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algol [13]
3 years ago
15

Tionna has saved dimes and quarters in her piggy bank. Define the variables

Mathematics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

Step-by-step explanation:

y = .10d + .25q

y is the total amount, if you have d dimes and q quarters

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I need help with this question plsss
olga55 [171]

Answer: maybe its 6

Step-by-step explanation:

i think its 6 because it has +3 so maybe adding 3+3

7 0
3 years ago
How would i approximate each irrational square root number to the nearest 0.05? 34, 82, 45, 104, -71, and -19
Digiron [165]

Using a calculator or a pencil, find each square root, including
three (3) decimal places.  A negative number has no real square
roots, so skip '-71' and '-19'.

For each square root . . .

-- If the digits in the 2nd and 3rd decimal place are 00 through 24, then
make the 2nd decimal place a '0' and discard the rest.

-- If the digits in the 2nd and 3rd decimal place are 25 through 74, then
make the 2nd decimal place a '5' and discard the rest.

-- If the digits in the 2nd and 3rd decimal place are 75 through 99, then
increase the 1st decimal place by 1, make the 2nd decimal place a '0',
and discard the rest.
4 0
4 years ago
What is the sum of ​​​​​​12,837.45 and 15,910.65? Enter your answer in the box below.
Minchanka [31]

Answer: 28748.1

Step-by-step explanation:

6 0
3 years ago
If f(x) = |(x2 − 9)(x2 + 1)|, how many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem?
ivann1987 [24]

Answer:

only one number c=0 in the interval [-1,1]

Step-by-step explanation:

Given : Function f(x) = |(x^2-9)(x^2 + 1)|   in the interval [-1,1]

To find : How many numbers in the interval [−1, 1] satisfy the conclusion of the mean value theorem.

Mean value theorem : If f is a continuous function on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a point 'c' in (a,b) such that f'(c)=\frac{f(b)-f(a)}{b-a}

Solution : f(x) is a function that satisfies all of the following :

1) f(x)  is continuous on the closed interval [-1,1]  

\lim_{x\to a} f(x)=f(a)

2) f(x) is differentiable on the open interval  (-1,1)

Then there is a number  c such that  f'(c)=\frac{f(b)-f(a)}{b-a}

f(a)=f(-1) = |(-1^2-9)(-1^2 + 1)|=|(-8)(2)|=16

f(b)=f(1) = |(1^2-9)(1^2 + 1)|=|(8)(2)|=16

f'(c)=\frac{f(b)-f(a)}{b-a}=\frac{16-16}{2}=0

f'(c)=0 ........[1]

Now, we find f'(x)

f(x) = |(x^2-9)(x^2 + 1)|

f(x) =x^4-8x^2-9

Differentiating w.r.t  x

f'(x) =4x^3-16x

In place of x we put x=c

f'(c) =4c^3-16c

f'(c) =4c^3-16c=0  (by [1], f'(c)=0)

4c(c^2-4)=0

4c=0,c^2-4=0

either c=0 or  c^2-4=0\rightarrow c=\pm2

we cannot take c=\pm2  because they don't lie in the interval [-1,1]

Therefore, there is only one number c=0 which lie in interval [-1,1] and satisfying the conclusion of the mean value theorem.



3 0
3 years ago
Please help! And 20 points! :3<br> I need help with these 3 questions- <br> -3-||
BaLLatris [955]

Answer:

where r the questions--?

Step-by-step explanation:

7 0
3 years ago
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