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Cloud [144]
3 years ago
5

Are the following lines parallel perpendicular or neither: x=6 and x=12

Mathematics
1 answer:
suter [353]3 years ago
5 0

Answer:

Not Parallel

Step-by-step explanation:

I think so sorry If I'm wrong

You might be interested in
Which of the following has a solution set of {x | x = 0}?
notka56 [123]

Answer:

(b)  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

Step-by-step explanation:

Here, the given expression is : {x | x = 0}

So, the ONLY element in the given set = {0}

Now, take each option and solve the given expression:

(a)  x + 1 < -1

Adding -1 BOTH sides, we get:

x + 1 -1 < -1  -1

or, x < - 2 ⇒ x = { -∞ , .... , -4,-3}

Also,   x + 1 < 1

Adding -1 BOTH sides, we get:

x + 1 -1 < 1  -1

or, x <0 ⇒ x = { -∞ , .... , -4,-3,-2,-1}

So, (x + 1 < -1) ∩ (x + 1 < 1) =  { -∞ , .... , -4,-3}∩ { -∞ , .... , -4,-3,-2,-1}  

=  { -∞ , .... , -4,-3}

⇒ (x + 1 < -1) ∩ (x + 1 < 1) ≠ {0}

Similarly, solving

(b) (x + 1 ≤ 1) ∩ (x + 1 ≥ 1)

(x + 1 ≤ 1) =  x≤ 0 =   { -∞ , .... , -4,-3,-2,-1, 0}

(x + 1 ≥ 1) =  x ≥ 0 =  {0,1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = {0}

(b)(x + 1 < 1) ∩ (x + 1 > 1)

(x + 1 < 1) =  x <  0 =   { -∞ , .... , -4,-3,-2,-1}

(x + 1 > 1) =  x  > 0 =  { 1,2,3,... ∞}

⇒ (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) = Ф

Hence,  (x + 1 ≤ 1) ∩ (x + 1 ≥ 1) =  {x | x = 0}

8 0
3 years ago
Read 2 more answers
1 | Find the equivalent to:
Gennadij [26K]

Answer:

B) \frac{7}{7^5}

Step-by-step explanation:

\frac{7^3*7^-2}{7^5} =\frac{1}{2401}

\frac{7}{7^5} =\frac{1}{2401}

hope this helped!

brainliest please!

7 0
2 years ago
"I think of a number, take away 1 and multiply the result by 3."​
iogann1982 [59]

Answer:

I got 6.

Step-by-step explanation:

I picked my number to be 3. I took away 1 to make it 2. I then multiplied 2 by 3, to get 6.

3 0
3 years ago
Read 2 more answers
Please help me asap<br>​
Lera25 [3.4K]

Answer:

Step-by-step explanation:

From the picture attached,

∠4 = 45°, ∠5 = 135° and ∠10 = ∠11

Part A

∠1 = ∠4 = 45°  [Vertically opposite angles]

∠1 + ∠3 = 180° [Linear pair of angles]

∠3 = 180° - ∠1

     = 180° - 45°

     = 135°

∠2 = ∠3 = 135°  [Vertically opposite angles]

∠8 = ∠5 = 135° [Vertically opposite angles]

∠5 + ∠6 = 180° [Linear pair of angles]

∠6 = 180° - 135°

∠6 = 45°

∠7 = ∠6 = 45° [Vertically opposite angles]

By triangle sum theorem,

m∠4 + m∠7 + m∠10 = 180°

45° + 45° + m∠10 = 180°

m∠10 = 180° - 90°

m∠10 = 90°

m∠10 = m∠12 = 90°  [Vertically opposite angles]

m∠10 = m∠11 = 90° [Given]

Part B

1). ∠1 ≅ ∠4  [Vertically opposite angles]

2). ∠7 + ∠5 = 180° [Linear pair]

3). ∠9 + ∠10 = 180° [Linear pair]

7 0
3 years ago
ALGEBRA 1
Pavlova-9 [17]
It would be helpful if you could post a photo so I can look at the context. Thank you
4 0
3 years ago
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