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3 years ago

8

How would u test whether 7 is a factor of 291

1 answer:

Nina [5.8K]3 years ago

7 0

Divide 291 by 7. If you get a whole number, its a factor. If it comes out as a decimal or fraction, its not a factor.

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What it 55/100 in simplest form

den301095 [7]

55/100 = (55:5)/(100:5) = 11/20

55/100 = 0.55

6 0

3 years ago

Read 2 more answers

If jamie has $5 and purchases a soda for $1.15 how much change does jamie get back

ArbitrLikvidat [17]

Jamie get back:5-1.15
=$3.85

5 0

3 years ago

The _______ of an observation is determined by how far the values of the independent variables are from their means.

stira [4]

Answer:

The leverage of an observation is determined by how far the values of the independent variables are from their means.

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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

Find the solutions to y = (x – 1)2 for x = –3, 0, and 3.

Harman [31]

Answer:

if the 2 is the power of (x-1) here is ur answer

$y = (x - 1 ){}^{2} \\if \: x \: is \: \: equal \: to \: - 3 \\ y =( - 3 - 1) {}^{2} \\ y = ( - 4) {}^{2} \\ y = 16$

$if \: x = 0 \\ y = (0 - 1) {}^{2} \\y = ( - 1) { }^{2} \\ y = 1$

$if \: x = 3 \\ y = (3 - 1) {}^{2} \\ y = 2 { }^{2} \\ y = 4$

(-3,16) (0,1) (3,4)

7 0

3 years ago