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3 years ago

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Mathematics

1 answer:

mezya [45]3 years ago

6 0

Answer:

Brainly B.Over what interval is the approximation accurate within 0.2 Roubd bounds to 4 decimal (s) while keeping the estimate within bounds Copy this write answer

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Find the tenth term in each sequence

ozzi

Where’s the sequence?

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What is the missing value?

madam [21]

the answer is 40

X 1 (+1) 2 (+1) 3 (+1) 4 (+1) 5

Y 4 (+9) 13 (+9) 22 (+9) 31 (+9) 40

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A training field is designed by joining a rectangle and two semicircles, as shown below. The rectangle is

crimeas [40]

You know that the rectangle is 94m so we multiply that by 2 and get 188. We would use 70m as the diameter of the circle. multiply the diameter by pi which is 219.91. Then you add 188 to 219.91 and you get you answer.

Answer: 407.91

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3 years ago

Which term describes a fraction that has a variable or variable expression in its numerator, its denominator, or both?

inna [77]

The right answers is B. Rational expression.

This is true given that the statement establishes a term that describes a fraction that has a variable or variable expression in its numerator, denominator or both, that is, this is the concept of the quotient of two polynomials called rational expression. Finally, we can get the following rational expressions that matches with this concept, namely:

 $(1) \ \frac{1}{x} \\ \\ (2) \ \frac{2x-1}{x+1} \\ \\ (3) \ \frac{x^{2}-1}{x^{2}+1}$

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3 years ago

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A yeast culture weighing 2 grams is removed from a refrigerator unit and is expected to grow at the rate of Upper W prime (t )eq

Naily [24]

Answer:

a) during the first 10 hours of growth the weight will increase 271.8% relative to the initial state ( or 2.718 g)

b) during the first 20 hours of growth the weight will increase 467.1 % relative to the initial state ( or 4.671 g)

Step-by-step explanation:

since the growing rate law is

W'(t) = 0.1 gr/hour * e^(0.1gr/hour* t) , W'(t) [gr/hour]

and following mathematical conventions: W'(t)= dW/dt

then

dW/dt=0.1 e^(0.1t)

∫dW =∫0.1 e^(0.1t) dt

W = e^(0.1t) + C

at the beginning, (time t=0) the weight is W=2 grams .Therefore

2 g = e^(0.1 g/h*0) + C → 2 g = 1 g + C → C = 1 g

then

W = e^(0.1t) + 1 g

at t= 10 hours

W = e^(0.1 g/h*10h) + 1 g = 3.718 g/h

therefore the weight will increase

ΔW = 3.718 g - 1 g = 2.718 g or 271.8% relative to the initial state

for t=20 hours

W = e^(0.1 g/h*20h) + 1 g = 8.389 g/h

thus, the from t= 10 hours to t= 20 hours the weight will increase

ΔW = 8.389 g/h - 3.718 g = 4.671 g or 467.1 %relative to the initial state

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3 years ago