80

Please help. I don't understand how to do number 14

murzikaleks [220]

This is quiet simple. i like to use a method my math teacher calls "plug and chug". we know that y= 4x +2
some people call this the substitution method but it has the same concept. now that we know what y equals, we plug it into teh equation:
(4x+2) -x=17
then we "chug it":
4x+2-x=17
then we have to isolate x. add the x's:
5x+2=17
in order to isolate x subtract 2 from both sides:
5x=15
now to isolate again divide both sides with 5:
x=3
so x is 3. bow we plug that into what y is;
y=4 (3) +2
y=14
so:
x=3
y=14
(to double check plug and chug into the first question . if x and y are correct, you will get 17)

8 0

3 years ago

Please please help please

fenix001 [56]

Answer:

the triangle must be assumed to be a right triangle
AB = 148.4 yards
segment AB is shorter than the curved path

Step-by-step explanation:

Part 1: The main assumption required is the one required to use the hint. The triangle must be assumed to be a right triangle.

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Part 2: The approximate distance AB is the hypotenuse of the assumed right triangle, so will satisfy the Pythagorean theorem. For distances in yards, ...

AB² = 80² +125² = 22025

AB = √22025 ≈ 148.4 . . . . . yards

The approximate length across the pond is 148.4 yards.

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Part 3: A straight line is the shortest path between two points. Segment AB is a straight line path, so the length of segment AB will be shorter than any curved path between A and B.

4 0

4 years ago

Evaluate the line integral ∫CF⋅dr, where F(x,y,z)=5sinxi+4cosyj−2xzk and C is given by the vector function r(t)=t3i−t2j+t1k , 0≤

pickupchik [31]

$\mathbf r(t)=x(t)\,\mathbf i+y(t)\,\mathbf j+z(t)\,\mathbf k$
$\implies\begin{cases}x(t)=t^3\\y(t)=-t^2\\z(t)=t\end{cases}$
$\implies\dfrac{\mathrm d\mathbf r}{\mathrm dt}=3t^2\,\mathbf i-2t\,\mathbf j+\mathbf k$

with $0\le t\le1$ . With this parameterization, the vector field can be written as

$\mathbf F(x,y,z)=\mathbf F(x(t),y(t),z(t))=5\sin(t^3)\,\mathbf i+4\cos(t^2)\,\mathbf j-2t^4\,\mathbf k$

Now the line integral is given by

$\displaystyle\int_C\mathbf F\cdot\mathrm d\mathbf r=\int_{t=0}^{t=1}\mathbf F(x(t),y(t),z(t))\cdot\dfrac{\mathrm d\mathbf r(t)}{\mathrm dt}\,\mathrm dt$
$=\displaystyle\int_0^1(15t^2\sin(t^3)-8t\cos(t^2)-2t^4)\,\mathrm dt$
$=-5\cos1-4\sin1+\dfrac{23}5$