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2 years ago

SOLVE ALL THESE EQUATION BY USING THE METHOD SUBSTITUTION ALGEBRAICALLY HERE'S ARE ALL THE EQUATION:

Mathematics

1 answer:

pav-90 [236]2 years ago

3 0

Answer:

See below ~

Step-by-step explanation:

Question 1

-7x - y = -1
y = 8x + 16

⇒ -7x - (8x + 16) = -1

⇒ -7x - 8x - 16 = -1

⇒ -15x = 15

⇒ x = -1

⇒ y = 8(-1) + 16 = 8

⇒ Solution = (-1, 8)

Question 2

3x + 4y = 0
y = -3x - 18

⇒ 3x + 4(-3x - 18) = 0

⇒ 3x - 12x - 72 = 0

⇒ -9x = 72

⇒ x = -8

⇒ y = -3(-8) - 18 = 6

⇒ Solution = (-8, 6)

Question 3 (not clear)

Question 4

-8x - 7y = 0
y = 6x

⇒ -8x - 7(6x) = 0

⇒ -8x - 42x = 0

⇒ -50x = 0

⇒ x = 0

⇒ y = 6(0) = 0

⇒ Solution = (0, 0)

Question 8

2x - 6y = -14
y = -5x - 19

⇒ 2x - 6(-5x - 19) = -14

⇒ 2x + 30x + 114 = -14

⇒ 32x = -128

⇒ x = -4

⇒ y = -5(-4) - 19 = 1

⇒ Solution = (-4, 1)

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Answer:

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Step-by-step explanation:

Step one:

given data

heigth/depth = 3.5m

radius= 9m

Step two:

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Hence the volume of the pool is

$V= \pi r^2h$

substitute

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Step-by-step explanation:

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Step-by-step explanation:

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Which equation represents a line perpendicular to the line shown on the graph?

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Answer:

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I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

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3 years ago