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3 years ago

12

Is the point (1, -3) a solution to the system? Show all work 50 points

2 answers:

Genrish500 [490]3 years ago

5 0

Answer:

yes

Step-by-step explanation:

u dont need work for that it bascially givs u the answer

Agata [3.3K]3 years ago

3 0

Answer:

Yes there is no work needed i just did it

Step-by-step explanation:

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tatyana61 [14]

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3 years ago

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evablogger [386]

The answer is D!! hope i’m not too late

4 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

What are the zeros of the quadratic function?<br> y = x^2 - 2x - 15

Pie

Answer:

$x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}$

Step-by-step explanation:

$-> 0=2x^2-2x-15$

-zeros means the solution aka x intercept, meaning y=0

$x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:2\left(-15\right)}}{2\cdot \:2}$

=> according to quadratic formula

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where $ax^2+bx+c=0$

=> $x_{1,\:2}=\frac{-\left(-2\right)\pm \:2\sqrt{31}}{2\cdot \:2}$

simplify

$x_1=\frac{-\left(-2\right)+2\sqrt{31}}{2\cdot \:2},\:x_2=\frac{-\left(-2\right)-2\sqrt{31}}{2\cdot \:2}$

$x=\frac{1+\sqrt{31}}{2},\:x=\frac{1-\sqrt{31}}{2}$

Cannot simplify farther

3 0

3 years ago

Which is the length of the hypotenuse of the right triangle? Round your answer to the nearest tenth of a centimeter.

Olegator [25]

Answer:

Length of hypotenuse of given triangle

= sqrt(356) (exact value)

= 18.87 (to 2 decimal places)

Step-by-step explanation:

Using the pythagorean Theorem,

The hypotenuse

c = sqrt(a^2+b^2)

= sqrt(16^2+10^2)

= sqrt(356) (exact value)

= 18.87 (to 2 decimal places)

8 0

3 years ago

Read 2 more answers