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3 years ago

Please help

Mathematics

2 answers:

mrs_skeptik [129]3 years ago

8 0

I believe the correct amount difference is 23,640 more than the baby

dexar [7]3 years ago

3 0

23,740 more pounds youre welcom

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If M=(2,5) is the midpoint of AB, and A=(0,-1), then the coordinates of B are _____.

faust18 [17]

Answer:

The point M(Mx, My) is mid-point of AB with A(Ax, Ay) and B(Bx, By).

=> 2*Mx = Ax + Bx => 2*2 = 0 + Bx => Bx = 4

=> 2*My = Ay + By => 2*(5) = -1 + By => By = 11

=> B(4, 11)

=> Option A is correct

Hope this helps!

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myrzilka [38]

The answer would be 54 since there is only 180 degrees in a triangle so you subtract 90 and 36 from 180. You get 54

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3 years ago

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Graph the line that has a slope of -1/6 and includes the point (0, 5).

sleet_krkn [62]

Answer:

y = -1/6x + 5

Step-by-step explanation:

y = mx + c

Gradient (m) = -1/6

Y-intercept (c) = 5

y = -1/6x + 5

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1 year ago

Worth brainliest!!

Ber [7]

Answer:

In the first law, an object will not change its motion unless a force acts on it.

In the second law, the force on an object is equal to its mass times its acceleration.

In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

Step-by-step explanation:

Hope this helped!

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3 years ago

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Solve the equation x^6+3x^5+x^4-5x^3-6x^2-2x=0 in the real number system

koban [17]

First, pull out a factor of $x$ .

$x^6+3x^5+x^4-5x^3-6x^2-2x=x(x^5+3x^4+x^3-5x^2-6x-2)$

Notice that when $x=-1$ (which you can arrive at via the rational root theorem), you have

$(-1)^5+3(-1)^4+(-1)^3-5(-1)^2-6(-1)-2=-1+3-1-5+6-2=0$

which means you can pull out a factor of $x+1$ . Upon dividing you get

$\dfrac{x^5+3x^4+x^3-5x^2-6x-2}{x+1}=x^4+2x^3-x^2-4x-2$

The rational root theorem will come in handy again, suggesting that $x=-1$ appears a second time as a root, which means

$\dfrac{x^4+2x^3-x^2-4x-2}{x+1}=x^3+x^2-2x-2$

Now this is more readily factored without having to resort to the rational root theorem. You have

$x^3+x^2-2x-2=x^2(x+1)-2(x+1)=(x^2-2)(x+1)$

so in fact, $x=-1$ shows up as a root for a third time.

So, you have

$x^6+3x^5+x^4-5x^3-6x^2-2x=x(x+1)^3(x^2-2)=0$

Two roots are obvious, $x=0$ and $x=-1$ (with multiplicity 3). The remaining two are given by

$x^2-2=0\implies x^2=2\implies x=\pm\sqrt2$

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4 years ago