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Schach [20]
3 years ago
5

80 girls and boys have planned for a movie. They are in a ratio of 6 girls to 2 boys. How many girls are there?

Mathematics
2 answers:
mina [271]3 years ago
4 0
There are 60 girls 20 boys in this equation.
Leno4ka [110]3 years ago
4 0
There are 60 girls and 20 boys.

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Brittany is making hair bows to sell at a craft show. Each hair bow requires 3/4 yard of ribbon. Brittany plans to make 50 bows.
Nadya [2.5K]
Brittany would use 37 and 1/2
7 0
3 years ago
Read 2 more answers
In 7 hours, Goran reads 56 chapters of a book.<br> What is his rate in chapters per hour?
tensa zangetsu [6.8K]

Answer:

His rate in chapters per hour is 8 chapters per hour

Step-by-step explanation:

If you divide 56 by 7 you get 8 also if you want to check to be sure you can just multiply 8 and 7 or 7 and 8

5 0
2 years ago
-<br><img src="https://tex.z-dn.net/?f=%20-%203%5Cpi%20%2B%20w%20%3D%202%5Cpi" id="TexFormula1" title=" - 3\pi + w = 2\pi" alt="
pantera1 [17]

Isolate the w. Note the equal sign. What you do to one side, you do to the other.

Add 3π to both sides

- 3π (+3π) + w = 2π (+3π)

w = 2π + (3π)

w = 5π

w = 5π is your answer

hope this helps

4 0
3 years ago
Fill in the blanks to explain how to solve 3/5 divided by 1/10.
MrMuchimi
You can use a common denominator and rewrite 3/5 divided by 1/10 as 6/10 divided by 1/10.
Than you can think, “How many ___s are in ___? (Sorry, i don’t know this part)
3/5 divided 1/10 is 6
7 0
3 years ago
For each vector field f⃗ (x,y,z), compute the curl of f⃗ and, if possible, find a function f(x,y,z) so that f⃗ =∇f. if no such f
butalik [34]

\vec f(x,y,z)=(2yze^{2xyz}+4z^2\cos(xz^2))\,\vec\imath+2xze^{2xyz}\,\vec\jmath+(2xye^{2xyz}+8xz\cos(xz^2))\,\vec k

Let

\vec f=f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k

The curl is

\nabla\cdot\vec f=(\partial_x\,\vec\imath+\partial_y\,\vec\jmath+\partial_z\,\vec k)\times(f_1\,\vec\imath+f_2\,\vec\jmath+f_3\,\vec k)

where \partial_\xi denotes the partial derivative operator with respect to \xi. Recall that

\vec\imath\times\vec\jmath=\vec k

\vec\jmath\times\vec k=\vec i

\vec k\times\vec\imath=\vec\jmath

and that for any two vectors \vec a and \vec b, \vec a\times\vec b=-\vec b\times\vec a, and \vec a\times\vec a=\vec0.

The cross product reduces to

\nabla\times\vec f=(\partial_yf_3-\partial_zf_2)\,\vec\imath+(\partial_xf_3-\partial_zf_1)\,\vec\jmath+(\partial_xf_2-\partial_yf_1)\,\vec k

When you compute the partial derivatives, you'll find that all the components reduce to 0 and

\nabla\times\vec f=\vec0

which means \vec f is indeed conservative and we can find f.

Integrate both sides of

\dfrac{\partial f}{\partial y}=2xze^{2xyz}

with respect to y and

\implies f(x,y,z)=e^{2xyz}+g(x,z)

Differentiate both sides with respect to x and

\dfrac{\partial f}{\partial x}=\dfrac{\partial(e^{2xyz})}{\partial x}+\dfrac{\partial g}{\partial x}

2yze^{2xyz}+4z^2\cos(xz^2)=2yze^{2xyz}+\dfrac{\partial g}{\partial x}

4z^2\cos(xz^2)=\dfrac{\partial g}{\partial x}

\implies g(x,z)=4\sin(xz^2)+h(z)

Now

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+h(z)

and differentiating with respect to z gives

\dfrac{\partial f}{\partial z}=\dfrac{\partial(e^{2xyz}+4\sin(xz^2))}{\partial z}+\dfrac{\mathrm dh}{\mathrm dz}

2xye^{2xyz}+8xz\cos(xz^2)=2xye^{2xyz}+8xz\cos(xz^2)+\dfrac{\mathrm dh}{\mathrm dz}

\dfrac{\mathrm dh}{\mathrm dz}=0

\implies h(z)=C

for some constant C. So

f(x,y,z)=e^{2xyz}+4\sin(xz^2)+C

3 0
3 years ago
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