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Brrunno [24]
3 years ago
11

Which function is shown in the graph below? On a coordinate plane, a function is shown. The curve starts in quadrant 4 and curve

s up to quadrant 1. It goes through (0.5, negative 0.4), (1, 0), and (6, 1). y = log Subscript one-sixth Baseline x y = log Subscript 0.5 Baseline x y = log Subscript 1 Baseline x y = log Subscript 6 Baseline x
Mathematics
2 answers:
klio [65]3 years ago
8 0

Answer:

c

Step-by-step explanation:

guajiro [1.7K]3 years ago
7 0

Answer:

3rd option on Edge

Step-by-step explanation:

Just did it and got it right

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35÷10=? in decimal form
spin [16.1K]

Answer:

3.5

Step-by-step explanation:

35 divided by 10 equals 3.5 .

4 0
3 years ago
Not sure what to do here can someone please help
kumpel [21]

Answer:

  x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

__

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

  \sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

  x = -1: √(-1+2) +1 = √(3(-1)+3)   ⇒   1+1 = 0 . . . . not true

  x = 2: √(2+2) +1 = √(3(2) +3)   ⇒   2 +1 = 3 . . . . true . . . x = 2 is the solution

__

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

  u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}

Solutions to this equation are ...

  u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

  x = u² -2 = 2² -2

  x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0
2 years ago
Is this a function, explain answer
likoan [24]

No. There are two spots in the vertical line. (Vertical line test)

5 0
3 years ago
Read 2 more answers
Kenny has a jar full of nickels and dimes. If he counted the money in his jar, how much does he have ?
shusha [124]

Answer:

Step-by-step explanation:

  it c

4 0
3 years ago
Read 2 more answers
Find the slope in the equation: y = 7x - 10 <br> 10<br> 7x<br> -10<br> 7
iogann1982 [59]

Answer:

7

Step-by-step explanation:

Remember this equation: y=mx+b where m=slope and b=y-intercept.

Therefore, the slope in the equation is 7.

4 0
3 years ago
Read 2 more answers
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