All categories

3 years ago

Which two-dimensional views are standard in an orthographic drawing?

Mathematics

1 answer:

elixir [45]3 years ago

5 0

I think its B I’m not really sure

You might be interested in

Determine the verticle asymptote and horizontal asymptote of x^2-25/2x^+13x+15?

In-s [12.5K]

Answer:

what are you doing here honey and what is this gonna be about me in a hurry for the first two

5 0

3 years ago

3x+4y=10 solve for y

Sever21 [200]

Answer:

y=5/2 - 3x/4

Step-by-step explanation:

7 0

3 years ago

Choices;<br> a) 21.8°<br> b) 35.2°<br> c) 45.1°

kifflom [539]

Answer:

$\displaystyle a) \: {x}^{ \circ} = 21.8 ^{ \circ}$

Step-by-step explanation:

we are given opposite and adjacent

we want to figure out x°

remember that,

$\displaystyle \: \tan( \theta) = \frac{opp}{adj}$

where our opp is 5 and adj is 12.5

let $\theta$ be x°

$\displaystyle \: \tan( {x}^{ \circ} ) = \frac{5}{12.5}$

$\displaystyle \: {x}^{ \circ} = \arctan \left( \frac{5}{12.5} \right)$

by using calculator

$\displaystyle \: {x}^{ \circ} = 21.8$

hence,

our answer is a

6 0

3 years ago

Find the line through

Kaylis [27]

Answer:

$(\frac{1}{2},\frac{-1}{2},\frac{1}{2})$

Step-by-step explanation:

We have been given the intersection coordinates:

(7,1,-6) and perpendicular to the line x=-1+t ,y=-2+t and z=-1+t

From the condition of perpendicularity we know:

$(7,1,-6)(x,y,z)$

$\Rightarrow 7x+y-6z=0$

Now, we have given x=-1+t , y=-2+t and z= -1+t

$7(-1+t)+(-2+t)-6(-1+t)=0$

$-7+7t-2+t+6-6t=0$

$\Rightarrow -3+2t=0$

$2t=3$

$t=\frac{3}{2}$

Now, substituting t in the given coordinates x,y and z we get:

$x=-1+\frac{3}{2}=\frac{1}{2}$

$y=-2+\frac{3}{2}=\frac{-1}{2}$

And $z=-1+\frac{3}{2}=\frac{1}{2}$

3 0

3 years ago

Read 2 more answers

Solving for angles - Geometry

Otrada [13]

Answer:

<BAC=19

Step-by-step explanation:

<AOB=<DOC=142

<BAC=<ABD

2<BAC=180-142

2<BAC=38

<BAC=38/2

<BAC=19

7 0

3 years ago

Read 2 more answers