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Masteriza [31]
3 years ago
11

Factorise this equation

Mathematics
1 answer:
pashok25 [27]3 years ago
8 0

x^{2}  - 2x +35

To factor we must first find <u>factors</u> of 35

The only factors of 35 are :<em> 5 & 7 </em>

-2x is what 5 and 7 make

but only if 7 is <u>negative </u>and 5 is <u>positive </u>

so our final factors are

(x - 7) (x  + 5)

______________________________________________

2x^{2} +7x +6

For the second equation, were going to multiple the coefficient of x^{2} ( which is 2 ) by 6

We get 12

What factors of 12 add up to 7?

<em>3 and 4 </em>

So our final factors are

(2x+3)(x+2)

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Step-by-step explanation:

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Javier had $305 in his bank account. His bank charges a fee of $7.50 each month that a balance is below $500. If he makes no oth
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Evaluate the following expression 4×[1+(19+26)÷9]
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4 0
4 years ago
Read 2 more answers
Show that the relation R consisting of all pairs(x, y)such that x and y are bit strings of length three or more that agree in th
Mice21 [21]

Answer:

See proof below

Step-by-step explanation:

An equivalence relation R satisfies

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  • Symmetry: For all x,y, if xRy then yRx.
  • Transitivity: For all x,y,z, If xRy and yRz then xRz.

Let's check these properties: Let x,y,z be bit strings of length three or more

The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.

If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.

If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.

5 0
3 years ago
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