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Norma-Jean [14]
3 years ago
9

Given ​ f(x)=x2+14x+40 ​. Enter the quadratic function in vertex form in the box.

Mathematics
1 answer:
sammy [17]3 years ago
8 0
Please indicate exponentiation using " ^ "   Thanks.

f(x)=x2+14x+40 => <span>f(x)=x^2+14x+40

Next, complete the square:

</span>f(x)=x^2+14x+ 49 - 49 +40 = (x+7)^2 - 9 

Write this in the form y = (x+7)^2 - 9  or y + 9 = (x+7)^2
Comparing this result to y = a(x-h)^2 + k, we see that h = -7 and k=-9

In vertex form, the equation is y + 9 = (x+7)^2 and the vertex is at (-7, -9).
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nalin [4]

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x = 4

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Step-by-step explanation:

6x + 3 = 7x - 1

7x - 6x = 3+ 1

6x + 3  + 7x + 1 = 56

x = 4

3 0
3 years ago
Please help they are separate questions please help me with this!
monitta

Answer: Mary makes $1160, Steven makes $142.50

Step-by-step explanation:

Facts

MARY

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8% of 14,500= 0.08x14,500= 1160

Therefore, Mary makes $1,160 in commision.

STEVEN

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6 0
2 years ago
What is the volume of this right rectangular prism. 7 feet length 2 feet width and 3.5 feet Height
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6 0
3 years ago
How do you find the x intercepts of 3x^(5/3) - 4x^(7/3)
kaheart [24]
Set the whole expression = to 0 and solve for x.

3x^(5/3) - 4x^(7/3) = 0.  Factor out x^(5/3):     x^(5/3) [3 - 4x^(2/3)] = 0

Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.

In the latter case, 4x^(2/3) = 3. 

To solve this:  mult. both sides by x^(-2/3).  Then we have

4x^(2/3)x^(-2/3) = 3x^(-2/3),            or 4 = 3x^(-2/3).  It'd be easier to work with this if we rewrote it as

4           3
---  = --------------------
1            x^(+2/3)

Then 

4
---  = x^(-2/3).  Then, x^(2/3) = (3/4), and x = (3/4)^(3/2).  According to my     3                      calculator, that comes out to x = 0.65 (approx.)

Check this result!  subst. 0.65 for x in the given equation.  Is the equation then true?

My method here was a bit roundabout, and longer than it should have been.  Can you think of a more elegant (and shorter) solution?
7 0
3 years ago
Read 2 more answers
Prove that the statement (ab)^n=a^n * b^n is true using mathematical induction.
mamaluj [8]

Answer:

see below

Step-by-step explanation:

      (ab)^n=a^n * b^n

We need to show that it is true for n=1

assuming that it is true for n = k;

(ab)^n=a^n * b^n

( ab) ^1 = a^1 * b^1

ab = a * b

ab = ab

Then we need to show that it is true for n = ( k+1)

or (ab)^(k+1)=a^( k+1) * b^( k+1)

Starting with

  (ab)^k=a^k * b^k    given

Multiply each side by ab

ab *  (ab)^k= ab *a^k * b^k

   ( ab) ^ ( k+1) = a^ ( k+1) b^ (k+1)

Therefore, the rule is true for every natural number n

5 0
3 years ago
Read 2 more answers
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