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Effectus [21]
3 years ago
14

Factor the trinomials 5x2+7x-6

Mathematics
1 answer:
Gnesinka [82]3 years ago
3 0
The answer is (5x - 3)(x + 2)
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Write an expression for 40% of 25
Montano1993 [528]

For this case we must find an expression to find 40% of 25. So, we make a rule of three:

25 -----------> 100%

x -------------> 40%

Where "x" represents the amount that determines 40% of 25.

x = \frac {40 * 25} {100}\\x = 10

40% of 25 is 10. And an expression to find it is:

x = \frac {40 * 25} {100}

Answer:

x = \frac {40 * 25} {100}


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3 years ago
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Plss help is this linear or non-linear??
skad [1K]

Answer:

linear

Step-by-step explanation:

The line is straight, it's not like going in any other direction at all or anything

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Question part points submissions used consider the following. g(x)={((x**2-a**2)/(x-a) text(if ) x != a,4 text(if ) x =
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For x\neq a, we have

\dfrac{x^2-a^2}{x-a}=\dfrac{(x-a)(x+a)}{x-a}=x+a

So for g(x) to be continuous at x=a, we require that the limit as x\to a is equal to 4.

\displaystyle\lim_{x\to a}g(x)=\lim_{x\to a}(x+a)=2a=4\implies a=2
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Find the least number which is exactly divisible by 15,25 and go without leaving a reminder.​
ladessa [460]
It would be 75, hope this helps!
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2 years ago
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Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If​ women's weights are
Mariulka [41]

Answer:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(173.6,49.8)  

Where \mu=173.6 and \sigma=49.8

We are interested on this probability

P(141.8

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(141.8

And we can find this probability with this difference and using the normal standard table:

P(-0.639

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0
3 years ago
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