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2 years ago

The equation of a vertical line passing through the point (-1,8) is

Mathematics

1 answer:

Afina-wow [57]2 years ago

8 0

Answer:

y=8 if you draw your points on the map the question asks a vertical line pasing through (-1,8) the answer would by y=8

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What is the area of the triangle given below?

daser333 [38]

Answer:

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Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

If p is the midpoint of line xy, xp=11-6b and py=6-b, find xp

MariettaO [177]

Since p is the midpoint then xp and pay must equal each other
So, 11-6b=6-b
Solve for b
11-6b=6-b
-6 -6
5-6b=-b
+6b +6b
5=5b
b=1

xp=11-6(1)
xp=5

5 0

3 years ago

Xy=10<br> 4x+2y=30 <br> I need help asap

emmasim [6.3K]

Answer:

x = 25/3; y = -5/3

Step-by-step explanation:

You did well so far by multiplying both sides of the first equation by 2.

Now notice that you have -2y in the new first equation and 2y in the second equation. -2y and 2y add to 0.

Now add the equations to eliminate y.

2x - 2y = 20

(+) 4x + 2y = 30

-------------------------

6x = 50

6x/6 = 50/6

x = 50/6 = 25/3

2x - 2y = 20

2(25/3) - 2y = 20

50/3 - 2y = 60/3

-2y = 60/3 - 50/3

-2y = 10/3

-2y/2 = (10/3)/(-2)

y = -5/3

Solution: x = 25/3; y = -5/3

4 0

1 year ago

evaluate the line integral ∫cf⋅dr, where f(x,y,z)=5xi−yj+zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

meriva

We have

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} \vec f(\vec r(t)) \cdot \dfrac{d\vec r}{dt} \, dt$

and

$\vec f(\vec r(t)) = 5\sin(t) \, \vec\imath - \cos(t) \, \vec\jmath + t \, \vec k$

$\vec r(t) = \sin(t)\,\vec\imath + \cos(t)\,\vec\jmath + t\,\vec k \implies \dfrac{d\vec r}{dt} = \cos(t) \, \vec\imath - \sin(t) \, \vec\jmath + \vec k$

so the line integral is equilvalent to

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (5\sin(t) \cos(t) + \sin(t)\cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (6\sin(t) \cos(t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \int_0^{\frac{3\pi}2} (3\sin(2t) + t) \, dt$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(-\frac32 \cos(2t) + \frac12 t^2\right) \bigg_0^{\frac{3\pi}2}$

$\displaystyle \int_C \vec f \cdot d\vec r = \left(\frac32 + \frac{9\pi^2}8\right) - \left(-\frac32\right) = \boxed{3 + \frac{9\pi^2}8}$

7 0

2 years ago

The decimal -2.97 is the solution to which subtraction problem?

Sladkaya [172]

Answer:

-6.1 - (-3.13) = -2.97

Step-by-step explanation:

two negative symbols = a positive.

7 0

2 years ago