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nekit [7.7K]
2 years ago
6

Find the difference. (5x2 + 2x + 11) − (7 + 4x − 2x2)

Mathematics
2 answers:
belka [17]2 years ago
7 0

Answer:

7x^2-2x+4

Step-by-step explanation:

(5x^2+2x+11)-(7+4x-2x^2)

Breaking the parentheses using the law of signs for addition / subtraction:

5x^2+2x+11-7-4x+2x^2

Combining like terms:

(5x^2+2x^2)+(2x-4x)+(11-7)

add/subtract like terms:

7x^2-2x+4

k0ka [10]2 years ago
6 0
<span>(5x^2 + 2x + 11) − (7 + 4x − 2x^2)
= </span>5x^2 + 2x + 11 − 7 - 4x +  2x^2
= 7x^2 - 2x + 4
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Answer:an expression to describe the total cost of a party is written as

y = 13.50p + 50

Step-by-step explanation:

The cost of renting the party room at a restaurant is $50.00. Snacks cost $13.50 per person. Assuming the total number of persons invited for the party is p, and the total cost of renting the party room is y. Therefore an expression to describe the total cost of a party is written as

y = 13.50p + 50

The $50 always remains constant

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Factor completely 2x^2+6x-80
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Answer:

x= -8 , x=5

Step-by-step explanation:

2x^2+6x-80=0

x^2+3x-40=0

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Ini apa we jawabannya pelasee​
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The Pacific halibut fishery has been modeled by the differential equation dy dt = ky 1 − y K where y(t) is the biomass (the tota
Dafna1 [17]

Answer:

a. The biomass weighs 2.30 * 10^7 kg after a year

b. It'll take 2.56 years to get to 4*10^7kg

Step-by-step explanation:

a.

k = 0.78,K = 6E7 kg

Given

dy/dt = ky(1- y/K)

Make ky dt the subject of formula

ky dt = dy/(1-y/K) --- make k dt the subject of formula

k dt = dy/(y(1-y/K))

k dt = K dy / y(K-y)

k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides

kt + c = ln(y/(K-y))

Ce^(kt) = y/(K-y)

Substitute the values of k and K

Ce^(0.78t) = y/(6*10^7 - y) ----- (1)

Given that y(0) = 2 * 10^7kg

(1) becomes

Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)

Ce° = (2*10^7)/(4*10^7

C = 2/7

Substitute 2/7 for C in (1)

2/7e^0.78t = y/(6*10^7 - y) ---(2)

We're to find the biomass a year later

So, t = 1

2/7e^0.78 = y/(6*10^7 - y)

0.62 = y/(6*10^7 - y)

y = 0.62(6*10^7 - y)

y = 0.62*6*10^7 - 0.62y

y + 0.62y = 0.62*6*10^7

1.62y = 0.62*6*10^7

1.62y = 3.72 * 10^7

y = 2.30 * 10^7kg.

Hence, the biomass weighs 2.30 * 10^7 kg after a year

b.

Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg

Substitute 4*10^7 for y in (2)

2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)

2/7e^0.78t = 4*10^7/2*10^7

2/7e^0.78t = 2

e^0.78t = (2*7)/2

e^0.78t = 2

t = 2 * 1/0.78

t = 2.56 years

Hence, it'll take 2.56 years to get to 4*10^7kg

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