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2 years ago

Solve for k 4.5+1.5k=18−3k

Mathematics

2 answers:

sertanlavr [38]2 years ago

5 0

Answer:

$\boxed{ \bold{ \huge{ \boxed{ \sf{k = 3}}}}}$

Step-by-step explanation:

$\sf{4.5 + 1.5k = 18 - 3k}$

Move 3k to left hand side and change it's sign

Similarly, move 4.5 to right hand side and change it's sign

$\longrightarrow{ \sf{1.5k + 3k = 18 - 4.5}}$

Collect like terms

$\longrightarrow{ \sf{4.5k = 18 - 4.5}}$

Subtract 4.5 from 18

$\longrightarrow{ \sf{4.5k = 13.5}}$

Divide both sides by 4.5

$\longrightarrow{ \sf{ \frac{4.5k}{4.5} = \frac{13.5}{4.5} }}$

Calculate

$\longrightarrow{ \sf{k = 3}}$

Hope I helped!

Best regards! :D

romanna [79]2 years ago

4 0

Answer:

k = 3

Step-by-step explanation:

Subtract 4.5 from both sides:

4.5 + 1.5k = 18 - 3k

-4.5 -4.5

1.5k = 13.5 -3k

Subtract -3k from both sides:

1.5k = 13.5 - 3k

+3k +3k

4.5k = 13.5

Now simplify:

4.5k = 13.5

k = 3

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2 years ago

Put the equation in standard form with a positive x coefficient. 2y+5=-4(x+2)

umka21 [38]

The standard form of equation is: 4x+2y = -13

Step-by-step explanation:

the standard form of a linear equation in two variables is given by:

$ax+by = c$

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The standard form of equation is: 4x+2y = -13

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3 years ago

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 :

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

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

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

= $\frac{3}{9680}$

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

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

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

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

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

Now taking natural logs on both the sides we get

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3 years ago

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Stella [2.4K]

The proportion looks like this
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8 0

3 years ago

Read 2 more answers