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3 years ago

How to solve a system of equations

Mathematics

1 answer:

enot [183]3 years ago

6 0

30: r = -32
38: n < -24
54: ?

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Simora [160]

I’m not sure I’m going to look for it on the internet for you buddy because I don’t know yet

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3 years ago

Determine whether −2a3b3+15c38z3 is a monomial, binomial, trinomial, or other polynomial.

Flauer [41]

Monomial is the answer

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2 years ago

PLEASE HELP ME IM STUCK ON THIS QUESTION.

Vesna [10]

Answer:

You got it right. Triangle DAC

Step-by-step explanation:

Line segment DA is a perpendicular bisector(it cuts line segment CE perfectly in half making line segment AE an CA congruent. Since the triangle share line segment DA that makes them congruent by the SAS postulate

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3 years ago

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tiny-mole [99]

Step-by-step explanation:

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3 years ago

Find the equation of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit.

leonid [27]

<u>Answer-</u>

The equations of the locus of a point that moves so that its distance from the line 12x-5y-1=0 is always 1 unit are

$12x-5y+14=0 \\ 12x-5y-14=0$

<u>Solution-</u>

Let a point which is 1 unit away from the line 12x-5y-1=0 is (h, k)

The applying the distance formula,

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{12^2+5^2}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{\sqrt{169}} \right |=1$

$\Rightarrow \left | \frac{12h-5k-1}{13} \right |=1$

$\Rightarrow 12h-5k-1=\pm 13$

$\Rightarrow 12h-5k=\pm 14$

$\Rightarrow 12h-5k=14,\ 12h-5k=-14$

$\Rightarrow 12h-5k-14=0,\ 12h-5k+14=0$

$\Rightarrow 12x-5y-14=0,\ 12x-5y+14=0$

Two equations are formed because one will be upper from the the given line and other will be below it.

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3 years ago