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Allushta [10]
2 years ago
15

Kyle has 105 baseball cards and must give 1/5 of his figures away. How many cards will Kyle give away? What fraction of his card

s does he have left?​
Mathematics
1 answer:
andreev551 [17]2 years ago
4 0

Answer:

I think it would be 21, because 105 / 5 is 21, and if the denominator is the holder spot, it should be 21

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The hypotenuse of a 45°-45°-90° triangle measures 128 cm. A right triangle is shown. The length of the hypotenuse is 128 centime
Aleks [24]

Answer:

64√2 or 64 StartRoot 2 EndRoot

Step-by-step explanation:

A 45-45-90 traingle is a special traingle.  Let's say one of the leg of the triangle is x. The other one is also x because of the isosocles triangle theorem.  Therefore, using the pytagorean theorem, you find that x^2+x^2=c^2.  2(x)^2=c^2.  You then square root both sides and get c= x√2.  

Therefore, the two legs are x and the hypotenuse is x√2.  x√2=128 because the question says that the hypotenuse is 128.  Solve for x by dividing both sides by √2.  X=128/√2.  You rationalize it by multiplying the numberator and denominator of the fraction by √2.  √2*√2= 2.

X=(128√2)/2= 64√2 cm.

Since X is the leg, the answer would be 64√2

3 0
3 years ago
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Will give brainliest
Rashid [163]

Answer:

thanks for the points ily

5 0
3 years ago
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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]

This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

4 0
3 years ago
Determine the interquartile range for the data.<br> 16, 19, 25, 20,<br> 22, 21, 17, 20
Roman55 [17]

Answer:

the answer is 15 subtract the highest number from the lowest

5 0
3 years ago
F(x) = 4x+3x<br> g(x) = 2x-5<br> 7. Find (f+9) (x) <br> 8. Find (f - g) (x)
notka56 [123]

Answer:

7. (f+g)(x) = 6x - 2

8. (f-g)(x) = 2x + 8

Step-by-step explanation:

We are given two function, f(x) and g(x).

We want to find their addition and their subtraction, and we do this combining like terms.

The functions are:

f(x) = 4x + 3

g(x) = 2x - 5

7. Find (f+9) (x)

4x + 3 + 2x - 5 = 4x + 2x + 3 - 5 = 6x - 2

So

(f+g)(x) = 6x - 2

8. Find (f - g) (x)

4x + 3 - (2x - 5) = 4x + 3 - 2x + 5 = 4x - 2x + 3 + 5 = 2x + 8

So

(f-g)(x) = 2x + 8

3 0
2 years ago
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