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postnew [5]
3 years ago
8

Select Is a Function or Is not a Function to correctly classify each relation. Title Is a Function Is not a Function {(2,2),(4,4

),(6,6),(8,8)} {(0,3),(3,5),(5,6),(8,4)} {(1,2),(3,3),(4,8),(6,3)} {(3,4),(5,2),(5,6),(7,3)}
Mathematics
2 answers:
maks197457 [2]3 years ago
6 0

Answer:

The answer is:

{(2,2),(4,4),(6,6),(8,8)} is a function

{(0,3),(3,5),(5,6),(8,4)} is a function

{(1,2),(3,3),(4,8),(6,3)} is a function

{(3,4),(5,2),(5,6),(7,3)} is not a function

Step-by-step explanation:

Sorry I don't have an explanation, but I know these are correct because I took the test and got this correct. ;)

Dvinal [7]3 years ago
3 0

Answer:

1 function {(2,2),(4,4),(6,6),(8,8)} written is a function the other following coordinates are not functions due to it not being a straight line as a 'proper' function should be.

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(4x^2 – 7x + 3) + (2x^2 – 2x – 5) i need this combined, im stuck
Gala2k [10]
6x^2 -9x -2 just add common variables
5 0
2 years ago
If f(x) = 2x ^ 2 - 5 and g(x) = x ^ 2 - 4x - 8 find (f - g)(x)
choli [55]

Answer:  (f-g)(x) is equivalent to f(x) - g(x)

So that would be (2x²-5) - (x²-4x-8) = 2x² - x² + 4x + 8 - 5 = x² + 4x + 3

Hope this helps.

5 0
2 years ago
1) Use Newton's method with the specified initial approximation x1 to find x3, the third approximation to the root of the given
neonofarm [45]

Answer:

Check below, please

Step-by-step explanation:

Hello!

1) In the Newton Method, we'll stop our approximations till the value gets repeated. Like this

x_{1}=2\\x_{2}=2-\frac{f(2)}{f'(2)}=2.5\\x_{3}=2.5-\frac{f(2.5)}{f'(2.5)}\approx 2.4166\\x_{4}=2.4166-\frac{f(2.4166)}{f'(2.4166)}\approx 2.41421\\x_{5}=2.41421-\frac{f(2.41421)}{f'(2.41421)}\approx \mathbf{2.41421}

2)  Looking at the graph, let's pick -1.2 and 3.2 as our approximations since it is a quadratic function. Passing through theses points -1.2 and 3.2 there are tangent lines that can be traced, which are the starting point to get to the roots.

We can rewrite it as: x^2-2x-4=0

x_{1}=-1.1\\x_{2}=-1.1-\frac{f(-1.1)}{f'(-1.1)}=-1.24047\\x_{3}=-1.24047-\frac{f(1.24047)}{f'(1.24047)}\approx -1.23607\\x_{4}=-1.23607-\frac{f(-1.23607)}{f'(-1.23607)}\approx -1.23606\\x_{5}=-1.23606-\frac{f(-1.23606)}{f'(-1.23606)}\approx \mathbf{-1.23606}

As for

x_{1}=3.2\\x_{2}=3.2-\frac{f(3.2)}{f'(3.2)}=3.23636\\x_{3}=3.23636-\frac{f(3.23636)}{f'(3.23636)}\approx 3.23606\\x_{4}=3.23606-\frac{f(3.23606)}{f'(3.23606)}\approx \mathbf{3.23606}\\

3) Rewriting and calculating its derivative. Remember to do it, in radians.

