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3 years ago

If p(q) = q^2 + 4q − 12,then what is p(−1)?

2 answers:

8 0

Answer: p(-1) = -15

Step-by-step:

Plug -1 into the equation in place of q.

p(-1) = (-1)^2 + 4(-1) - 12

p(-1) = (-1)^2 - 4 - 12

p(-1) = (-1)^2 -16

p(-1) = 1 - 16

p(-1) = -15

Ganezh [65]3 years ago

3 0

1): "q2" was replaced by "q^2".

Step by step solution :

STEP
1
:
Trying to factor by splitting the middle term

1.1 Factoring q2-4q-12

The first term is, q2 its coefficient is 1 .
The middle term is, -4q its coefficient is -4 .
The last term, "the constant", is -12

Step-1 : Multiply the coefficient of the first term by the constant 1 • -12 = -12

Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is -4 .

-12 + 1 = -11
-6 + 2 = -4 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and 2
q2 - 6q + 2q - 12

Step-4 : Add up the first 2 terms, pulling out like factors :
q • (q-6)
Add up the last 2 terms, pulling out common factors :
2 • (q-6)
Step-5 : Add up the four terms of step 4 :
(q+2) • (q-6)
Which is the desired factorization

Equation at the end of step
1
:

(q + 2) • (q - 6) = 0
STEP
2
:
Theory - Roots of a product

2.1 A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation:

2.2 Solve : q+2 = 0

Subtract 2 from both sides of the equation :
q = -2

Solving a Single Variable Equation:

2.3 Solve : q-6 = 0

Add 6 to both sides of the equation :
q = 6

Supplement : Solving Quadratic Equation Directly

Solving q2-4q-12 = 0 directly

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The function h(t) = –16t2 + 96t + 6 represents an object projected into the air from a cannon. The maximum height reached by the

jolli1 [7]

If you've started pre-calculus, then you know that the derivative of h(t)
is zero where h(t) is maximum.

The derivative is      h'(t) = -32 t + 96 .

At the maximum ... h'(t) = 0

   32 t = 96 sec

   t =  3 sec .
___________________________________________

If you haven't had any calculus yet, then you don't know how to
take a derivative, and you don't know what it's good for anyway.

In that case, the question GIVES you the maximum height.
Just write it in place of h(t), then solve the quadratic equation
and find out what 't' must be at that height.

   150 ft = -16 t² + 96 t + 6

Subtract 150ft from each side: -16t² + 96t - 144 = 0 .

Before you attack that, you can divide each side by -16,
making it a lot easier to handle:

     t² - 6t + 9 = 0

I'm sure you can run with that equation now and solve it.
The solution is the time after launch when the object reaches 150 ft.
It's 3 seconds.
(Funny how the two widely different methods lead to the same answer.)

The answer is from AL2006

6 0

4 years ago

Patulong naman sa business math HAHHA

igor_vitrenko [27]

English mam ahahahhahahahahahha

5 0

4 years ago

Read 2 more answers

The venticles of a right angle are a[0,4], b[3,-2] c[-3, -4] Find It's area.

givi [52]

Answer:

The area of this triangle is about 21.2132 square units.

Step-by-step explanation:

First, find the lengths of the legs AB and BC.

Length of AB ===

Find the difference in position vertically:

$-2-4=-6$

The points are 6 units apart vertically.

Find the difference in position horizontally:

$3-0=3$

The points are 3 units apart horizontally.

These lengths form a right triangle with the distance between the points as the hypotenuse, so you can use the pythagorean theorem to solve:

$a^2+b^2=c^2\\3^2+6^2=c^2\\9+36=c^2\\45=c^2\\c\approx6.7082$

AB is about 6.7082 units long.

Length of BC ===

Same process as above.

Find the vertical distance:

$-4--2=-2$

2 units apart vertically.

Find the horizontal distance:

$-3-3=-6$

6 units apart horizontally.

Use the pythagorean theorem:

$2^2+6^2=c^2\\4+36=c^2\\40=c^2\\c=6.3246$

BC is about 6.3246 units long.

Area ===

Finally, you can use these to find the area of the triangle. The area of a right triangle is just half the area of a rectangle with the same base and height:

$A=\frac{b\times h}{2}\\\\A=\frac{6.7082\times6.3246}{2}\\\\A=\frac{42.4264}{2}\\\\A=21.2132$