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Daniel [21]
3 years ago
5

the girls decided to go on a hike On the Sunny Oaks Forest Trails the girls want to hike as a group but they all walk at differe

nt speeds 5 girls walk 2.5 miles per hour 8 girls walk 3 miles per hour and two girls walk 3.8 miles per hour what is the average pace? * please take picture of work *
Mathematics
1 answer:
snow_tiger [21]3 years ago
5 0
The answer would be 4.1 as the average. (Sorry I could not take a picture of my work, but here's a brief explanation of how I did it. 1st I added all three numbers, 2.5, 3.8, and 3, then divide by 3 because there are 3 numbers to add, when done dividing, you should have your answer! (I did not do the work on paper, I did this in my mind seeing that you just add those 3 numbers then divide by 3)
 
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Please please help me with this
Lemur [1.5K]

Answer:

c

Step-by-step explanation:

c appears to be the better description as the line on the left has a solid circle at x = 1 indicating it is defined for x ≤ 1

While the line on the right has an open circle at x = 1 indicating it is defined for x > 1

Both line segments have a negative slope, thus

y = - 2x + 1 x > 1

y = - x + 2 : x ≤ 1

7 0
3 years ago
Someone please help correct answers only
sergeinik [125]
I think the correct answer is P=3
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8 0
2 years ago
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What is the 4(4x)+(x) equal!?
e-lub [12.9K]

4(4x)+(x)

= 16x+x

Combine like terms

16x+x

(16x+x)

= 17x

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I hope that's help ! Good night .

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3 years ago
(Please help with this, I'm in the middle of a quiz)
astraxan [27]

Answer:

D)16 ft (feet)

Step-by-step explanation:

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Then you square root that and get 16.

Hope this helps.

6 0
3 years ago
F⃗ (x,y)=−yi⃗ +xj⃗ f→(x,y)=−yi→+xj→ and cc is the line segment from point p=(5,0)p=(5,0) to q=(0,2)q=(0,2). (a) find a vector pa
DerKrebs [107]

a. Parameterize C by

\vec r(t)=(1-t)(5\,\vec\imath)+t(2\,\vec\jmath)=(5-5t)\,\vec\imath+2t\,\vec\jmath

with 0\le t\le1.

b/c. The line integral of \vec F(x,y)=-y\,\vec\imath+x\,\vec\jmath over C is

\displaystyle\int_C\vec F(x,y)\cdot\mathrm d\vec r=\int_0^1\vec F(x(t),y(t))\cdot\frac{\mathrm d\vec r(t)}{\mathrm dt}\,\mathrm dt

=\displaystyle\int_0^1(-2t\,\vec\imath+(5-5t)\,\vec\jmath)\cdot(-5\,\vec\imath+2\,\vec\jmath)\,\mathrm dt

=\displaystyle\int_0^1(10t+(10-10t))\,\mathrm dt

=\displaystyle10\int_0^1\mathrm dt=\boxed{10}

d. Notice that we can write the line integral as

\displaystyle\int_C\vecF\cdot\mathrm d\vec r=\int_C(-y\,\mathrm dx+x\,\mathrm dy)

By Green's theorem, the line integral is equivalent to

\displaystyle\iint_D\left(\frac{\partial x}{\partial x}-\frac{\partial(-y)}{\partial y}\right)\,\mathrm dx\,\mathrm dy=2\iint_D\mathrm dx\,\mathrm dy

where D is the triangle bounded by C, and this integral is simply twice the area of D. D is a right triangle with legs 2 and 5, so its area is 5 and the integral's value is 10.

4 0
2 years ago
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