Ask question

Ask question

All categories

2 years ago

5

The length of a rectangle is three times the width. Ten area of the rectangle is 108 sq in. What is the width of the triangle? A

.6in B.9in C.12in D.36in

1 answer:

zubka84 [21]2 years ago

4 0

the answer would be a. 6"

You might be interested in

A hose filling a pool discharges water at a rate of 4L per minute. How long in hours does it take to fill a pool with a 300L cap

zimovet [89]

Answer:

no se de que hablas

Step-by-step explanation:

hablas español o no

7 0

2 years ago

2. Which of the following points lies in the solution set of the inequality y>/=3x+10?

Alex_Xolod [135]

Sorry if I am wrong I am not 100% sure but I think it’s 3

3 0

2 years ago

Please answer it now in two minutes

oksano4ka [1.4K]

Answer:

3.9

Step-by-step explanation:

Pythagorean theorem:

a^2 + b^2 = c^2

a^2 + 1^2 = 4^2

a^2 + 1 = 16

a^2 = 15

a = sqrt(15)

a = 3.9

Answer a = 3.9 yards

3 0

2 years ago

Read 2 more answers

I can't figure out how to find the proper area of a prism, can anyone help?

andrey2020 [161]

Answer:

Search Results

Featured snippet from the web

A right prism is made up of two identical, parallel bases on the ends, and the faces are perpendicular to the bases. The formula for finding surface area is 2B + hP (where B is the area of one of the bases, h is the prism's height and P is the perimeter of the base)

Step-by-step explanation:

hope this helps

8 0

3 years ago

A 24.6 g marble sliding to the right at 62.0 cm/s overtakes and collides elastically with a 12.3 g marble moving in the same dir

Softa [21]

Answer:

Velocity of the 24.6 g marble is 31 cm/s.

Step-by-step explanation:

Given:

Marble 1:

Mass of the marble $(m_1)$ = 12.3 gm

Velocity of $m_1$ before collision $v_1_i$ = 15.5 cm/s

Velocity of $m_1$ after collision $v_1_f$ = 77.5 cm/s

Marble 2:

Mass of the marble $(m_2)$ = 24.6 gm

Velocity of $m_2$ before collision $v_2_i$ = 62 cm/s

Velocity of $m_2$ after collision $v_2_f$ = ?

From the law of conservation of momentum, we know that momentum before equals the momentum after :

So,

⇒ $P(i)=P(f)$

⇒ $(m_1\times v_1_i )+(m_2\times v_2_i) =(m_1\times v_1_f)+(m_2\times v_2_f)$

⇒ Plugging the values.

⇒ $(12.3\times 15.5) +(24.6\times 62)=(12.3\times 77.5)+(24.6\times v_2_f)$

⇒ $190.65+1525.2=953.25+24.6\times v_2_f$

⇒ $1715.85=953.25+24.6\times v_2_f$

⇒ $1715.85-953.25=24.6\times v_2_f$ ...subtracting both sides 953.25

⇒ $762.6=24.6\times v_2_f$

⇒ $\frac{762.6}{24.6}\ =v_2_f$ ...dividing both sides with 24.6

⇒ $31=v_2_f$

⇒ $v_2_f =31$ cm/s

The velocity of the 24.6 g marble after the collision is 31 cm/s and it will move opposite to of the 12.3 g marbles that is towards left.

6 0

3 years ago