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myrzilka [38]
3 years ago
15

PLEASE HELP ME !!!!!!!!!!! Will mark Brianliest correct answer !!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
xxMikexx [17]3 years ago
3 0

Answer:

9 inches

Step-by-step explanation:

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3^-6x(3^3 divide 3^0)^2
Greeley [361]

Answer:

{3}^{2}

Step-by-step explanation:

3^{ - 6}  \times (3^4  \div  3^0)^2

\frac{1}{  {3}^{6} }  \times ( {3}^{4}  \div  {3}^{0}  {)}^{2}

\frac{1}{729}  \times ( {3}^{4}  \div  {3}^{0}  {)}^{2}

\frac{1}{729}  \times ( {3}^{4}  {)}^{2}

\frac{1}{729}  \times 6561

\frac{1}{729}  \times (729(9))

9

{3}^{2}

<h3>Hope it is helpful...</h3>
8 0
3 years ago
3+3+3<br> Please Help....
NeX [460]

Answer:

9

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
HELP PLEASE NEED AN ANSWER
Reptile [31]
B. y= -1/2x + 2

Hope that helped
5 0
4 years ago
Read 2 more answers
Which statement describes the inverse of m(x) = x2 – 17x?
stealth61 [152]

Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

y = x² - 17·x

Therefore;

0 = x² - 17·x - y

From the general solution of a quadratic equation, 0 = a·x² + b·x + c we have;

x = \dfrac{-b\pm \sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

By comparison to the equation,0 = x² - 17·x - y, we have;

a = 1, b = -17, and c = -y

Substituting the values of a, b and c into the formula for the general solution of a quadratic equation, we have;

x = \dfrac{-(-17)\pm \sqrt{(-17)^{2}-4\times (1) \times (-y)}}{2\times (1)} = \dfrac{17\pm \sqrt{289+4\cdot y}}{2}

Which can be simplified as follows;

x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

And further simplified as follows;

x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

We note that the maximum or minimum point of the function, m(x) = x² - 17·x found by differentiating the function and equating the result to zero, gives;

m'(x) = 2·x - 17 = 0

x = 17/2

Similarly, the second derivative is taken to determine if the given point is a maximum or minimum point as follows;

m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

8 0
3 years ago
Ian invested$1500 dollars in the bank. After 2 years he withdrew all his money. He received $1590. What interest rate was he get
dybincka [34]

Answer:

either 45$ a year or divide it to days/months

Step-by-step explanation:

6 0
3 years ago
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