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3 years ago

14

The difference of two trinomials is x2 - 10x + 2. If one of the trinomials is 3x? - 11x - 4, then which expression could be the

othe
trinomial?
0 27-x-2
2x2 + x + 6
4x2 + 21x + 6
0
4x2
21x-2

1 answer:

klio [65]3 years ago

4 0

Answer:

$2x^2-x-6$

or

$4x^2-21x-2$

Step-by-step explanation:

The difference of two trinomials $P(x)$ and $Q(x)$ is $x^2 - 10x + 2.$ This means that

$P(x)-Q(x)=x^2-10x+2$

or

$Q(x)-P(x)=x^2-10x+2$

If one of the trinomials is $P(x)=3x^2-11x - 4,$ then

$(3x^2-11x-4)-Q(x)=x^2-10x+2$

or

$Q(x)-(3x^2-11x-4)=x^2-10x+2$

1. If $(3x^2-11x-4)-Q(x)=x^2-10x+2,$ then

$Q(x)=(3x^2-11x-4)-(x^2-10x+2)\\ \\Q(x)=(3x^2-x^2)+(-11x+10x)+(-4-2)\\ \\Q(x)=2x^2-x-6$

2. If $Q(x)-(3x^2-11x-4)=x^2-10x+2,$ then

$Q(x)=(x^2-10x+2)+(3x^2-11x-4)\\ \\Q(x)=(x^2+3x^2)+(-10x-11x)+(2-4)\\ \\Q(x)=4x^2-21x-2$

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Option C

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<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$

Substituting we get,

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{2}{3}\right)^{2} \times\left(\frac{2}{3}\right)$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}$

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{8}{27}$

Hence, the ratio of volume of two cylinders is 8 : 27

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