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baherus [9]
3 years ago
9

Determine whether the graph table or equation represents a linear nonlinear function.

Mathematics
1 answer:
Fofino [41]3 years ago
3 0

Answer:

nonlinear

Step-by-step explanation:

x(x + 3) =  {x}^{2}  + 3  \\ \: it \: will \: be \: a \: parabola \: because \: of \: the \:  {x}^{2}

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According to data, as education level increases the unemployment rate decreases. This is an example of
Crank
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2 years ago
Need Help Badly<br>What is the solution
Kamila [148]
Sqrt(1-3x)=x+3
[sqrt(1-3x)]^2=(x+3)^2
1-3x=(x+3)(x+3)
1-3x=x^2+6x+9
-3x=x^2+6x+8
0=x^2+9x+8

The answers are -1 and -8 BUT, we have to plug them back into the original equation to make sure we don't get a negative under the square root sign.

After doing this, we realize that only -1 works, so the answer is x=-1

Sorry that this took forever to answer. I was thinking of a good way to explain this, and if you need any further explanation, message me:)

Best wishes:)


5 0
2 years ago
Look at pic 10 pts will mark brainilest
MissTica

Answer:

The answer is B or 8

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
The coordinates of the endpoints of AB and CD are A(2,
Ratling [72]

Answer:

Option 1: CD is a perpendicular bisector of AB

Step-by-step explanation:

Let us find out the slopes of various line segments and the Distances and then we will draw the conclusions accordingly.

Formula to find slope

m= \frac{y_2-y_1}{x_2-x_1}

Formula to Find Distance between two points

D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

mAB ( represents , Slope of AB )

1. mAC= \frac{3-2}{2-5}=\frac{1}{-3}=-\frac{1}{3}

2. mBC=\frac{2-1}{5-8}=\frac{1}{-3}=-\frac{1}{3}

3. mCD=\frac{5-2}{6-5}=\frac{3}{1}=3

4. AC=\sqrt{(3-2)^2+(2-5)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

5. BC=\sqrt{(2-1)^2+(5-8)^2} =\sqrt{(1)^2+(-3)^2}=\sqrt{1+9}=\sqrt{10}

mAC = mBC  , and C is common point , hence these three are collinear points  making a straight line whole slope is -\frac{1}{3}

mAB=-\frac{1}{3}

mCD=3

mAB \times mCD = -\frac{1}{3} \times 3 = -1

Hence CD ⊥ AB

Also

From Point 4 and point 5 above , we see that

AC = CB

Hence CD bisect AB at C, also CD ⊥ AB

There fore

CD is a perpendicular bisector of AB

Therefor option 1 is true

4 0
3 years ago
Can an expert help with this
SVEN [57.7K]

The formula of an area of a circle:

A_O=\pi r^2

We have

r = 1.8 m

π ≈ 3.14

Substitute:

A_O=\pi\cdot1.8^2=3.24\pi\ m^2\approx3.24\cdot3.14=10.1736\ m^2\approx10.17\ m^2

7 0
2 years ago
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