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Mice21 [21]
2 years ago
11

Chapter : EXPONENTS AND POWERS Can anyone give me answer for this please

Mathematics
1 answer:
Aneli [31]2 years ago
5 0

The value of given expression is: 3

<h3>What is exponents and powers?</h3>

Exponent refers to the number of times a number is used in a multiplication. Power can be defined as a number being multiplied by itself a specific number of times.

9*(15)³/45 * 5^{-2}/9

= 3²* 3³*5³/5² * 3² * 3² * 5

= 3^{2+3-4} * 5^{3-3}

= 3*1

=3

Learn more about exponents and power here:

brainly.com/question/15722035

#SPJ1

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In the number 432.785 which number is the hundredths digit
dimulka [17.4K]
The number in the hundredths digit is 8.
7 0
3 years ago
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Plz plz plz plz help me​
Mrac [35]

Answer:

Solution given:

perpendicular [P]=15m

base[B]=8m

hypotenuse [H]=17m

rate :4 plants per square metre

no of plants=?

we have

Area of triangular garden: ½(P*B)=½(15*8)=60m²

Now

Total no of plants =rate ×Area =4×60=240

Michael needs <u>240</u> plants in the garden

3 0
3 years ago
What is the product? (4y - 3) (2y^2 +3y - 5)
igomit [66]

(4y-3)(2y²+3y-5)

First , let's start with "4y"

4y*2y² = 8y³

4y*3y =  12y²

4y*-5 = -20y

Next, let's multiply by "-3"

-3*2y² = -6y²

-3*3y = -9y

-3*-5 = 15

Now, let's combine all of our values.

8y³+12y²-6y²-20y-9y+15 = 8y³+6y²-29y+15

8 0
3 years ago
If f(x)=2x-4 find f(q+1)
musickatia [10]
Is there any other information given on this?

5 0
3 years ago
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
4 years ago
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