is given as

 [tex]X=\triangle b+\triangle a[/tex]

 [tex]X=4+8/3[/tex]

 [tex]X=6.67ft/s[/tex]

Answer:

A

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