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wlad13 [49]
3 years ago
7

The original equation is this:

Mathematics
1 answer:
likoan [24]3 years ago
3 0

Answer:

x=0, x=5

Step-by-step explanation:

x²-5x+8=8

x²-5x=0

x(x-5)=0

x=0 , x-5=0

x=5

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<img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%20x%2B2" id="TexFormula1" title="f(x)=\sqrt x+2" alt="f(x)=\sqrt x+2" align
Andrew [12]

Answer:

(x-2)^2

Step-by-step explanation:

To find the inverse of this function, you need to replace the location of the x and the y.

You start with y=\sqrt{x}+2. Now, replace the x and the y, and try to isolate the y.

x=\sqrt{y}+2\\x-2=\sqrt{y}\\y=(x-2)^2

Hope this helps!

8 0
3 years ago
Helpppppppppppp plzzzzzzzzzzzzzzzzzzz
ZanzabumX [31]

Answer:

Step-by-step explanation:

\left[\begin{array}{ccc}7&31\\142&19\end{array}\right] = 7(19) - 142(31) = <em>- 4269</em>

\left[\begin{array}{ccc}7&11&21\\55&-8&2\\-16&-1&-9\end{array}\right] = 7(- 8)(- 9) + 11(2)(- 16) + 21(55)(- 1) - 21(- 8)(- 16) - 7(2)(- 1) - 11(55)(- 9) = <em>1768</em>

\left[\begin{array}{ccc}9.07&6.02&2.01\\-30.7&2.5&3.5\\3.55&-1.1&2.35\end{array}\right] = 9.07(2.5)(2.35) + 6.02(3.5)(3.55) + 2.01(- 30.7)(- 1.1) - 2.01(2.5)(3.55) - 9.07(3.5)(- 1.1) - 6.02(- 30.7)(2.35) = <em>647.3561</em>

4 0
3 years ago
Simplify your answer<br> 8c^3d^2/<br> 4cd^2
forsale [732]

Answer:

the answer is 42

Step-by-step explanation:

6 0
4 years ago
The Holy Roman Empire was a group of regions in Central Europe that was held together by a political union. The Holy Roman Empir
VladimirAG [237]

Answer: x


Step-by-step explanation:

Given: The Holy Roman Empire was a group of regions in Central Europe that was held together by a political union. The Holy Roman Empire was formed in the year 962.

Let x represent any year.

Now, the inequality in terms of x and 962 that is true only for values of x that represent years before the year that the Holy Roman Empire formed is given by :-

x

7 0
4 years ago
Read 2 more answers
Which of the following equations is used to find the value of c (the distance to the foci) for an ellipse? 
Reika [66]

The equation of ellipse:

\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1

The formula of c (the distance to the foci)

c^2=a^2-b^2\qquad|\text{add}\ b^2\ \text{to the both sides}\\\\c^2+b^2=a^2

<h3>Answer: c² + b² = a²</h3>
4 0
3 years ago
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