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2 years ago

-5x-y= 8 -4x+y=2 solve the system

Mathematics

1 answer:

yan [13]2 years ago

7 0

$\huge\fcolorbox{blue}{aqua}{Answer:}$

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1. $- 5x - y = 8$

<h3>Solve for x</h3>

$-5x-y=8$

$-5x=8+y$

$-5x=y+8$

$\frac{-5x}{-5}=\frac{y+8}{-5}$

$x=\frac{y+8}{-5}$

$x=\frac{-y-8}{5}$

<h3>Solve for y</h3>

$-5x-y=8$

$-y=8+5x$

$-y=5x+8$

$\frac{-y}{-1}=\frac{5x+8}{-1}$

$y=\frac{5x+8}{-1}$

$y=-5x-8$

────────────────────

2. $-4x-y=2$

<h3>Solve for x</h3>

$-4x-y=2$

$-4x=2+y$

$-4x=y+2$

$\frac{-4x}{-4}=\frac{y+2}{-4}$

$x=\frac{y+2}{-4}$

$x=-\frac{y}{4}-\frac{1}{2}$

<h3>Solve for y</h3>

$-4x-y=2$

$-y=2+4x$

$-y=4x+2$

$\frac{-y}{-1}=\frac{4x+2}{-1}$

$y=\frac{4x+2}{-1}$

$y=-4x-2$

════════════════════

Explanation:

#Carry on learning

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Answer:

1) ∠A=84°

2) ∠C=20°

Step-by-step explanation:

First, find ∠C:

<em>(I'm assuming the exterior angle of 126° makes a straight line with ∠C)</em>

The angles on a straight line always add up to 180. Therefore:

∠C+126=180

∠C=180-126

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Then find ∠B:

We also know that all the angles in a triangle add up to 180. Therefore:

∠A+∠B+∠C=180

∠A+∠B+54=180

∠A+∠B=126

2(∠B)+∠B=126

3(∠B)=126

∠B=42

Now, find ∠A:

∠A=2(∠B)

∠A=2(42)

∠A=84°

First, find ∠B:

<em>(Again, I'm assuming the exterior angle of 100° makes a straight line with ∠B)</em>

The angles on a straight line always add up to 180. Therefore:

∠B+100=180

∠B=180-100

∠B=80

Then find ∠A:

We also know that all the angles in a triangle add up to 180. Therefore:

∠A+∠B+∠C=180

∠A+80+∠C=180

∠A+∠C=100

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3 years ago

18 cm 18 cm 9 cm Area:

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Answer:

2916 cm

Step-by-step explanation:

In area you have to multiply

So all I did was

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What is the value of the expression i 0 × i 1 × i 2 × i 3 × i 4?

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ANSWER

The value of the expression is
$- 1$

EXPLANATION

Method 1: Rewrite as product of
${i}^{2}$

The expression given to us is,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4}$

We use the fact that
${i}^{2} = - 1$
to simplify the above expression.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{1} \times {i}^{3} \times {i}^{2} \times {i}^{4}$

This implies,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2} \times {i}^{2}$

We substitute to obtain,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times - 1 \times - 1 \times - 1\times - 1 \times - 1$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = 1\times 1 \times 1 \times - 1 = - 1$

Method 2: Use indices to solve.

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{0 + 1 + 2 + 3 + 4}$

This implies that,

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = {i}^{10}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( {{i}^{2}} )^{5}$

${i}^{0} \times {i}^{1} \times {i}^{2} \times {i}^{3} \times {i}^{4} = ( - 1 )^{5} = - 1$

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