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3 years ago

7

there was a total of six football games a month. the season is played for five months. how many football games are in the season

s? answer

1 answer:

Dvinal [7]3 years ago

6 0

Answer:

There are 30 games a season

Step-by-step explanation:

To find the number of games in a season multiply the number of games in a month times the number of months in a season:

1month= 28, 30, or 31 days(does not matter currently)

6 games a month

played 5 months=season

6•5=30 games in a season

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7x+14<br> 2x2 + x-6<br> What is the GCF

fenix001 [56]

i will solve it first i need full form

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3 years ago

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Can someone please explain with answer?

ira [324]

Using: A^2+b^2=c^2

A&b= (√10)^2=10

10+10=c^2

20=c^2

(take the square root of 20)

c=2√5

Your answer is D

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Ana tried to prove that an exterior triangle angle measure equals the sum of the measures of the two

maxonik [38]

The first mistake of Ana is sum of interior is 180° that is given by the definition. Then the correct option is A.

The statements, reasons, and options are attached below.

<h3>What is the triangle?</h3>

A triangle is a three-sided polygon with three angles. The angles of the triangle add up to 180 degrees.

Linear angle – If the total of two angles is 180 degrees, they are said to be linear angles.

Ana tried to prove that an exterior triangle angle measure equals the sum of the measures of the two interior angles not adjacent to it.

∠1 + ∠2 + ∠3 = 180° (By definition of triangle)

∠3 + ∠4 = 180° (By definition of linear angle)

∠1 + ∠2 + ∠3 = ∠3 + ∠4 (Substitution)

∠1 + ∠2 = ∠4 (Subtraction on both side by ∠3)

The first mistake of Ana is sum of interior is 180° that is given by the definition.

Then the correct option is A.

More about the triangle link is given below.

brainly.com/question/25813512

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2 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

4 years ago

1 + 2 + (-5) + 4 <br><br> A . -6<br> B . -5 <br> C . 0<br> D . 2<br> E . 4<br> F . 12

Dovator [93]

Answer:

Option D. 2

Step-by-step explanation:

1+2+(-5)+4

= 1+2-5+4

= 1+2+4-5

= 7-5

= 2

5 0

3 years ago