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2 years ago

Which undefined term can contain parallel lines

Mathematics

1 answer:

Dahasolnce [82]2 years ago

5 0

The undefined term that can contain parallel lines are line.

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-5+22=r-4+3r+5 I need help please and thank u

Brilliant_brown [7]

Answer:

r = 4

General Formulas and Concepts:

Pre-Alg

Order of Operations: BPEMDAS
Equality Properties

Step-by-step explanation:

Step 1: Define equation

-5 + 22 = r - 4 + 3r + 5

Step 2: Solve for r

Combine like terms: 17 = 4r + 1
Subtract 1 on both sides: 16 = 4r
Divide 4 on both sides: 4 = r
Rewrite: r = 4

Step 3: Check

Plug in r to verify it's a solution.

Substitute: -5 + 22 = 4 - 4 + 3(4) + 5
Add/Subtract: 17 = 3(4) + 5
Multiply: 17 = 12 + 5
Add: 17 = 17

6 0

2 years ago

Each small square in the graph paper represents 1 square unit

mrs_skeptik [129]

Answer:

But the graph paper is not here...

4 0

3 years ago

Alison does 8 math problems in 2 minutes. How long will it take her to do a set of 160 problems?

erastovalidia [21]

Answer:

40 min

Step-by-step explanation:

find unit rateeee p=problems

8 p= 2 min

8/2=4

so Alison can do 4 math problems per minute

so 160/4= 40 minutes

hope this helps

5 0

2 years ago

Read 2 more answers

Can you figure out the solution? Options: a) 8 b) 6 c) 2 d) 5

Nataly [62]

Answer:

Step-by-step explanation:

the answer is correct yep

7 0

2 years ago

Read 2 more answers

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

3 years ago