Ask question

Ask question

All categories

3 years ago

7

Which lines, if any, must be parallel based on the given diagram?

2 answers:

garik1379 [7]3 years ago

7 0

m and n are parallel and p and q are parallel from one another (parallel is when they are side by side and can never intertwine) i think the answer is C

Semmy [17]3 years ago

6 0

I cant really see the letters but the ones that are going straight up. I believe this is the answer because as you can see it's going up without intersecting between each other.

You might be interested in

Colin invests £1100 into his bank account.

lutik1710 [3]

Answer:

165 intreset 1265 total

Step-by-step explanation:

I = P*R*T

1100*.05*3

8 0

3 years ago

Read 2 more answers

Mrs.Storz is driving at 48.5 mph for 1.2 hours. What will be the distance travelled by Mrs.Storz?

Anna35 [415]

Answer:

58.2

Step-by-step explanation:

48.5*1.2=58.2

7 0

3 years ago

Which quadratic equation defines the function that has zeros at -8 and 6?

Citrus2011 [14]

Answer is D.) Hope I could help have a good day

7 0

3 years ago

(WILL GIVE BRAINLIEST IF CORRECT)

Alex

Answer:

C. 40

Step-by-step explanation:

20 + 4x = 180

subtract 20 from both sides

4x = 160

Divide both sides by

x = 40

6 0

3 years ago

Read 2 more answers

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago