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3 years ago

3. What are the intersection points of the line whose equation is y=3x +3 and the

Mathematics

1 answer:

Gre4nikov [31]3 years ago

8 0

Answer:

(-2 , -3) and (0.6 , 4.8)

Step-by-step explanation:

y=3x +3

(x - 2)² + y² = 25

(x - 2)² + (3x +3)² = 25

x²-4x+4+9x²+18x+9-25=0

10x²+14x-12=0

5x²+7x-6=0

(x+2)(5x-3)=0

x = -2 or x = 3/5 (-0.6)

y = -3 or y = 4 4/5 (4.8)

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2(x-3)=5(x-3)+10<br><br><br> Need urgent help

ArbitrLikvidat [17]

Step-by-step explanation:

2(x-3)=5(x-3)+10

=> 2x - 6 = 5x - 15 + 10

=> -6 + 15 -10 = 5x - 2x

=> 5x -2x = 15 - 6 - 10

=> 3x = 15 - 16

=> 3x = -1

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2 years ago

Given that p=9i+12j and q=-6i-8j. Evaluate |p-q|-{|p|-|q|}

krok68 [10]

Answer:

$|p-q|-(|p|-|q|) = 20$

Step-by-step explanation:

First let's find the value of 'p-q':

$p - q = 9i + 12j - (-6i - 8j)\\p - q = 9i + 12j + 6i + 8j\\p - q = 15i + 20j\\$

To find |p-q| (module of 'p-q'), we can use the formula:

$|ai + bj| = \sqrt{a^{2}+b^{2}}$

Where 'a' is the coefficient of 'i' and 'b' is the coefficient of 'j'

So we have:

$|p - q| = |15i + 20j| = \sqrt{15^{2}+20^{2}} = 25$

Now, we need to find the module of p and the module of q: $|p| = |9i + 12j| = \sqrt{9^{2}+12^{2}} = 15$

$|q| = |-6i - 8j| = \sqrt{(-6)^{2}+(-8)^{2}} = 10$

Then, evaluating |p-q|-{|p|-|q|}, we have:

$|p-q|-(|p|-|q|) = 25 - (15 - 10) = 25 - 5 = 20$

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2 years ago

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OLEGan [10]

Answer:

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Step-by-step explanation:

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2 years ago

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pashok25 [27]

Answer:

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3 years ago

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OLEGan [10]

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