3 years ago

Solve the equation x^3 + 2x^2 - 11x -12 = 0

Mathematics

1 answer:

vagabundo [1.1K]3 years ago

4 0

Answer: there are 4 solutions

x = -2

x = -1/2 = -0.500

x =(3-√5)/2= 0.382

x =(3+√5)/2= 2.618

Step-by-step explanation:

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I need help is it A B C or D

sweet-ann [11.9K]

All you have to do is plug it into slope-intercept form, which is y = mx + b

y = (-2)x + 4
y = -2x + 4

the answer is A

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3 years ago

Prove that 3 power p + 1 - 3 power P upon 2 multiply 3 power p is equal to 1

Anit [1.1K]

Answer:

see explanation

Step-by-step explanation:

note that $3^{p+1}$ = $3^{p}$ × 3 ( when multiplying exponents are added )

Given

$\frac{3^{p+1}-3^{p} }{2.3^{p} }$ ← factor out $3^{p}$ on numerator

= $\frac{3^{p}(3-1) }{2.3^{p} }$

= $\frac{2.3^{p} }{2.3^{p} }$

= 1

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3 years ago

Shortly after their arrival, Europeans began introducing pigs to the Americas as a source of food. Some escaped, and others were

natka813 [3]

Answer: About 800 thousand

The more accurate value is 807,528 but this is also an approximation.

=======================================================

Work Shown:

$t = \text{number of years since the year 2000}$

$P_0 = \text{population (in millions) in the year 2000}$

$P = P_0(1.2)^t\\\\5 = P_0(1.2)^{10}\\\\5 \approx P_0*6.1917364224\\\\P_0 \approx \frac{5}{6.1917364224}\\\\P_0 \approx 0.80752791444922\\\\P_0 \approx 0.807528\\\\$

That's the rough population of wild pigs (in millions) for the year 2000.

Multiply by $10^6$ to get it in terms of units instead.

$0.807528\times10^6 = 807,528$

There were roughly 800 thousand wild pigs in the year 2000.

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2 years ago

Of the 15 cars in a parking lot, 5 are green, 2 are gold, and the rest are black. What are the odds against the nest car leaving

Degger [83]

8/15 or 0.5333 or 53.3333%

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3 years ago

Ally works in a shop for $18 per hour <br>

Naddika [18.5K]

Answer:

You have to finish the question lol

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers