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3 years ago

Solve equations using addition x+y=5 and x-y=3

1 answer:

8 0

X + y = 5 ---------- (1)
x - y = 3 ----------- (2)

(1) + (2):
2x = 8
x = 4 ----------- sub into (1)

4 + y = 5
y = 5 - 4
y = 1

Ans: x = 4, y = 1

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What is the slope of a line that is perpendicular to the line 2y – 3x = 8?

GarryVolchara [31]

Answer:

$= \frac{3}{2}$

Step-by-step explanation:

$y = mx + c$

Here,

<h2>m => slope</h2><h3>c => intercept</h3>

In this equation ,

$2y - 3x = 8$

to find the value of m or the value of slope we have to solve for y

Let's solve,

$2y - 3x = 8 \\ 2y = 8 + 3x \\ \frac{2y}{2} = \frac{8 + 3x}{2} \\ y = 4 + \frac{3x}{2} \\ y = \frac{3x}{2} + 4$

So, the slope is,

$= \frac{3}{2}$

6 0

3 years ago

Which of the following is equivalent to tan2θcos(2θ) for all values of θ for which tan2θcos(2θ) is defined?

Aloiza [94]

Answer:

2sin²θ - tan²θ

Step-by-step explanation:

Given

tan²θcos(2θ)

Required

Simplify

We start by simplifying cos(2θ)

cos(2θ) = cos(θ+θ)

From Cosine formula

cos(A+A) = cosAcosA - sinAsinA

cos(A+A) = cos²A - sin²A

By comparison

cos(2θ) = cos(θ+θ)

cos(2θ) = cos²θ - sin²θ ----- equation 1

Recall that cos²θ + sin²θ = 1

Make sin²θ the subject of formula

sin²θ = 1 - cos²θ

Substitute sin²θ = 1 - cos²θ in equation 1

cos(2θ) = cos²θ - (1 - cos²θ)

cos(2θ) = cos²θ - 1 +cos²θ

cos(2θ) = cos²θ + cos²θ - 1

cos(2θ) = 2cos²θ - 1

Substitute 2cos²θ - 1 for cos(2θ) in the given question

tan²θcos(2θ) becomes

tan²θ(2cos²θ - 1)

Open brackets

2cos²θtan²θ - tan²θ

------------------------

Simplify tan²θ

tan²θ = (tanθ)²

Recall that tanθ = sinθ/cosθ

So, we have

tan²θ = (sinθ/cosθ)²

tan²θ = sin²θ/cos²θ

------------------------

Substitute sin²θ/cos²θ for tan²θ

2cos²θtan²θ - tan²θ becomes

2cos²θ(sin²θ/cos²θ) - tan²θ

Open bracket (cos²θ will cancel out cos²θ) to give

2(sin²θ) - tan²θ

2sin²θ - tan²θ

Hence, the simplification of tan²θcos(2θ) is 2sin²θ - tan²θ

Option E is correct

7 0

3 years ago