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Shalnov [3]
4 years ago
11

What is -4 = s-6/-4 equaled to

Mathematics
1 answer:
zmey [24]4 years ago
7 0

Move the negative sign to the left

-4 = s - (-6/4)

Simplify 6/4 to 3/2

-4 = s - (-3/2)

Simplify brackets

-4 = s + 3/2

Subtract 3/2 from both sides

-4 - 3/2 = s

Simplify -4 - 3/2 to -11/2

-11/2 = s

Switch sides

<u>s = -11/2</u>

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Select "Growth" or "Decay" to classify each function.
allochka39001 [22]

1. The growth rate equation has a general form of:

y = A (r)^t

The function is growth when r≥1, and it is a decay when r<1. Therefore:

y=200(0.5)^2t                    --> Decay

y=1/2(2.5)^t/6                  --> Growth

y=(0.65)^t/4                       --> Decay

 

2. We rewrite the given equation (1/3)^d−5 = 81

 

Take the log of both sides:

(d – 5) log(1/3) = log 81

d – 5 = log 81 / log(1/3)

d – 5 = - 4

 

Multiply both sides by negative 1:

- d + 5 = 4

So the answer is D

5 0
3 years ago
Simplify: 7^6 ÷ 7^2<br><br> A.) 7^3<br> B.) 7^4<br> C.) 7^8<br> D.) 7^12
juin [17]
7^6/7^2
(7*7*7*7*7*7)/(7*7)
the answer is b. 7^4
6 0
4 years ago
Read 2 more answers
Help please!! It would be nice if you could add an explanation, but you don't have to. I just need the answer.
Firlakuza [10]
The answer is B I solved this my self
4 0
3 years ago
Suppose m = 2 + 6i, and | m + n | = 3√10, where n is a complex number.
Ksju [112]

Answer: a) √50

b) n = 1 + 7i

Step-by-step explanation:

first, the modulus of a complex number z = a + bi is

IzI = √(a^2 + b^2)  

The fact that n is complex does not mean that n doesn't has a real part, so we must write our numbers as:

m = 2 + 6i

n = a  + bi

Im + nI = 3√10

Im + n I = √(a^2 + b^2 + 2^2 + 6^2)= 3√10

            = √(a^2 + b^2 + 40) = 3√10

             a^2 + b^2 + 40 = 3^2*10 = 9*10 = 90

             a^2 + b^2 = 90 - 40 = 50

            √(a^2 + b^2 ) = InI = √50

The modulus of n must be equal to the square root of 50.

now we can find any values a and b such a^2 + b^2 = 50.

for example, a = 1 and b = 7

1^2 + 7^2 = 1 + 49  = 50

Then a possible value for n is:

n = 1 + 7i

6 0
3 years ago
Factor completely. <br> <img src="https://tex.z-dn.net/?f=x%5E%7B8%7D-%5Cfrac%7B1%7D%7B81%7D" id="TexFormula1" title="x^{8}-\fra
Eduardwww [97]

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

6 0
3 years ago
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