The answer is choice A.
We're told that the left and right walls of the cube (LMN and PQR) are parallel planes. Any line contained in one of those planes will not meet another line contained in another plane. With choice A, it's possible to have the front and back walls be non-parallel and still meet the initial conditions. If this is the case, then OS won't be paralle to NR. Similarly, LP won't be parallel to MQ.
Answer:
x = 5
y = 4
Step-by-step explanation:
1. 2x + 2y = 18
2. x + 3y = 17
Multiply the first equation by 1 and the second equation by 2 to eliminate x
We have
2x + 2y = 18
2x + 6y = 34
Subtract equation 2 from equation 1
-4y = -16
Divide both sides by -4 to isolate y
-4y/-4 = -16/-4
y = 4
Now substitute 4 for y in either equation to get x. Using equation 2 we have
x + 3y = 17
x + 3 x 4 = 17
x + 12 = 17
Subtract 12 from both sides
x + 12 - 12 = 17 - 12
x = 5
x = 5 and
y = 4
Answer:
a
=
4/
1
−
|
x
|
h
=
0
Step-by-step explanation:
1. 9^2
2. 15^3
3. 14^3
4. 11^5
5. 16^4
We know that
cos A=adjacent side angle A/hypotenuse
adjacent side angle A=24 units
hypotenuse=26 units
cos A=24/26-----> 12/13
cos B=adjacent side angle B/hypotenuse
adjacent side angle B=10 units
hypotenuse=26 units
cos B=10/26------> 5/13
the answers are
cos A=12/13
cos B=5/13
cot A=adjacent side angle A/opposite side angle A
adjacent side angle A=24 units
opposite side angle A=10 units
cot A=24/10------> cot A=12/5
cot B=adjacent side angle B/opposite side angle B
adjacent side angle B=10 units
opposite side angle B=24 units
cot B=10/24------> cot B=5/12