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3 years ago

Names two planes that do not intersect

Mathematics

2 answers:

weeeeeb [17]3 years ago

8 0

Answer:

parellel planes do not intersect

non-coplanar also dont

Luden [163]3 years ago

8 0

Answer:

y=x+0z

y=x+0z-5

Step-by-step explanation:

By far the easiest way to do it is using linear equations

we know that

y=x and

y=x-5

don't ever touch

so we can graph it on 3D and we get

and like the lines the planes don't touch

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If angle 2 has a measurement of (x+4) and angle 3 has a measurement of 3x what does x equal?

deff fn [24]

Answer:

x = 44 degrees.

Step-by-step explanation:

Hi!

Since m<2 and m<3 equal 180 degrees (sum of flat line), we can form an equation

(x+4) + (3x) = 180

x + 4 + 3x = 180

4x + 4 = 180

4x = 176

Divide by 4 on both sides

x = 44 degrees

Let me know if you have any questions

6 0

3 years ago

Write the equation of the line that includes the point (1,5) and has a slope of 3 in standard form.

Ymorist [56]

Answer:

-3x + y = 2

Step-by-step explanation:

The slope is 3. So the coefficient of x is 3.

when x = 1, y = 5.

So 3*1 + c = 5

So c is 2.

So y = 3x + 2

Therefore, taking 3x to the LHS, we get,

-3x + y = 2

Hope it helps ;)

7 0

3 years ago

Read 2 more answers

nikitadnepr [17]

Answer:

i am not so sure but I think its 10x = 33

3 0

3 years ago

If f(x) = 2 + 1, then f(3) =

Agata [3.3K]

F(3) is 3 since the equation of the line is f(x)=3

4 0

2 years ago

Find an equation of the plane through the point (−5,−1,2) with normal vector ????=⟨−1,−5,2⟩.

Ganezh [65]

Answer:

$x + 5y - 2z + 14 = 0$

Step-by-step explanation:

A plane that passes through the point $(x_{0}, y_{0}, z_{0})$ with a normal vector of $(a,b,c)$ has the following equation, initially:

$a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$

After we solve this, we have

$ax + by + cz + d = 0$

In this problem, we have that:

$(x_{0}, y_{0}, z_{0}) = (-5,-1,2)$

$(a,b,c) = (−1,−5,2)$

$a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$

$-1(x - (-5)) - 5(y - (-1)) + 2(z - 2) = 0$

$-(x + 5) - 5(y + 1) + 2(z - 2) = 0$

$-x - 5 - 5y - 5 + 2z - 4 = 0$

$-x - 5y + 2z - 14 = 0$

Multplying everything by -1

$x + 5y - 2z + 14 = 0$

3 0

4 years ago