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3 years ago

Two brown and 8 red birds are flying. What is the probability that two brown sit together?

1 answer:

6 0

1/5? I believe! I am assuming because there are 2 brown and 8 red, so there’s a total of 10 birds. Then you take 2/10, reduce that, giving you 1/5

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Y=x+2 y=-3X <br><br>solve system of equations algebraically <br>

goblinko [34]

Answer:

(x,y) = -1/2 , 3/2

Step-by-step explanation:

Let y = x +2 ----- (1)

y = -3x ------ (2)

Put equal both sides x +2 = -3x

x = -3x -2

x +3x = -2

4x = -2

x = -1/2 Put it in 2nd equation

Y = -3 (-1/2)

y = 3/2

so (x,y) = -1/2 , 3/2

Hope you got it.

Please Mark as Brainliest.

8 0

3 years ago

0.5(60) x + 5 experiment with Solutions

8_murik_8 [283]

Answer:

150

Step-by-step explanation:

Multiply 0.5 by 60 first (30), then multiply 30 by 5(150)

3 0

3 years ago

Will mark brainlest help me please

Allushta [10]

Answer:

no le entiendo por qué estás en inglés

7 0

3 years ago

2logbase4(x)-logbase4(5)=125

Goryan [66]

$D:x>0\\\\ 2\log_4x-\log_45=125\\ \log_4x^2-\log_45=\log_44^{125}\\ \log_4\dfrac{x^2}{5}=\log_44^{125}\\ \dfrac{x^2}{5}=4^{125}\\ x^2=5\cdot4^{125}=5\cdot2^{250}=5\cdot(2^{125})^2\\ x=\sqrt{5\cdot2^{250}} \vee x=-\sqrt{5\cdot2^{250}}\\ x=\sqrt{5\cdot4^{125}} \vee x=-\sqrt{5\cdot4^{125}}\\ x=\sqrt{5\cdot(2^{125})^2} \vee x=-\sqrt{5\cdot(2^{125})^2}\\ x=\sqrt5 \cdot\sqrt{(2^{125})^2} \vee x=-\sqrt5 \cdot\sqrt{(2^{125})^2} \\ x=2^{125}\sqrt5 \vee x=-2^{125}\sqrt5$

$x=-2^{125}\sqrt5 \not \in D\Rightarrow \boxed{x=2^{125}\sqrt5}$

7 0

3 years ago

Use Cramer’s Rule to solve system of equation.

alisha [4.7K]

$\left\{\begin{array}{ccc}5x+2y=4\\3x+4y+2z=6\\7x+3y+4z=29\end{array}\right\\\\A=\left[\begin{array}{ccc}5&2&0\\3&4&2\\7&3&4\end{array}\right]\\\\\det A=5\cdot4\cdot4+3\cdot3\cdot0+2\cdot2\cdot7-7\cdot4\cdot0-3\cdot2\cdot5-3\cdot2\cdot4=54\\W_x=\left[\begin{array}{ccc}4&2&0\\6&4&2\\29&3&4\end{array}\right]\\\det W_x=4\cdot4\cdot4+2\cdot2\cdot29+6\cdot3\cdot0-29\cdot4\cdot0-3\cdot2\cdot4-6\cdot2\cdot4=108$

$W_y=\left[\begin{array}{ccc}5&4&0\\3&6&2\\7&29&4\end{array}\right]\\\det W_y=5\cdot6\cdot4+3\cdot29\cdot0+4\cdot2\cdot7-7\cdot6\cdot0-29\cdot2\cdot5-3\cdot4\cdot4=-162\\W_z=\left[\begin{array}{ccc}5&2&4\\3&4&6\\7&3&29\end{array}\right]\\\det W_z=5\cdot4\cdot29+3\cdot3\cdot4+6\cdot2\cdot7-7\cdot4\cdot4-3\cdot6\cdot5-3\cdot2\cdot29=324\\\\x=\dfrac{\det W_x}{\det A}=\dfrac{108}{54}=2\\\\y=\dfrac{\det W_y}{\det A}=\dfrac{-62}{54}=-3\\\\z=\dfrac{\det W_z}{\det A}=\dfrac{324}{54}=6$

5 0

3 years ago