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Digiron [165]
3 years ago
13

PLEASE HELP Which line has an x-intercept of -5 and a y-intercept of 3?

Mathematics
1 answer:
anygoal [31]3 years ago
5 0

Answer:

the first answer or a

Step-by-step explanation:

just believe meeeee

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Describe how to draw a line that passes through the origin and has a slope of -2/3
Bas_tet [7]
Place a dot on the origin.
A line that has a slope of \frac{-2}{3} goes by this formula:
\frac{rise}{run}
So "rise -2" and "run 3."
Rise -2 is basically go down 2, so go down 2 units from the origin. Run 3 means go right 3 units from (0, -2).
Draw a line through (3, -2) and (0, 0).
5 0
3 years ago
Find 12 ÷ 3/5. Show or explain your reasoning.
scZoUnD [109]
20 because when you divide 12 by 3/5 you get 20 use a calculator
3 0
3 years ago
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Let S be the surface defined by x 2 + 2y 3 + 3z 4 = 6. Let T be the surface defined parametrically by r(u, v) = (1+ln u, 2e v+u−
aleksandrvk [35]

The tangent to C through (1, 1, 1) must be perpendicular to the normal vectors to the surfaces S and T at that point.

Let f(x,y,z)=x^2+2y^3+3z^4. Then S is the level curve f(x,y,z)=6. Recall that the gradient vector is perpendicular to level curves; we have

\nabla f(x,y,z)=(2x,6y,12z^2)

so that the gradient of f at (1, 1, 1) is

\nabla f(1,1,1)=(2,6,12)

For the surface T, we have

\begin{cases}1+\ln u=1\\2e^v+u-2=1\\uv+1=1\end{cases}\implies u=1,v=0

so that \vec r(1,0)=(1,1,1). We can obtain a vector normal to T by taking the cross product of the partial derivatives of \vec r(u,v), and evaluating that product for u=1,v=0:

\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=\left(u-2ve^v,-1,\dfrac{2e^v}u\right)

\left(\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right)(1,0)=(1,-1,2)

Now take the cross product of the two normal vectors to S and T:

(2,6,12)\times(1,-1,2)=(24,8,-8)

The direction of vector (24, 8, -8) is the direction of the tangent line to C at (1, 1, 1). We can capture all points on the line containing this vector by scaling it by t\in\Bbb R. Then adding (1, 1, 1) shifts this line to the point of tangency on C. So the tangent line has equation

\vec\ell(t)=(1,1,1)+t(24,8,-8)=(1+24t,1+8t,1-8t)

7 0
3 years ago
Its positive or negative? <--------------
dangina [55]
I'm new to this but, I believe its negative. I might be wrong, but please correct me.
7 0
3 years ago
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I need the answer please
adelina 88 [10]

H because that would equal 2

8 0
3 years ago
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