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irina1246 [14]
3 years ago
7

What equation reoulte from completing the square and then factoring? x2 + 4x-5

Mathematics
1 answer:
tangare [24]3 years ago
3 0

Answer:

Completing the square: (x^2+4x)-5=0

Factored form: (x+5)(x-1)=0

Step-by-step explanation:

1.

Completing the square is a method used to rewrite a quadratic equation in vertex form. First one groups the linear and the quadratic terms. Then one will factor out the coefficient of the quadratic term. After doing so, one makes the grouped equation a perfect square trinomial by introducing a term, don't forget to balance the equation. Finally one simplifies the equation. The result is a quadratic equation in vertex form.

The quadratic equation:

x^2+4x-5=0

Group the linear and quadratic terms:

(x^2+4x)-5=0

The quadratic term doesn't have a coefficient, so one doesn't need to factor the group. Now, one has to add a term to make the group a perfect square trinomial. Remember to balance the equation:

(x^4+4x-4)-5-4=0

Simplify,

(x-2)^2-9=0

2.

Factoring a quadratic equation is a method of rewriting a quadratic equation as the product of two linear equations. One splits the constant term up into factors, such that the sum of the factors is equal to the coefficient of the linear term.

x^2+4x-5=0

Factor:

(x+5)(x-1)=0

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Divide:<br><br><br> 27÷77,868<br> Help
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8 0
3 years ago
if there are two containers of sugar solution, the first is 4% concentration and second is 8% concentration. how much of each sh
Elina [12.6K]

Answer:

<u>30 gallons solution 4% and 10 gallons solution 8%</u>

Step-by-step explanation:

<h3>x•4%+y•8%=40•5% </h3>

<em><u>4x+8y=40•5 and x+y=40</u></em>

x=40-y

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5 0
2 years ago
One hundred items are simultaneously put on a life test. Suppose the lifetimes
romanna [79]

Answer:

a) \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b) \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

Step-by-step explanation:

Given:

The lifetimes of the individual items are independent exponential random variables.

Mean = 200 hours.

Assume, Ti be the time between ( i-1 )st and the ith failures. Then, the T_{i} are independent with \mathrm{T}_{\mathrm{i}} being exponential with rate \frac{(101-i)}{200} .

Therefore,

a) E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]

=\sum_{i=1}^{5} \frac{200}{101-i}

\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}

b)

The variance is given by, \mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]

\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}

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3 years ago
Write the given expression as a single trigonometric function.
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Answer:

Just did it, had a good guess.

Step-by-step explanation:

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