Plz help

Mathematics

1 answer:

mixas84 [53]3 years ago

7 0

$x2y + z = 9 \\ \\ 1. \: x + 2y + z + - 2y = 9 + - 2y \\ x + z = - 2y + 9 \\ 2. \: x + z + - z = - 2y + 9 + - z \\ x = - 2y - z + 9 \\ \\ \\ x + y - z = 5 \\ \\ 1. \: x + y - z + - y = 5 + - y \\ x - z = - y + 5 \\ 2. \: x - z + z = - y + 5 + z \\ x = - y + z + 5 \\ \\ \\ 3x - y + 2z = 12 \\ 1. \: 3x - y + 2z + y = 12 + y \\ 3x + 2z = y + 12 \\ 2. \: 3x = y - 2z + 12 \\ 3. \: \frac{3x}{3} = \frac{y - 2z + 12}{3} \\ x = \frac{1}{3} y + \frac{ - 2}{3} z + 4$

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In the diagram, what is the measure of angle 1?

VLD [36.1K]

We know that
angle (3x) and angle (9x) are supplementary angles
so
3x+9x=180°------> 12x=180°------> x=180°/12-----> x=15°

angle (9x) and angle (1) are supplementary angles
so
9x+∡1=180---------> 9*15+∡1=180
∡1=180-9*15---------> ∡1=180-135------> ∡1=45°

the answer is
∡1 is 45°

alternative method
angle 1 = angle 3x----------> vertical angles
∡1=3x-----> 3*15-----> 45°

7 0

3 years ago

I’m lost again can somebody help asap ??

serious [3.7K]

Fully detailed and equipped information process right below. The answer for this question is "25" (simplified form).

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(5^{- 5} \times 2^8 \times 1 \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(2^8 \times 1 \times \dfrac{1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{2^8 \times 1}{5^5} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(1 \times \dfrac{256}{3125} \Bigg)^{- 2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \Bigg(\dfrac{3125}{256} \Bigg)^2}$

$\mathbf{2^{28} \times \Bigg(\dfrac{5^{- 2}}{2^3} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1}{2^3 \times 5^2} \Bigg)^4 \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(2^3 \times 5^2)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \Bigg(\dfrac{1^4}{(5^2)^4 \times (2^3)^4} \Bigg) \times \dfrac{3125^2}{256^2}}$

$\mathbf{2^{28} \times \dfrac{1}{2^{12} \times 5^8} \times \dfrac{3125^2}{256^2}}$

$\mathbf{\dfrac{3125^2 \times 1 \times 2^{28}}{256^2 \times 5^8 \times 2^{12}}}$

$\mathbf{\dfrac{2^{(28 - 12)} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 3125^2}{256^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times (5^5)^2}{(2^8)^2 \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10}}{2^{16} \times 5^8}}$

$\mathbf{\dfrac{2^{16} \times 5^{10 - 8}}{2^{16}}}$

$\mathbf{\dfrac{5^{10 - 8}}{1}}$

$\mathbf{\therefore \quad 5^2}$

$\mathbf{\therefore \quad 25}$

$\boxed{\mathbf{\underline{\therefore \quad Final \: \: Answer \: \: is: \: 25}}}$

Hope it helps.

7 0

3 years ago

You roll a standard number cube. Find P (number greater than 4). A. 2/3. B. 1/2. C. 3/5. D. 1/3

damaskus [11]

Total Outcome = 6

Number greater than 4 = 2 (5&6)

So, probability would be 2/6 = 1/3

(D)OPTION D IS YOUR ANSWER

4 0

3 years ago

Read 2 more answers

I need some help please