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FrozenT [24]
3 years ago
10

Malia created a correct function table using the equation g = 4f. Which table could be Malia's? A. B. C. D. 

Mathematics
2 answers:
sasho [114]3 years ago
3 0
I can not answer this question because  I do not have access to the graphs.

yKpoI14uk [10]3 years ago
3 0
Can you show a picture of the graphs??
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Show that a discrete metric on a vector space X≠{0} cannot be obtained from a norm (the pair (X,d) is called the discrete metric
topjm [15]

You know that the discrete metric only takes values of 1 and 0. Now suppose it comes from some norm ||.||. Then for any α in the underlying field of your vector space and x,y∈X, you must have that

∥α(x−y)∥=|α|∥x−y∥.

But now ||x−y|| is a fixed number and I can make α arbitrarily large and consequently the discrete metric does not come from any norm on X.

Step-by-step explanation:

hope this helps

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2 years ago
Sketch one cycle of the cosine function y= cos 2θ
Liula [17]
Based on function transformations, the graph of y= cos(2θ), is the same as the graph of <span>y= cos(θ), with a component B of 2, in which case, it means the period is the only thing that differs

well the period for this function will then be  </span>\bf \cfrac{2\pi }{B}\iff \cfrac{2\pi }{2}\implies \pi

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Estimate the area of the parallelogram.
lisov135 [29]
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5 0
3 years ago
Read 2 more answers
Carry out three steps of the Bisection Method for f(x)=3x−x4 as follows: (a) Show that f(x) has a zero in [1,2]. (b) Determine w
ELEN [110]

Answer:

a) There's a zero between [1,2]

b) There's a zero between [1.5,2]

c) There's a zero between  [1.5,1.75].

Step-by-step explanation:

We have f(x)=3^x-x^4

A)We need to show that f(x) has a zero in the interval [1, 2]. We have to see if the function f is continuous with f(1) and f(2).

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(2)=3^2-(2)^4=9-16=(-7)

We can see that f(1) and f(2) have opposite signs. And f(1)>f(2) and the function is continuous, this means that exists a real number c between the interval [1,2] where f(c)=0.

B)We have to repeat the same steps of A)

For the subinterval [1,1.5]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1) and f(1.5) have the same signs, this means there's no zero in the subinterval [1,1.5].

For the subinterval [1.5,2]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.5) and f(2) have opposite signs, this means there's a zero between the subinterval [1.5,2].

C)We have to repeat the same steps of A)

For the subinterval [1,1.25]:

f(x)=3^x-x^4\\\\f(1)=3^1-(1)^4=3-1=2\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5

f(1) and f(1.25) have the same signs, this means there's no zero in the subinterval [1,1.25].

For the subinterval [1.25,1.5]:

f(x)=3^x-x^4\\\\f(1.25)=3^1^.^2^5-(1.25)^4=3.94-2.44=1.5\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13

f(1.25) and f(1.5) have the same signs, this means there's no zero in the subinterval [1.25,1.5].

For the subinterval [1.5,1.75]:

f(x)=3^x-x^4\\\\f(1.5)=3^1^.^5-(1.5)^4=5.19-5.06=0.13\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)

f(1.5) and f(1.75) have opposite signs, this means there's a zero between the subinterval [1.5,1.75].

For the subinterval [1.75,2]:

f(x)=3^x-x^4\\\\f(1.75)=3^1^.^7^5-(1.75)^4=6.83-9.37=(-2.54)\\\\f(2)=3^2-(2)^4=9-16=(-7)

f(1.75) and f(2) have the same signs, this means there isn't a zero between the subinterval [1.75,2].

The graph of the function shows that the answers are correct.

6 0
3 years ago
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Tamiku [17]

Answer:

the answer is 28 if you multiply if not 30

3 0
3 years ago
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