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2 years ago

Laura has 30 cupcakes she wants to give each friend 1/6 of her cupcakes how many cupcakes will Each friend get

Mathematics

2 answers:

sdas [7]2 years ago

7 0

Answer:

5 cupcakes

Step-by-step explanation:

=1x 30/ 6

I hope it helps

Thank U

Mariana [72]2 years ago

5 0

Answer:

No Of cup cakes laura has : 30

No of c Cupcakes She wants to give her friends:1/6

Cupcakes each friend got : 30 × 1/6

: 5

There fore Number of cupcakes each friend gets: 5 Cupcakes

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Find the total surface area of the cube-shaped gift below

garri49 [273]

Answer:

Step-by-step explanation:

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2 years ago

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"Suppose an object falling in the atmosphere has mass m=15kg and the drag coefficient is γ=9kg/s. Recall that the differential e

Art [367]

Answer:

a. v(t)= -6.78 $e^{-16.33t}$ + 16.33 b. 16.33 m/s

Step-by-step explanation:

The differential equation for the motion is given by mv' = mg - γv. We re-write as mv' + γv = mg ⇒ v' + γv/m = g. ⇒ v' + kv = g. where k = γ/m.Since this is a linear first order differential equation, We find the integrating factor μ(t)= $e^{\int\limits^ {}k \, dt }$ = $e^{kt}$ . We now multiply both sides of the equation by the integrating factor.

μv' + μkv = μg ⇒ $e^{kt}$ v' + k $e^{kt}$ v = g $e^{kt}$ ⇒ [v $e^{kt}$ ]' = g $e^{kt}$ . Integrating, we have

∫ [v $e^{kt}$ ]' = ∫g $e^{kt}$

v $e^{kt}$ = $\frac{g}{k}$ $e^{kt}$ + c

v(t)= $\frac{g}{k}$ + c $e^{-kt}$ .

From our initial conditions, v(0) = 9.55 m/s, t = 0 , g = 9.8 m/s², γ = 9 kg/s , m = 15 kg. k = y/m. Substituting these values, we have

9.55 = 9.8 × 15/9 + c $e^{-16.33 * 0}$ = 16.33 + c

c = 9.55 -16.33 = -6.78.

So, v(t)= 16.33 - 6.78 $e^{-16.33t}$ . m/s = - 6.78 $e^{-16.33t}$ + 16.33 m/s

b. Velocity of object at time t = 0.5

At t = 0.5, v = - 6.78 $e^{-16.33 x 0.5}$ + 16.33 m/s = 16.328 m/s ≅ 16.33 m/s

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2 years ago

Pre-Calc: Find all the zeros of the function.

uysha [10]

The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)