One number is 1/4 of another number. The sum of the two numbers is 5. Find the two numbers. Use a comma to separate your answer

im struggling with the same one

Answer:

is it going to be 10.5

Step-by-step explanation:

I do not have any explanation

Answer: 0 (zero)

Step-by-step explanation:

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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:

[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:

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°

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:

64 Bottles

Step-by-step explanation:

that is the procedure above

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

Step-by-step explanation:

Unit circle

Answer:

A

≈

16.4

Hello!

[tex]\large\boxed{P = 300m}[/tex]

Use the following formula for the perimeter:

P = 2l + 2w, where:

l = length

w = width

Therefore:

P = 2(90) + 2(60)

Simplify:

P = 180 + 120 = 300 m

Answer:

well how about you use common sense 100 yards long on each side 200 yards then add 5o yards since the the that is how wide it is then add another 50 and you get 300 yards then convert that to meters

Answer:

14=-2x-4

18=-2x

x=-9

Hope This Helps!!!

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]

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