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3 years ago

Prove the sum of two rational numbers is rational where a, b, c, and d are integers and b and d cannot be zero.

Mathematics

1 answer:

uranmaximum [27]3 years ago

6 0

Step-by-step explanation:

just look what happened :

every term was multiplied by a factor with the value 1 (to keep the original value) but to establish a common denominator.

a/b was multiplied by d/d, and c/d was multiplied by b/b, so that the common denominator is then bd.

so, the first answer option is correct.

although I am a bit unhappy with its phrasing.

but the other options are definitely wrong.

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Is (x-4) a factor of f(x) = x3 - 2x2 + 5x + 1? Use either the remainder theorem or the factor theorem to explain your reasoning.

Virty [35]

Answer:

(x-4) is not a factor of f(x)=x³-2x²+5x+1

Step-by-step explanation:

Uding the remainder theorem,(x-4) is a factor if the remainder is 0

Plug in x=4

(4)³-2(4)²+5(4)+1

64-32+20+1

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3 years ago

What does z equal? <br><br> Equation: 9 (z - 82) = 90

matrenka [14]

Answer:

Z=92

Step-by-step explanation:

Distribute, 9z-738=90

Add 738, 9z=828

Divide, z=92

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3 years ago

Read 2 more answers

Someone please be awesome and help me please :(

solong [7]

Answer:

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Step-by-step explanation:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.

$x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

Now we are read to write that one part (the first three terms together) as a square:

$(x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2=0$

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.

$(x+\frac{b}{2a})^2+\frac{4ac}{4a^2}-\frac{b^2}{4a^2}=0$

They put it in ( )

$(x+\frac{b}{2a})^2+(\frac{4ac}{4a^2}-\frac{b^2}{4a^2})=0$

I'm going to go ahead and combine those fractions now:

$(x+\frac{b}{2a})^2+(\frac{-b^2+4ac}{4a^2})=0$

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):

$(x+\frac{b}{2a})^2-(\frac{b^2-4ac}{4a^2})=0$

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:

$(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}$

I'm going to square root both sides to rid of the square on the x+b/(2a) part:

$x+\frac{b}{2a}=\pm \sqrt{\frac{b^2-4ac}{4a^2}}$

$x+\frac{b}{2a}=\pm \frac{\sqrt{b^2-4ac}}{2a}$

Now subtract b/(2a) on both sides:

$x=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$

Combine the fractions (they have the same denominator):

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

6 0

3 years ago

What's the prime factorization of 78

Masja [62]

Answer:

2,3 &13

Step-by-step explanation:

78=2times39=3times13

therefore the prime factors of 78=2,3,13

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3 years ago

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Murrr4er [49]

Answer:

Step-by-step explanation:x

−

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3 years ago