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swat32
3 years ago
7

PLEASE SOLVE! Also how do you properly solve them ( please include steps)

Mathematics
2 answers:
Marizza181 [45]3 years ago
6 0
1) 2x+3 = 5
-3 -3
2x =2
x=1

2) -x + 8 = 5
-8 -8
-x = -3
X=3


Virty [35]3 years ago
4 0
Here is the answer I hate typing so I just wrote it and took the picture

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As x approaches 4, the value of 4x + 3 approaches<br> Gri
gogolik [260]

Answer:

19

Step-by-step explanation:

4 \times x + 3 = 4 \times 4 + 3 = 16 + 3 = 19

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Find the value of the trig function indicated
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Answer:

Is this the full question?

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I need help with #11 and #12 how do I do this?
USPshnik [31]

Answer:

  11. x = 6; y = 6.5

  12. x = 2; y = 4

Step-by-step explanation:

In each figure, the hash marks on the lines indicate the parallel lines (marked with red arrows) are equally-spaced. That means the two expressions involving x are equal to each other, and the two expressions involving y are equal to each other.

__

11. 2x +1 = x +7

  x = 6 . . . . . . . subtract (x+1) from both sides

and

  3y -8 = y +5

  2y = 13 . . . . add (8-y) to both sides

  y = 6.5 . . . . divide by the coefficient of y

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12. x +3 = 3/2x +2

  1 = 1/2x . . . . . subtract (x+2)

  2 = x . . . . . . . multiply by 2

and

  2y -1 = 3y -5

  4 = y . . . . . . add (5 -2y)

8 0
3 years ago
Read 2 more answers
A climatologist wants to determine the thickness of a glacier with a surface area of about 2 km². He selected 10 locations at ra
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Answer:

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Step-by-step explanation:

I want to hep but i think i got the wrong answer

8 0
3 years ago
Find the distance from the origin to the graph of 7x+9y+11=0
Cerrena [4.2K]
One way to do it is with calculus. The distance between any point (x,y)=\left(x,-\dfrac{7x+11}9\right) on the line to the origin is given by

d(x)=\sqrt{x^2+\left(-\dfrac{7x+11}9\right)^2}=\dfrac{\sqrt{130x^2+154x+121}}9

Now, both d(x) and d(x)^2 attain their respective extrema at the same critical points, so we can work with the latter and apply the derivative test to that.

d(x)^2=\dfrac{130x^2+154x+121}{81}\implies\dfrac{\mathrm dd(x)^2}{\mathrm dx}=\dfrac{260}{81}x+\dfrac{154}{81}

Solving for (d(x)^2)'=0, you find a critical point of x=-\dfrac{77}{130}.

Next, check the concavity of the squared distance to verify that a minimum occurs at this value. If the second derivative is positive, then the critical point is the site of a minimum.

You have

\dfrac{\mathrm d^2d(x)^2}{\mathrm dx^2}=\dfrac{260}{81}>0

so indeed, a minimum occurs at x=-\dfrac{77}{130}.

The minimum distance is then

d\left(-\dfrac{77}{130}\right)=\dfrac{11}{\sqrt{130}}
4 0
4 years ago
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