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blagie [28]
3 years ago
15

ice cream is served for dessert if Gio serves himself and 3 friends half a cup of ice cream how many total pints did Gio and his

friends eat?
Mathematics
1 answer:
Assoli18 [71]3 years ago
7 0

Answer:

0.25 pints

Step-by-step explanation:

To find how many pints they ate, find the number of pints in half a cup.

In 1 cup, there are 0.5 pints.

This means in half a cup, there will be 0.25 pints.

So, Gio and his friends ate 0.25 pints

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suppose the circumference of a crop circle is 150.7968 hectometers (hm) what's the radius of the circle? (usett=3.1416)​
gulaghasi [49]

Answer:

The radius of the circle is 24\ hm

Step-by-step explanation:

we know that

The circumference of a circle is equal to

C=2\pi r

In this problem we have

C=150.7968\ hm

\pi=3.1416

substitute and solve for r

150.7968=2(3.1416)r

r=150.7968/(2(3.1416))=24\ hm

7 0
3 years ago
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If f (x) = 8(5x-2) find f -1 (x).
saw5 [17]

Step-by-step explanation:

f(x)=8(5x-2)

interchanging the role of x&y

x=8(5y-2)

  • x=40y-16
  • x+16=401y
  • y=(x+16)/40

f-1(x)=(x+16)/10

<h3>stay safe healthy and happy.</h3>
6 0
3 years ago
Need help in this ASAP
Artemon [7]

Answer:

The m∠YXW is 125°.

Addition equation: m∠YXW = m∠YXZ + m∠ZXW = 85° + 40° = 125°

Step-by-step explanation:

To find the m∠YXW you would need to add the m∠YXZ and m∠ZXW together:

m∠YXW = m∠YXZ + m∠ZXW = 85° + 40° = 125°

4 0
3 years ago
What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
emmasim [6.3K]
\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

So, the flux is given by

\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA
\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du
\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv
\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv

where t=-\sin v in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi
5 0
3 years ago
The following table shows the expressions that represent the sales of two companies:
arlik [135]

Answer: Company B (1.80)

Step-by-step explanation:

1.04= 4% Increase

1.80= 80% Increase

8 0
3 years ago
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