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blagie [28]
3 years ago
8

What is the answer to this algebra problem

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
5 0
The problem said : -10,8 =-2 and -15,9=-6

kiruha [24]3 years ago
4 0
First find the slope of the two coordinates.

Slope is represented by the equation:

M = y2 - y1 / x2 - x1

Substitute in the coordinates.

M = 9 - 8 / -15 - (-10)

M = 9 - 8 / -15 + 10

M = 1 / -5

M = - 1/5


Then use the equation a line:

y - y1 = m(x - x1)

y - 8 = -1/5(x - (-10)

y - 8 = -1/5(x + 10)

y - 8 = -1/5x - 2

y = -1/5x + 6 is the solution
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A cylindrical water tank has a radius of 6 feet and a height of 20 feet. A larger tank with the same height has a volume that is
likoan [24]

Answer:

Radius of large cylinder = 10.39 feet\\

Step-by-step explanation:

Volume of the larger tank is three times the volume of smaller tank

Volume of smaller tank

= \pi r^{2} h\\= (3.14) * 6^{2} * 20\\= 2260.8 ft^{3} \\

Volume of larger tank

= 3 * 2260.8 ft^3\\= 6782.4 ft^3\\

Or it can also be written as

\pi r^{2} h = 6782.4 ft^3\\(3.14) * ( r^2) 20 = 6782.4 ft^3\\r^2 = 108\\r = 10.39

6 0
3 years ago
PLEASE HELP
stepan [7]

Answer:

to big but i will slove it for u ok so see solution c one is correct

5 0
3 years ago
Read 2 more answers
Help please...........-
vovangra [49]

Answer:c

Step-by-step explanation:

c

8 0
3 years ago
Read 2 more answers
A line has the equation -5x -14y +168 = 0
valkas [14]

Answer:

y = -5/14x + 12

Step-by-step explanation:

add 14y to both sides to isolate variables and divide everything by 14 to get 1y and boom

6 0
2 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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