5\cos(x)-x-1=0 \:and f'(x)=-5\sin(x)-1

x_{1}=1\\x_{2}=1-\frac{f(1)}{f'(1)}=1.13471\\x_{3}=1.13471-\frac{f(1.13471)}{f'(1.13471)}\approx 1.13060\\x_{4}=1.13060-\frac{f(1.13060)}{f'(1.13060)}\approx 1.13059\\x_{5}= 1.13059-\frac{f( 1.13059)}{f'( 1.13059)}\approx \mathbf{ 1.13059}

For the second root, let's try -1.5

x_{1}=-1.5\\x_{2}=-1.5-\frac{f(-1.5)}{f'(-1.5)}=-1.71409\\x_{3}=-1.71409-\frac{f(-1.71409)}{f'(-1.71409)}\approx -1.71410\\x_{4}=-1.71410-\frac{f(-1.71410)}{f'(-1.71410)}\approx \mathbf{-1.71410}\\

For x=-3.9, last root.

x_{1}=-3.9\\x_{2}=-3.9-\frac{f(-3.9)}{f'(-3.9)}=-4.06438\\x_{3}=-4.06438-\frac{f(-4.06438)}{f'(-4.06438)}\approx -4.05507\\x_{4}=-4.05507-\frac{f(-4.05507)}{f'(-4.05507)}\approx \mathbf{-4.05507}\\

5) In this case, let's make a little adjustment on the Newton formula to find critical numbers. Remember their relation with 1st and 2nd derivatives.

x_{n+1}=x_{n}-\frac{f'(n)}{f''(n)}

f(x)=x^6-x^4+3x^3-2x

\mathbf{f'(x)=6x^5-4x^3+9x^2-2}

\mathbf{f''(x)=30x^4-12x^2+18x}

For -1.2

x_{1}=-1.2\\x_{2}=-1.2-\frac{f'(-1.2)}{f''(-1.2)}=-1.32611\\x_{3}=-1.32611-\frac{f'(-1.32611)}{f''(-1.32611)}\approx -1.29575\\x_{4}=-1.29575-\frac{f'(-1.29575)}{f''(-4.05507)}\approx -1.29325\\x_{5}= -1.29325-\frac{f'( -1.29325)}{f''( -1.29325)}\approx  -1.29322\\x_{6}= -1.29322-\frac{f'( -1.29322)}{f''( -1.29322)}\approx  \mathbf{-1.29322}\\

For x=0.4

x_{1}=0.4\\x_{2}=0.4\frac{f'(0.4)}{f''(0.4)}=0.52476\\x_{3}=0.52476-\frac{f'(0.52476)}{f''(0.52476)}\approx 0.50823\\x_{4}=0.50823-\frac{f'(0.50823)}{f''(0.50823)}\approx 0.50785\\x_{5}= 0.50785-\frac{f'(0.50785)}{f''(0.50785)}\approx  \mathbf{0.50785}\\

and for x=-0.4

x_{1}=-0.4\\x_{2}=-0.4\frac{f'(-0.4)}{f''(-0.4)}=-0.44375\\x_{3}=-0.44375-\frac{f'(-0.44375)}{f''(-0.44375)}\approx -0.44173\\x_{4}=-0.44173-\frac{f'(-0.44173)}{f''(-0.44173)}\approx \mathbf{-0.44173}\\

These roots (in bold) are the critical numbers

3 0
3 years ago
You pay $1.56 for 4 pounds of bananas. Write and solve an equation to find out how much it costs for 1 pound of bananas.
Tom [10]

Answer:

1.56 / 4 = 0.39

1 Pound of bananas cost $0.39

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
If f(x) is a linear function, what is the value of n?
lana [24]
The generic equation for a linear function can be expressed in the slope intercept form f(x) = mx + b, where m is the slope and b is the y intercept. For this problem we can first find the equation of the line. Then we substitute x = 7 to get the f(x) value, which is n at the point x = 7.

To find the equation of the linear function we first find the slope. Slope is defined as the change in f(x) divided by the change in x. As we are given a linear function, the slope at every point is the same. We can pick any two points known to find the slope.

Let's pick (3, 7) and (9, 16). The slope m is m = (16-7)/(9-3) = 9/6 = 3/2.

Now that we have the slope, we can plug in the slope and one of the points to find b. Let's use the point (3, 7).
f(x) = mx + b
7 = (1/2)(3) + b
b = 11/2

Now we can write the equation
f(x) = (1/2)x + 11/2

Plugging in x = 7 we find that f(7) = 9. n = 9
4 0
3 years ago
Read 2 more answers
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