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3 years ago

What is x+5x+2-4x=+2

Mathematics

2 answers:

34kurt3 years ago

6 0

Answer:

the solution is this:

x= -2

frez [133]3 years ago

6 0

Answer:
x=0

Explanation:

Cancel 2 on both sides leaving you with

x+5x−4x=0

Then Simplify x+5x−4x to 2x.

2x=0

Divide both sides by 2

x=0

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Aldon and Jamal raised the same amount of money for the school fundraiser. Aldon

Naddika [18.5K]

Answer: $5

Step-by-step explanation: This can be represented by the equation

40 + 12T = 25 + 15T

We just solve for T.

Subtract 12T from both sides

40 = 25 + 3T Subtract 25

15 = 3T Divide by 3

5 = T

7 0

3 years ago

Can y’all explain what I have to do and z help me

zysi [14]

Answer:

Step-by-step explanation:

The answer is D. Both are negative because they're both elevations below 0. However, -7 is higher, or closer to the regular level, because -10 is simply farther down.

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3 years ago

After travelling 2/5 of a Journey

Trava [24]

Answer:

= 100 km

Step-by-step explanation:

Total length = 5/5

Covered length = 2/5

Remaining length = 5/5 - 2/5 = 3/5

If 3/5 = 60 km

What about 5/5 = ?

= 5/5 x 60 x 5/3 (reciprical)

= 60 x 5/3

= 300/3

= 100 km

Covered distance = 100 - 60 = 40 km

6 0

3 years ago

PLEASE HELP QUICK 3t-2=-11

valentina_108 [34]

-3
step by step
add 2 to -11 and divide-9 by 3

5 0

3 years ago

Read 2 more answers

Consider a rabbit population P(t) satisfying the logistic equation StartFraction dP Over dt EndFraction equals aP minus bP squa

maria [59]

Solution:

Given :

$\frac{dP}{dt}= aP-bP^2$ .............(1)

where, B = aP = birth rate

D = $bP^2$ = death rate

Now initial population at t = 0, we have

$P_0$ = 220 , $B_0$ = 9 , $D_0$ = 15

Now equation (1) can be written as :

$\frac{dP}{dt}=P(a-bP)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$ .................(2)

Now this equation is similar to the logistic differential equation which is ,

$\frac{dP}{dt}=kP(M-P)$

where M = limiting population / carrying capacity

This gives us M = a/b

Now we can find the value of a and b at t=0 and substitute for M

$a_0=\frac{B_0}{P_0}$ and $b_0=\frac{D_0}{P_0^2}$

So, $M=\frac{B_0P_0}{D_0}$

= $\frac{9 \times 220}{15}$

= 132

Now from equation (2), we get the constants

k = b = $\frac{D_0}{P_0^2} = \frac{15}{220^2}$

= $\frac{3}{9680}$

The population P(t) from logistic equation is calculated by :

$P(t)= \frac{MP_0}{P_0+(M-P_0)e^{-kMt}}$

$P(t)= \frac{132 \times 220}{220+(132-220)e^{-\frac{3}{9680} \times132t}}$

$P(t)= \frac{29040}{220-88e^{-\frac{396}{9680} t}}$

As per question, P(t) = 110% of M

$\frac{110}{100} \times 132= \frac{29040}{220-88e^{\frac{-396}{9680} t}}$

$220-88e^{\frac{-99}{2420} t}=200$

$e^{\frac{-99}{2420} t}=\frac{5}{22}$

Now taking natural logs on both the sides we get

t = 36.216

Number of months = 36.216

8 0

4 years ago