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const2013 [10]
3 years ago
6

Is x^2+13x-4=0 a quatratic function? Why?

Mathematics
1 answer:
zysi [14]3 years ago
8 0
Yes it is a quadratic function. A quadratic function contains the term x^2 and x^2 has the highest exponent in the whole function of a quadratic function. What I mean by this is for example x^3 + x^2 + 2x + 3 is not a quadratic. It includes x^2 but x^3 has the largest exponent. That makes that equation a cubic. Just for more info the largest exponent is referred to as the degree of the function so the degree of a quadratic function is always 2. Hope this helps!
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0.6 + 15b + 4 = 25.6 all equivelant
Mila [183]

Answer:

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

Step-by-step explanation:

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

0.6 + 15b + 4 = 25.6 all equivelant

3 0
3 years ago
Compare and Contrast how you would factor x^2−9 and x^2+9. For full credit, make sure you explain why they are factored differen
marishachu [46]

The factored form of the expression x^2 - 9 is (x+3)(x-3)

The expression x^2 + 9 has no real roots hence cannot be factored

Quadratic equations are equations with a degree of 2.

Given the quadratic function x^2-9 \ and \ x^2+9

For the function x^2 - 9

=x^2-9\\=x^2-3^2

Using the difference of two squares:

a^2-b^2=(a-b)(a+b)\\x^2-3^2=(x-3)(x+3)

For the expression x^2 + 9

The expression x^2 + 9 has no real roots hence cannot be factored

Learn more on factorization here: brainly.com/question/20293447

7 0
3 years ago
Prove that sinxtanx=1/cosx - cosx
maks197457 [2]

Answer:

See below

Step-by-step explanation:

We want to prove that

\sin(x)\tan(x) = \dfrac{1}{\cos(x)} - \cos(x), \forall x \in\mathbb{R}

Taking the RHS, note

\dfrac{1}{\cos(x)} - \cos(x) = \dfrac{1}{\cos(x)} - \dfrac{\cos(x) \cos(x)}{\cos(x)} = \dfrac{1-\cos^2(x)}{\cos(x)}

Remember that

\sin^2(x) + \cos^2(x) =1 \implies 1- \cos^2(x) =\sin^2(x)

Therefore,

\dfrac{1-\cos^2(x)}{\cos(x)} = \dfrac{\sin^2(x)}{\cos(x)} = \dfrac{\sin(x)\sin(x)}{\cos(x)}

Once

\dfrac{\sin(x)}{\cos(x)} = \tan(x)

Then,

\dfrac{\sin(x)\sin(x)}{\cos(x)} = \sin(x)\tan(x)

Hence, it is proved

5 0
3 years ago
A house that was originally listed for sale at $450,000 has decreased the asking price by 8%. By how much has the price been red
matrenka [14]
450000 * .08
= 36,000
They reduce $36,000 from the original price.
5 0
3 years ago
The graph represents this system of equations:
Kipish [7]

Answer:

(2.5, - 2)

Step-by-step explanation:

The solution to a system of equations given graphically is at the point of intersection of the 2 lines, that is

(2.5, - 2 ) ← point of intersection

3 0
3 years ago
Read 2 more answers
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