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3 years ago

6

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

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