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3 years ago

Find three solutions for y=-2x-1

Mathematics

2 answers:

Pani-rosa [81]3 years ago

5 0

Answer:

(1, -3)

(2, -5)

(3, -7)

Step-by-step explanation:

Randomly choose three values for "x". Substitute each of them separately into the formula to find the "y". Then write the solution as an ordered pair (x, y).

I will choose x=1, x=2 and x=3.

When x=1:

y = -2x - 1

y = -2(1) - 1

y = -2 - 1

y = -3

(1, -3)

When x=2:

y = -2x - 1

y = -2(2) - 1

y = -4 - 1

y = -5

(2 , -5)

When x=3:

y = -2x - 1

y = -2(3) - 1

y = -6 - 1

y = -7

(3, -7)

Phoenix [80]3 years ago

3 0

(1,-3), (2,-5), and (3,-7)

An easy way to find this is set you a definite y value. So try -3 as the value for y. Solve for x as normal so -3=-2x-1. Add 1 on both sides to get -2=-2x. Divide on both sides by -2 and get 1=x. Since the slope is -2 you keep subtracting -2 from -3 and you’ll get -5 when x=2 and so on.

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2 years ago

A stack of logs has 32 logs on the bottom layer. Each subsequent layer has 6 fewer logs than the previous layer. If the top laye

Svetlanka [38]

Answer: The total number of logs in the pile is 6.

Step-by-step explanation: Given that a stack of logs has 32 logs on the bottom layer. Each subsequent layer has 6 fewer logs than the previous layer and the top layer has two logs.

We are to find the total number of logs in the pile.

Let n represents the total number of logs in the pile.

Since each subsequent layer has 6 fewer logs then the previous layer, so the number of logs in each layer will become an ARITHMETIC sequence with

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We know that

the n-th term of an arithmetic sequence with first term a and common difference d is

$a_n=a+(n-10)d.$

Since there are n logs in the pile, so we get

$a+(n-1)d=2\\\\\Rightarrow 32+(n-1)(-6)=2\\\\\Rightarrow 32-6n+6=2\\\\\Rightarrow 6n=36\\\\\Rightarrow n=6.$

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6 0

3 years ago

Can someone please help with this math question:)

Bingel [31]

Solution:

The rationalisation factor for $\frac{1}{ a - \sqrt{b} }$ is $a + \sqrt{b}$

So, let us apply it here.

$\frac{1}{5 - \sqrt{2} }$

The rationalising factor for 5 - √2 is 5 + √2.

Therefore, multiplying and dividing by 5 + √2, we have

$= \frac{1}{5 - \sqrt{2} } \times \frac{5 + \sqrt{2} }{5 + \sqrt{2} } \\ = \frac{5 + \sqrt{2} }{(5 - \sqrt{2})(5 + \sqrt{2} ) } \\ = \frac{5 + \sqrt{2} }{ {(5)}^{2} - ( \sqrt{2})^{2} } \\ = \frac{5 + \sqrt{2} }{25 - 2} \\ = \frac{5 + \sqrt{2} }{23}$