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4 years ago

Two sides of a triangle have lengths 10 and 18. Which inequalities describe the values that possible lengths for the third side?

1 answer:

4 0

Well I don't know where inequalities would come in but the third length would be solved as follows.
a^2+b^2=c^2
10^2+18^2=424
$\sqrt{424}$

Answer; $2 \sqrt{106}$
I hope this is correct, its what I would do if I had this equation.

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matrenka [14]

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Step-by-step explanation:

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3 years ago

If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.

zmey [24]

Answer:

$a=3$ and $b=5$ .

Step-by-step explanation:

So I believe the problem is this:

$\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}$

where we are asked to find values for $a$ and $b$ such that the equation holds for any $x$ in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

$\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)$

$2(4x+3)=a(x+7)+b(x-3)$

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

$2(4\cdot -7+3)=a(-7+7)+b(-7-3)$

$2(-28+3)=a(0)+b(-10)$

$2(-25)=0-10b$

$-50=-10b$

Divide both sides by -10:

$\frac{-50}{-10}=b$

$5=b$

Now we have:

$2(4x+3)=a(x+7)+b(x-3)$ with $b=5$

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

$2(4\cdot 3+3)=a(3+7)+b(3-3)$

$2(12+3)=a(10)+b(0)$

$2(15)=10a+0$

$30=10a$

Divide both sides by 10:

$\frac{30}{10}=a$

$3=a$

So $a=3$ and $b=5$ .

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2 years ago

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