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3 years ago

Show that one zero of 8x230x27 is the square of the other.

Mathematics

1 answer:

murzikaleks [220]3 years ago

7 0

Answer:

8x2-30x+27 has one root (9/4) the square of the other (3/2)

Step-by-step explanation:

8x2+30x+27 = (2*x+3)*(4*x+9) => solution = {-3/2, -4/9}

One is not the square of the other because squares are positive.

8x2-30x+27 = (2*x-3)*(4*x-9) =>

solution = {3/2, 4/9} since (4/9 = (3/2)^2, therefore the preceding question is correct.

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While conducting an inventory in their bicycle shop, the owner noticed the number of bicycles is 2 fewer than 10 times the numbe

mr Goodwill [35]

Answer:

The number of bicycles in the store is $178\ bicycles$

Step-by-step explanation:

Let

x------> the number of bicycles

y-----> the number of tricycles

we know that

The bicycles has two wheels and the tricycles has three wheels

$2x+3y=410$ ------> equation A

$x=10y-2$ ------> equation B

substitute equation B in equation A and solve for y

$2(10y-2)+3y=410$

$20y-4+3y=410$

$23y=414$

$y=18\ tricycles$

Find the value of x

$x=10(18)-2=178\ bicycles$

3 0

3 years ago

Read 2 more answers

How to evaluate question number 4

ANTONII [103]

There are 2 ways that spring to mind.

One is to use the definition of the derivative at a point:

$f'(c)=\displaystyle\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$

In this case, $c=0$ and $f(x)=\sqrt{5-x}$ . Then

$f'(x)=-\dfrac1{2\sqrt{5-x}}\implies f'(0)=\displaystyle\lim_{x\to0}\frac{\sqrt{5-x}-\sqrt5}x=-\dfrac1{2\sqrt5}$

- - -

If you don't know about derivative yet (I would think you do, considering this is for a midterm in AP Calc, but I digress), the other way is to rely on algebraic manipulation. Multiply the numerator and denominator by the conjugate of the numerator to get

$\dfrac{\sqrt{5-x}-\sqrt5}x\cdot\dfrac{\sqrt{5-x}+\sqrt5}{\sqrt{5-x}+\sqrt5}=\dfrac{\left(\sqrt{5-x}\right)^2-\left(\sqrt5\right)^2}{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac x{x\left(\sqrt{5-x}+\sqrt5\right)}=-\dfrac1{\sqrt{5-x}+\sqrt5}$