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3 years ago

Solve the quadratic equation below by completing the square. What are the solutions?

Mathematics

1 answer:

Elden [556K]3 years ago

8 0

Answer:c

Step-by-step explanation:

Apex

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What is the value of y

8090 [49]

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3 years ago

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Forensic specialists can estimate the height of a deceased person from the lengths of the person's bones. These lengths are subs

Lera25 [3.4K]

Answer:

$130.2845\leq h\leq 137.7245$

Step-by-step explanation:

Given an inequality that relates the height h, in centimeters, of an adult female and the length f, in centimeters, of her femur by the equation

$|h - (2.47f + 54.10)| \leq 3.72$

If an adult female measures her femur as 32.25 centimeters, we can determine the possible range of her height by plugging f = 32.25cm into the modelled equation as shown:

$|h - (2.47(32.25) + 54.10)| \leq 3.72\\|h - (79.9045 + 54.10)| \leq 3.72\\|h - (134.0045)| \leq 3.72\\$

If the modulus function is positive then:

$h - 134.0045 \leq 3.72\\h \leq 3.71+134.0045\\h\leq 137.7245$

If the modulus function is negative then:

$-(h - 134.0045) \leq 3.72\\-h+134.0045 \leq 3.72\\-h\leq 3.72-134.0045\\-h\leq -130.2845\\$

multiply through by -1

$-(-h)\geq -(-130.2845)\\h\geq 130.2845\\130.2845\leq h$

combining the resulting inequalities, the estimate of the possible range of heights will be $130.2845\leq h\leq 137.7245$

8 0

3 years ago

When applying for a loan, you should _____.

ivanzaharov [21]

You should also have a little money saved aside so that you can pay them the money you saved and then work for the rest of the money you need to pay off

sorry if im incorrect

5 0

3 years ago

Mathematics-Session 1: No Calculator

IRINA_888 [86]

Answer:

Think it is D I am not sure

Step-by-step explanation:

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3 years ago

Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

6 0

3 years ago