Mathematics, 14.07.2019 13:30 nrosario33

Urgent. i need with these questions. you don’t have to answer all of them, but it would be appreciated. you so much!

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Answer from: montgomerykarloxc24x

$x(x+1)(x+8)=x(x^2+9x+8)=x^3+9x^2+8x$

$gcf(x^3, 9x^2, 8x)=x$

each term has a common factor of x. it's 1x because gcf(8,9)=1.

g(x) is f(x) reflected in the x axis, y=0

g(x) = - f(x)

$g(x) = - 3^x$

choice b

3 is the base for exponentiation; each unit increase in x is a factor of 3 in y.

33.

$x^2 + 12x - 64$

we need two numbers which add to 12 and multiply to -64. that's 16 and -4.

$x^2+12x-64 = (x+16)(x-4)$

how do we know it's right? well we could multiply the binomials to check that we get the original quadratic. we could test the implied zeros of x=-16 and x=4, i.e. when x=4, y=4^2+12(4)-64=16+48-64=0, good, so (x-4) is a factor.

Answer from: Quest

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

$ab^2+ac^2=bc^2$

$ab^2+ac^2=x^2$ )

triangle abd and cba are similar,

$\frac{ab}{bc}=\frac{bd}{ab}$

$ab^2=bc\cdot bd$

$25=bc\cdot bd$ (2)

triangle cad and cba are similar,

$\frac{ac}{bc}=\frac{cd}{ac}$

$ac^2=bc\cdot dc$

$7\times 7=bc\cdot dc$

$49=bc\cdot dc$ )

using (1), (2) and (3), we get

$25+49=x^2$

$74=x^2$

hence proved.

therefore option 2 is correct.

Answer from: Quest

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v

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