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2 years ago

If the sum of squares roots of the equation x2+ x -k=0 is 10

Mathematics

1 answer:

OlgaM077 [116]2 years ago

8 0

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Help I forgot how to do this

lakkis [162]

Answer:

number 5 is 11 and number 6 is 16

Step-by-step explanation:

you substitute a for 4 bc a equals 4 and than finish the equation 4+7=11

6. you substitute b for 2 and c for 8 and multiply which than is 16

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3 years ago

Find an equation in rectangular coordinates for the spherical equation ρ = 7 csc ( ϕ ) csc ( θ )

irinina [24]

$\rho=7\csc\phi\csc\theta\implies\rho\sin\phi\sin\theta=7\iff y=7$

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3 years ago

At the end of the summer, liam discovers that his radiator antifreeze solution has dropped below the safe level. If the radiator

attashe74 [19]

2 gallons 25% of 4 is 1 so there is about 1 gallon of pure in there no so adding two will make 3 gallons of antifreeze bringing total to 6 gallons 3/6 = .5 boom bam 50%

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3 years ago

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

Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
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.

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3 years ago

Syd chooses two different primes, both of which are greater than $10,$ and multiplies them. The resulting product is less than $

Mars2501 [29]

Answer:

Step-by-step explanation:

There are 7 primes between 10 and 35: 11, 13, 17, 19, 23, 29, 31.

The product of 11 and any of the others will be less than 350, 6 products.

The product of 13 and any below 26 will be less than 350, 3 more products.

The product of 17 and any below 20 will be less than 350, 1 more product.

There are a total of 10 different products below 350 possible.

_____

11·13, 11·17, 11·19, 11·23, 11·29, 11·31, 13·17, 13·19, 13·23, 17·19

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3 years ago