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yaroslaw [1]
3 years ago
7

Gulnaz plans to use less than 26 eggs while baking. She uses 5 eggs for each cake that she bakes, and 3 eggs for each quiche tha

t she bakes.
Write an inequality that represents the number of cakes (C)left parenthesis, C, right parenthesis and quiches (Q)left parenthesis, Q, right parenthesis Gulnaz can bake according to her plan.
Mathematics
1 answer:
zepelin [54]3 years ago
6 0

Answer:

5(x) +3(y)<26

Step-by-step explanation:

Let x represent the number of cakes she will bake and let you know represent the nymber of quiche she will bake.

She will use less than 26 eggs while baking and 5 eggs for each cake and 3 eggs for each quiche.

The inequality representing the above statement iz given below.

5(x) +3(y)<26

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Fofino [41]

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vi = initial velocity

t = time

Plug in the given values:

a = \frac{65-35}{5}\\\\a = 30 / 5  \\\\=6 m/s^{2}

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3 years ago
A bag contains 6 black marbles and 4 white marbles. Sally takes out a black marble and does not put it back. What is the probabi
Neko [114]

Answer:

\frac{5}{9}

Step-by-step explanation:

Total\ marbles=10\\\\Black=6\\\\White=4

After removing one black marble:

Total\ remaining\ marbles=9\\\\Remaining\ Black=5\\\\P(Black)=\frac{Remaining\ Black\ marbles}{Total\ remaining\ marbles}\\\\P(Black)=\frac{5}{9}

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Please help me with this question.
kow [346]
They are equivalent. combine like terms on the first one and 9x-4x+25 can be simplified to 5x+25.
6 0
2 years ago
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Leokris [45]

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3 0
3 years ago
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

8 0
3 years ago
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