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Mnenie [13.5K]
3 years ago
11

Draw the graphs of the pair of linear equations : x + 2y = 5 and 2x - 3y = -4 Also find the points where the lines meet the x -

axis .
Mathematics
1 answer:
aliina [53]3 years ago
6 0

Answer:

(1, 2)

Step-by-step explanation:

Given the equation of the lines x + 2y = 5 and 2x - 3y = -4

First we need to make x the subject of the formulas

For x+2y = 5

x = 5 - 2y ... 1

For 2x - 3y = -4

2x = -4+3y

x = (-4+3y)/2 ... 2

Equate 1 and 2

5 - 2y =  (-4+3y)/2

2(5-2y) = -4+3y

10 - 4y = -4+3y

-4 -3y = -4-10

-7y = -14

y = 14/7

y = 2

Substitute y = 2 into 1

x = 5 = 2y

x = 5 - 2(2)

x = 5 - 4

x = 1

Hence the point where the lines meet will be at (1, 2)

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Rob ran 27.2 miles less than Jasmine last week. Rob ran 10 miles. How many miles did Jasmine run?
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Answer:

Jasmine ran last week 37.2 miles

Step-by-step explanation:

Let

x ----> number of miles Rob ran last week

y ----> number of miles Jasmine ran last week

we know that

x=y-27.2 ----> equation A

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3 years ago
(cotx+cscx)/(sinx+tanx)
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Answer:   \bold{\dfrac{cot(x)}{sin(x)}}

<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

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\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

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Slope:
(y₂ - y₁) ÷ (x₂ - x₁)
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(-2 - 1) ÷ (-2 + 3)
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The answer is a.
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