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geniusboy [140]
3 years ago
15

Simplify the expression 1+4.25n+3/2p-3+(-2p)+5/4n

Mathematics
1 answer:
ale4655 [162]3 years ago
4 0

The answer is 5.5n-0.5p-4

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How much gold foil did it take to cover the trophy including the bottom?
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Answer:

i  need mesurements

Step-by-step explanation:

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3 years ago
How to calculate concentration with just the absorbance and % transmittance measurements from a spectrometer? ​
Natali [406]

Answer:

A=ε*l*c

A= 2- log₁₀ % T

Step-by-step explanation:

There is a linear relationship between the concentration of a sample and absorbance according to Beer-Lambert Law.

A=ε*l*c

where;

A=absorbance

ε=absorption coefficient

l=path length

c=concentration

Because % transmittance is transmittance value multiplied by 100 then, the equation that will allow us calculate absorbance from % transmittance value will be;

A= 2- log₁₀ % T where T is transmittance.

8 0
3 years ago
The isosceles triangle has a base that measures 14 units.
Basile [38]

Answer:

Between 7 and 14

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt
brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

f'(x) = \dfrac{(-x + 2)^4}{dx}  = -4 \cdot (-x + 2)^3 = 0

∴ (-x + 2)³ = 0

x = 2

f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx}  = -12 \cdot (-x + 2)^2

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0
3 years ago
Solve the inequality |x − 6| − 10 &lt; 1 Write the answer as a compound inequality
Tju [1.3M]
|x-6| < 1 + 10 (move 10 to the right side)
|x-6| < 11

1)x-6 < 11 (here x >= 6)
2)-x+6 < 11 (here x < 6)

1)x<17
2)x>-5

so we get -5 < x < 17

the answer: (-5; 17)
5 0
3 years ago
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