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3 years ago

Can you answer this math homework? Please!

Mathematics

1 answer:

inessss [21]3 years ago

8 0

Answer:

m = -3 and b = -3

Step-by-step explanation:

No solution at all when you graph the two equations

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A certain animated movie earned $ 1 . 1 ? 1 0 9 $1.1?10 9 in revenues at the box office. The movie lasts 9 . 1 ? 1 0 1 9.1?10 1

Leokris [45]

Answer:

$ $1.21\times 10^{7}$ per minute.

Step-by-step explanation:

We have been given that a certain animated movie earned $ $1.1\times10^{9}$ in revenues at the box office. The movie lasts $9.1\times10^{1}$ minutes.

To find the revenue earned per minute of the movie we will will divide $1.1\times10^{9}$ by $9.1\times10^{1}$ .

$\text{The revenue earned per minute by the movie}=\frac{1.1*10^{9}}{9.1*10^{1}}$

Using quotient rule of exponents we will get,

$\text{The revenue earned per minute by the movie}=\frac{1.1}{9.1}*10^{(9-1)}$

$\text{The revenue earned per minute by the movie}=\frac{1.1}{9.1}*10^{8}$

$\text{The revenue earned per minute by the movie}=0.1208791208791209\times 10^{8}$

We can write 0.1208791208791209 as $1.208791208791209\times10^{-1}$ .

$\text{The revenue earned per minute by the movie}=1.208791208791209\times 10^{-1} \times 10^{8}$

Using product rule of exponents we will get,

$\text{The revenue earned per minute by the movie}=1.208791208791209\times 10^{-1+8}$

$\text{The revenue earned per minute by the movie}=1.208791208791209\times 10^{7}$

Upon rounding our answer to two decimal places we will get,

$\text{The revenue earned per minute by the movie}=1.21\times 10^{7}$

Therefore, the movie had earned a revenue of $ $1.21\times 10^{7}$ per minute.

7 0

4 years ago

Read 2 more answers

Find the area of the circle

navik [9.2K]

Answer:

The answer is 28.26

Step-by-step explanation:

A=pir^2

A=pi(3)^2

A=9pi

A=28.26

Hope this helps :D

3 0

3 years ago

Read 2 more answers

Evaluate the surface integral ∫sf⋅ ds where f=⟨2x,−3z,3y⟩ and s is the part of the sphere x2 y2 z2=16 in the first octant, with

skad [1K]

Parameterize S by the vector function

$\vec s(u,v) = \left\langle 4 \cos(u) \sin(v), 4 \sin(u) \sin(v), 4 \cos(v) \right\rangle$

with 0 ≤ u ≤ π/2 and 0 ≤ v ≤ π/2.

Compute the outward-pointing normal vector to S :

$\vec n = \dfrac{\partial\vec s}{\partial v} \times \dfrac{\partial \vec s}{\partial u} = \left\langle 16 \cos(u) \sin^2(v), 16 \sin(u) \sin^2(v), 16 \cos(v) \sin(v) \right\rangle$

The integral of the field over S is then

$\displaystyle \iint_S \vec f \cdot d\vec s = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \vec f(\vec s) \cdot \vec n \, du \, dv$

$\displaystyle = \int_0^{\frac\pi2} \int_0^{\frac\pi2} \left\langle 8 \cos(u) \sin(v), -12 \cos(v), 12 \sin(u) \sin(v) \right\rangle \cdot \vec n \, du \, dv$