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katrin2010 [14]
2 years ago
10

PLEASE HELP I WILL GIVE BRAINLIEST I NEED THIS ASAP

Mathematics
1 answer:
Sophie [7]2 years ago
5 0

Answer:

I got this.

Step-by-step explanation:

A+B=C+D. SO YOU WANT TO DATE ME?

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Which ordered pair is not a solution to the equation y = 2x?
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Answer:

Step-by-step explanation:

Y=1/2y

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To solve the equation 12x = –72, you ?
IgorLugansk [536]

12x=-72 //:12

12x:12=-72:12

x=-6

6 0
3 years ago
Which set of data represents a linear function?
AVprozaik [17]
The answer is d because the x value does not repeat. Hopes this helps.
5 0
3 years ago
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BRAINLIESTTT ASAP! PLEASE HELP ME :)<br><br> Find the inverse of :
muminat

Answer:

h^{-1}(x) = \dfrac{5x - 6}{2}

Step-by-step explanation:

h(x)= \dfrac{2x + 6}{5}

(a) Rename h(x) as y  

y = \dfrac{ 2x+ 6}{5}

(b) Solve for x :  

\begin{array}{rcl}y& = &\dfrac{2x + 6}{5}\\\\5y & = & 2x + 6\\5y - 6 & = & 2x\\x & = & \mathbf{\dfrac{5y - 6}{2}}\end{array}

(d) Switch x and y  

y = \dfrac{5x -6}{2}

(e) Rename y as the inverse function  

h^{-1}(x) = \mathbf{\dfrac{5x - 6}{2}}

The graphs of inverse functions are reflections of each other across the line y = x.

In the diagram, the graph of h⁻¹(x) is the reflection of h(x) about the line y = x.

5 0
2 years ago
Read 2 more answers
the length of a rectangle is 8cm more than the width and its area is 172cm2 .find the width of the rectabgle
nikdorinn [45]
l-the\ length\\l+8-the\ width\\A=172\ cm^2\\\\A=lengtf\cdot width\\\\subtitute\\\\l(l+8)=172\\\\l^2+8l=172\ \ \ |subtract\ 172\ from\ both\ sides\\\\l^2+8l-172=0\\a=1;\ b=8;\ c=-172\\\\\Delta=b^2-4ac\\\\\Delta=8^2-4\cdot1\cdot(-172)=64+688=752 > 0\\\\l_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ l_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\\sqrt\Delta=\sqrt{752}=\sqrt{16\cdot47}=4\sqrt{47}\\\\l_1=\dfrac{-8-4\sqrt{47}}{2\cdot1}=-4-2\sqrt{47} < 0\\\\l_2=\dfrac{-8+4\sqrt{47}}{2\cdot1}=-4+2\sqrt{47} > 0\\\\The\ width:l+8=-4+2\sqrt{47}+8=4+2\sqrt{47}\\\\Answer:\boxed{length=(2\sqrt{47}-4)cm;\ width=(4+2\sqrt{47})cm}
length\approx9.71cm;\ width\approx17.71cm
4 0
3 years ago
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