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worty [1.4K]
3 years ago
8

What is y+6=5x please show steps to solve

Mathematics
1 answer:
NemiM [27]3 years ago
5 0
We can solve for x or solve for y,
for x
y+6=5x
divide both sides by 5
\frac{y+6}{5}=x

for  y
y+6=5x
minus 6 from both sides
y=5x-6
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if it takes 10 men 10 days to dig a hole how long will it take 5 men to dig a hole that's half as big
Nataly [62]

Answer:

10 days

Step-by-step explanation:

It will take 10 men 5 days to dig a hole half as big.

It will take 5 men 10 days to dig a hole half as big.

( Remember, the less people, the more time it will take. )

8 0
2 years ago
Read 2 more answers
1. Derive the half-angle formulas from the double
lilavasa [31]

1) cos (θ / 2) = √[(1 + cos θ) / 2], sin (θ / 2) = √[(1 - cos θ) / 2], tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) (x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°). The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

<h3>How to apply trigonometry on deriving formulas and transforming points</h3>

1) The following <em>trigonometric</em> formulae are used to derive the <em>half-angle</em> formulas:

sin² θ / 2 + cos² θ / 2 = 1                      (1)

cos θ = cos² (θ / 2) - sin² (θ / 2)           (2)

First, we derive the formula for the sine of a <em>half</em> angle:

cos θ = 2 · cos² (θ / 2) - 1

cos² (θ / 2) = (1 + cos θ) / 2

cos (θ / 2) = √[(1 + cos θ) / 2]

Second, we derive the formula for the cosine of a <em>half</em> angle:

cos θ = 1 - 2 · sin² (θ / 2)

2 · sin² (θ / 2) = 1 - cos θ

sin² (θ / 2) = (1 - cos θ) / 2

sin (θ / 2) = √[(1 - cos θ) / 2]

Third, we derive the formula for the tangent of a <em>half</em> angle:

tan (θ / 2) = sin (θ / 2) / cos (θ / 2)

tan (θ / 2) = √[(1 - cos θ) / (1 + cos θ)]

2) The formulae for the conversion of coordinates in <em>rectangular</em> form to <em>polar</em> form are obtained by <em>trigonometric</em> functions:

(x, y) → (r · cos θ, r · sin θ), where r = √(x² + y²).

3) Let be the point (x, y) = (2, 3), the coordinates in <em>polar</em> form are:

r = √(2² + 3²)

r = √13

θ = atan(3 / 2)

θ ≈ 56.309°

The point (x, y) = (2, 3) is equivalent to the point (r, θ) = (√13, 56.309°).

Let be the point (r, θ) = (4, 30°), the coordinates in <em>rectangular</em> form are:

(x, y) = (4 · cos 30°, 4 · sin 30°)

(x, y) = (2√3, 2)

The point (r, θ) = (4, 30°) is equivalent to the point (x, y) = (2√3, 2).

4) Let be the <em>linear</em> function y = 5 · x - 8, we proceed to use the following <em>substitution</em> formulas: x = r · cos θ, y = r · sin θ

r · sin θ = 5 · r · cos θ - 8

r · sin θ - 5 · r · cos θ = - 8

r · (sin θ - 5 · cos θ) = - 8

r = - 8 / (sin θ - 5 · cos θ)

The <em>linear</em> function y = 5 · x - 8 is equivalent to the function r = - 8 / (sin θ - 5 · cos θ).

To learn more on trigonometric expressions: brainly.com/question/14746686

#SPJ1

4 0
2 years ago
find the area of a parrellelogram whose base is of length 25 cm and the corresponding altitude is of length 8.4 cm​
Free_Kalibri [48]

Answer:

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6 0
2 years ago
What is w for -3w = -7(w+1)
Alchen [17]
-3w= -7w-7 =
-3w+7w = -7w+7w - 7 =
4w= -7
4w/4 = -7/4 =
w = -7/4

3 0
3 years ago
What’s the equation for a line passing through points (-1, 4.5) and (6, 8)?
san4es73 [151]

Answer:

<em>Given </em><em>points </em><em>(</em><em> </em><em>-</em><em>1</em><em> </em><em>,</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>)</em><em> </em><em>and </em><em>(</em><em> </em><em>6</em><em> </em><em>,</em><em> </em><em>8</em><em> </em><em>)</em>

Step-by-step explanation:

<em>first </em><em>we </em><em>should </em><em>calculate </em><em>slope </em>

<em>Slope </em><em>(</em><em> </em><em>m</em><em>) </em>

<em>=</em><em> </em><em>(</em><em> </em><em>y2 </em><em>-</em><em> </em><em>y1</em><em>) </em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>x2</em><em> </em><em>-</em><em> </em><em>x1</em><em>) </em>

<em>=</em><em> </em><em>(</em><em> </em><em>8</em><em> </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>)</em><em> </em><em>/</em><em> </em><em>(</em><em> </em><em>6</em><em> </em><em> </em><em>+</em><em> </em><em>1</em><em> </em><em>)</em>

<em>=</em><em> </em><em>3</em><em>.</em><em>5</em><em>/</em><em>7</em>

<em>=</em><em> </em><em>0</em><em>.</em><em>5</em>

<em>Now </em><em>the </em><em>equation </em><em>of </em><em>line </em><em>is </em><em>given </em><em>by </em>

<em>y </em><em>-</em><em> </em><em>y1 </em><em>=</em><em> </em><em>m </em><em>(</em><em> </em><em>x </em><em>-</em><em> </em><em>x</em><em>1</em><em> </em><em>)</em>

<em>y </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>(</em><em> </em><em>x </em><em>+</em><em> </em><em>1</em><em> </em><em>)</em>

<em>(</em><em> </em><em>y </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em> </em><em>)</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>(</em><em> </em><em>x </em><em>+</em><em> </em><em>1</em><em>)</em>

<em>y </em><em>-</em><em> </em><em>4</em><em>.</em><em>5</em><em> </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>0</em><em>.</em><em>5</em>

<em>y </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>0</em><em>.</em><em>5</em><em> </em><em>+</em><em> </em><em>4</em><em>.</em><em>5</em>

<em>y </em><em>=</em><em> </em><em>0</em><em>.</em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>5</em>

3 0
3 years ago
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