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3 years ago

What is the equation of the coordinate A and B if they are (4,7), (6,3) respectively. What is the equation of the straight line.

Mathematics

1 answer:

Nesterboy [21]3 years ago

3 0

Answer: y+2x=15

Step-by-step explanation:

Given

Two points A and B are (4,7) and (6,3) respectively

Using two-point form

$\Rightarrow \dfrac{y-y_1}{x-x_1}=\dfrac{y_2-y_1}{x_2-x_1}$

Putting values

$\Rightarrow \dfrac{y-7}{x-4}=\dfrac{3-7}{6-4}\\\\\Rightarrow \dfrac{y-7}{x-4}=\dfrac{-4}{2}=-2\\\\\Rightarrow y-7=-2x+8\\\\\Rightarrow y+2x=15$

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Answer:

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(2) FALSE.

(3) FALSE.

(4) TRUE.

(5) FALSE.

Step-by-step explanation:

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$2^{\frac{5}{2} } = (\sqrt{2} )^5 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 4\sqrt{2}\\\\\sqrt{32} = \sqrt{16 \ \times \ 2}\ = \ \sqrt{16} \ \times \ \sqrt{2} \ = \ 4\sqrt{2}$

Thus, the equation is TRUE.

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(3) $4^{\frac{1}{2} } = \sqrt[4]{64}$

$4^{\frac{1}{2} }= \sqrt{4} = 2\\\\\sqrt[4]{64} = (64)^{\frac{1}{4} } = (2^6)^{\frac{1}{4} }= 2^{\frac{6}{4} } = 2^{\frac{3}{2} }=(\sqrt{2} )^3 = (\sqrt{2} \times \sqrt{2} \times \sqrt{2} ) = 2\sqrt{2}$

Thus, the equation is FALSE.

(4) $2^8 = (\sqrt[3]{16} )^6$

$2^8 = 256\\\\ (\sqrt[3]{16} )^6 = (16)^{\frac{6}{3} } = (2^4)^{\frac{6}{3} } = (2)^{\frac{24}{3} } = 2^8 = 256$

Thus, the equation is TRUE.

(5) $(\sqrt{64} )^{\frac{1}{3} } = 8^{\frac{1}{6} }\\\\$

$8^{\frac{1}{6} } = (2^3)^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} } = \sqrt{2} \\\\(\sqrt{64} )^{\frac{1}{3} } = (2^6)^{\frac{1}{3} } = 2^{\frac{6}{3} } = 2^2 = 4$

Thus, the equation is FALSE.

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