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3 years ago

Solve. x 2 −5x−8=0 x2−5x−8=0 Enter your answers, as exact values, in the boxes. x = or x =

Mathematics

2 answers:

BARSIC [14]3 years ago

6 0

Answer:

Solve.

x2−5x−8=0

Step-by-step explanation:

Sidana [21]3 years ago

5 0

Answer:

x2-5x-8=0

Step-by-step explanation:

Two solutions were found :

x =(5-√57)/2=-1.275

x =(5+√57)/2= 6.275

Reformatting the input :

Changes made to your input should not affect the solution:

(1): "x2" was replaced by "x^2".

Step by step solution :

Step 1 :

Trying to factor by splitting the middle term

1.1 Factoring x2-5x-8

The first term is, x2 its coefficient is 1 .

The middle term is, -5x its coefficient is -5 .

The last term, "the constant", is -8

Step-1 : Multiply the coefficient of the first term by the constant 1 • -8 = -8

Step-2 : Find two factors of -8 whose sum equals the coefficient of the middle term, which is -5 .

-8 + 1 = -7

-4+ 2 = -2

-2+ 4 = 2

-1 + 8 = 7

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Add: (-3p6 + 5) + 9p6

timama [110]

Answer:

6p^6+ 5

Step-by-step explanation:

Step by step solution :

Step 1 :

Equation at the end of step 1 :

((0 - (3 • (p6))) + 5) + 32p6

Step 2 :

Equation at the end of step 2 :

((0 - 3p6) + 5) + 32p6

Step 3 :

Trying to factor as a Sum of Cubes :

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4 0

3 years ago

A scientist claims that 7% 7 % of viruses are airborne. If the scientist is accurate, what is the probability that the proportio

Reptile [31]

Answer:

The probability is $P(|p-\^{p}| > 0.03) = 0.0040$

Step-by-step explanation:

From the question we are told that

The population proportion is $p = 0.07$

The mean of the sampling distribution is $\mu_p = 0.07$

The sample size is n = 600

Generally the standard deviation is mathematically represented as

$\sigma_p = \sqrt{\frac{p (1 -p)}{n} }$

=> $\sigma_p = \sqrt{\frac{0.07(1 -0.07)}{600} }$

=> $\sigma_p = 0.010416$

Generally the probability that the proportion of airborne viruses in a sample of 600 viruses would differ from the population proportion by greater than 3% is mathematically represented as

$P(|p-\^{p}| > 0.03) = 1 - P(|p -\^{p}| \le 0.03)$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-0.03 \le p -\^{p} \le 0.03 )$

Now add p to both side of the inequality

=> $P(|p-\^{p}| > 0.03) = 1 - P( 0.07-0.03 \le \^{p} \le 0.03+ 0.07 )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(0.04 \le \^{p} \le 0.10 )$

Now converting the probabilities to their respective standardized score

=> $P(|p-\^{p}| > 0.03) = 1 - P(\frac{0.04 - 0.07}{0.010416} \le Z \le \frac{0.10 -0.07}{0.010416} )$

=> $P(|p-\^{p}| > 0.03) = 1 - P(-2.88 \le Z \le 2.88 )$

=> $P(|p-\^{p}| > 0.03) = 1 - [P(Z \le 2.88) - P(Z \le -2.88)]$

From the z-table

$P(Z \le 2.88) = 0.9980$

and

$P(Z \le -2.88) = 0.0020$

$P(|p-\^{p}| > 0.03) = 1 - [0.9980 - 0.0020]$

=> $P(|p-\^{p}| > 0.03) = 0.0040$

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4 years ago