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dimaraw [331]
2 years ago
7

Find the quotient 3/4÷1 1/10

Mathematics
2 answers:
disa [49]2 years ago
5 0
The answer is 15/22 can i get brainliest???? huh
Scorpion4ik [409]2 years ago
4 0

Answer:

15/22

Step-by-step explanation:

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(2h³j^-3 k⁴)/3jk<br>Just look at pic​
Troyanec [42]
The answer is: 2•h^3•k^3 / 3•j^4
4 0
2 years ago
Multiply. (2x2 + 4x - 3)(x2 - 2x + 5)
BigorU [14]
For a moment, let y=x^2-2x+5. Then by the distributive property,

(2x^2+4x-3)p=2x^2p+4xp-3p


Again by the distributive property,

2x^2p=2x^2(x^2-2x+5)=2x^4-4x^3+10x^2

4xp=4x(x^2-2x+5)=4x^3-8x^2+20x

-3p=-3(x^2-2x+5)=-3x^2+6x-15

Taking everything together, we get

(2x^2+4x-3)(x^2-2x+5)=(2x^4-4x^3+10x^2)+(4x^3-8x^2+20x)+(-3x^2+6x-15)
(2x^2+4x-3)(x^2-2x+5)=2x^4-x^2+26x-15
7 0
3 years ago
Read 2 more answers
1,15,29,43 formula sequence
irga5000 [103]
All you’re doing is adding 14 to each
1+14=15
15+14=29
29+14=42
can i get brainliest answer ?
6 0
3 years ago
Read 2 more answers
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
Drag each pair of numbers to show if their greatest common factor (GCF) is less than 2, equal to 2, equal to 3, or greater than
TEA [102]

Answer:

GCF < 2 : 2 and 7, 12 and 13

GCF = 2 : 2 and 6

GCF = 3 : 3 and 6

GCF > 3 : 4 and 12, 6 and 18

Step-by-step explanation:

The greatest common Factor (GCF) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.

(a) 2 and 7: Both 2 and 7 are prime numbers, therefore the GCF of 2 and 7 is 1 which is less than 2.

(b) 2 and 6: Both 2 and 6 are even numbers, therefore the GCF of 2 and 6 is 2.

(c) 4 and 12: Both 4 and 12 are even numbers and multiple of 4, therefore the GCF of 4 and 12 is 4 which is greater than 3.

(d) 12 and 13: 13 is a prime number, therefore the GCF of 12 and 13 is 1 which is less than 2.

(e) 6 and 18: Both 6 and 18 are even numbers and multiple of 6, therefore the GCF of 6 and 18 is 6 which is greater than 2 and greater than 3.

(f) 3 and 6: Both are multiple of 3, therefore the GCF of 3 and 6 is 3.

8 0
3 years ago
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