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3 years ago

Solve for v 209=40-v

Mathematics

1 answer:

andrew-mc [135]3 years ago

6 0

Answer:

-169

Step-by-step explanation:

We want to isolate v.

Method 1: subtract 40 from both sides, obtaining 169 = -v. Then change the sign of both sides, arriving at v = -169.

Method 2: Add v to both sides. Then v + 209 = 40.

Now subtract 209 from both sides, obtaining -169.

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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]

Answer-

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

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

Solution-

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

The applying the distance formula,

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

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

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

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

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

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3 years ago

Order the numbers from greatest to least 68.02 , 68.2 , 68.022

DanielleElmas [232]

Answer:

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3 years ago

7/8 × 16. Use a model to solve.

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3 years ago

Read 2 more answers

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vodomira [7]

Given $f(x) = x^2 + 6x + 9, g(x) = x^2 - 9$ :

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$(f-g)(x) = (x^2 + 6x+9)-(x^2-9) =6x+18=6(x+3)$ , which is the first second expression

$(fxg)(x) = (x^2+6x+9)(x^2-9) = (x^4-9x^2)+(6x^3-54x)+(9x^2-81)=x^4+6x^3-54x-81$ , which is the fourth expression

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3 years ago