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3 years ago

Select the two binomials that are factors of this trinomial.

Mathematics

2 answers:

lana [24]3 years ago

8 0

for $x^2-x-30$

find what 2 numbers multiply to get -30 and add to get -1

-6 and 5

factors to (x-6)(x+5)

so answers are A and C

const2013 [10]3 years ago

5 0

Answer: A & C

Step-by-step explanation:

x² - x - 30

∧

1 -30 = -29

2 -15 = -13

3 -10 = -7

5 -6 = -1 This is equal to the b-value of -1 so THIS WORKS!

(x + 5) (x - 6)

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What are the solutions of x^2-2x+5=0

sweet [91]

The solution of $x^{2}-2 x+5=0$ are 1 + 2i and 1 – 2i

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Given, equation is $x^{2}-2 x+5=0$

We have to find the roots of the given quadratic equation

Now, let us use the quadratic formula

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$ --- (1)

Let us determine the nature of roots:

Here in $x^{2}-2 x+5=0$ a = 1 ; b = -2 ; c = 5

$b^2 - 4ac = 2^2 - 4(1)(5) = 4 - 20 = -16$

Since $b^2 - 4ac < 0$ , the roots obtained will be complex conjugates.

Now plug in values in eqn 1, we get,

$x=\frac{-(-2) \pm \sqrt{(-2)^{2}-4 \times 1 \times 5}}{2 \times 1}$

On solving we get,

$x=\frac{2 \pm \sqrt{4-20}}{2}$

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$\begin{array}{l}{\mathrm{x}=\frac{2 \pm 4 i}{2}} \\\\ {\mathrm{x}=1 \pm 2 i}\end{array}$

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3 years ago

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elena55 [62]

<h2>Answer with explanation:</h2>

Let $\mu$ be the average life of light bulbs.

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Null hypothesis : $H_0 : \mu =5400$

Alternative hypothesis : $H_a : \mu >5400$

Since $H_a$ is right-tailed and population standard deviation is also known, so we perform right-tailed z-test.

Formula for Test statistic : $z=\dfrac{\overlien{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

where, n= sample size

$\overline{x}$ = sample mean

$\mu$ = Population mean

$\sigma$ =population standard deviation

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$z=\dfrac{5483-5400}{\dfrac{500}{\sqrt{95}}}=1.61796786124\approx1.6180$

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[We reject the null hypothesis when critical value is less than the test statistic value .]

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