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2 years ago

The "solutions" to a Quadratic Equation are where it is equal to zero. They are also

Mathematics

1 answer:

Cerrena [4.2K]2 years ago

5 0

Answer:

One of them is the root

im not sure about the other one: I'd just call it the value of x

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Which of the following functions represents an exponential decay with a 15% rate of change where A represents the initial value

melisa1 [442]

The function that represent this exponential decay is given as y = A(0.85)ˣ

<h3>What is an exponential function?</h3>

An exponential function is in the form:

y = abˣ

Where a is the initial value and b is the multiplication factor.

Given that A is the initial value, there is a decay of 15%, hence:

b = 100% - 15% = 0.85

The function that represent this exponential decay is given as y = A(0.85)ˣ

Find out more on exponential function at: brainly.com/question/12940982

3 0

1 year ago

Find an equation for the line that passes through the points (-5,-5) and (1,4)

Andreas93 [3]

The general line through (a,b) and (c,d) is

(c-a)(y-b)=(d-b)(x-a)

Here that's

(1 - - 5)(y - - 5) = (4 - -5)(x - - 5)

6(y+5)=9(x+5)

6y + 30 = 9x + 45

-15 = 9x - 6y

3x - 2y = -5

Answer: 3x - 2y = -5

We could put that in slope intercept form,

2y = 3x + 5

y = (3/2) x + 5/2

Answer: y = (3/2) x + 5/2

3 0

2 years ago

What is the value of (14)5?

valina [46]

Answer:

Step-by-step explanation:

7 0

1 year ago

Simplify: cos2x-cos4 all over sin2x + sin 4x

GrogVix [38]

Answer:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

Step-by-step explanation:

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}$

Apply formula:

$\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$ and

$\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)$

We get:

$=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}$

$=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}$

$=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}$

$=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}$

$=\frac{\sin\left(x\right)}{\cos\left(x\right)}$

$=\tan\left(x\right)$

Hence final answer is

$\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)$

6 0

2 years ago

Hey! i’ll give brainliest please help.

ruslelena [56]

Answer:

respect the traditions and beliefs

3 0

2 years ago

Read 2 more answers