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2 years ago

7

Help me please!!!!!!!

1 answer:

True [87]2 years ago

5 0

I'll give you answer for 48 once you give me the complete version of it

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I need help on all of them pleaseeee

dmitriy555 [2]

Step-by-step explanation:

90÷0.018=5000

90÷0.18= 500

90÷1.8 = 50

90÷18 =5

90÷180 =0.5

5 0

2 years ago

Find the x-intercepts of the parabola with vertex (1,-108) and y-intercept (0,-105). Write your answer in this form: (X1,y1), (x

Stolb23 [73]

Answer:

(7, 0) and (-5, 0)

Step-by-step explanation:

<u>Vertex form</u>

$y=a(x-h)^2+k$

(where (h, k) is the vertex)

Given:

vertex = (1, -108)

$\implies y=a(x-1)^2-108$

Given:

y-intercept = (0, -105)

$\implies a(0-1)^2-108=-105$

$\implies a(-1)^2=-105+108$

$\implies a=3$

Therefore:

$\implies y=3(x-1)^2-108$

The x-intercepts are when y = 0

$\implies 3(x-1)^2-108=0$

$\implies 3(x-1)^2=108$

$\implies (x-1)^2=36$

$\implies x-1=\pm \sqrt{36}$

$\implies x=1\pm 6$

$\implies x=7, x=-5$

Therefore, the x-intercepts are (7, 0) and (-5, 0)

7 0

2 years ago

-8x+3=8x+27 I need help plz I’m trash at math

belka [17]

Answer:

-1.5

Step-by-step explanation:

<u>See steps below:</u>

-8x+3=8x+27

-8x + 3 - 8x - 3 = 8x+ 27 - 8x - 3 ⇒ add -8x - 3 to both sides in order to collect like terms in the different sides of equation
-16x = 24 ⇒ simplified
x= 24/-16 ⇒ divide
x= - 1.5 ⇒ answer

3 0

2 years ago

If a wheel with a radius of 70 inches spins at a rate of 600 revolutions per minute, find the approximate linear velocity in mil

Ainat [17]

This is the concept of speed:
ω=600 rev/min=600*2π rad/min=1200 rad/min
v=rω=70*1200π =84,000 in/min
converting the above in miles per hour we get:
84000 in/min
=(84000*1.6*10^(-5)*60)
=80.64 mph

8 0

3 years ago

1. Consider the population model dP dt = 0.2P 1 − P 135 , where P(t) is the population at time t. (a) For what values of P is th

Kryger [21]

Answer:

a

$P = 0$ OR $P = 135$

b

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

c

Generally the carrying capacity is can be defined as the highest amount of population and environment can support for an unlimited duration or time period

d

$P = 67.5$

Step-by-step explanation:

From the question we are told that

The population model is $\frac{dP}{dt} = 0.2P(1 - \frac{P}{135} )$

Generally at equilibrium

$\frac{dP}{dt} = 0$

So

$0.2P = 0$

=> $P = 0$

Or

$(1 - \frac{P}{135} ) = 0$

=> $P = 135$

Thus at equilibrium P = 0 or P = 135

Generally when the population is increasing we have that

$\frac{dP}{dt} > 0$

So

$0.2P > 0$

=> $P > 0$

and

$(1 - \frac{P}{135} ) > 0$

$P < 135$

Now when the first value of P i.e $P< 0$ for $\frac{dP}{dt} > 0$

$P_2 > 135$

So when population increasing the values of P are

$P > 0$ and $P < 135$

OR

$P > 0$ and $P < 135$

So to obtain initial values of P where the population converge to the carrying capacity as $t \to [\infty]$

The rate equation can be represented as

$\frac{dP}{dt} = \frac{1}{5}P (1 - \frac{P}{135} )$

So we will differentiate the equation again we have that

$\frac{d^2 P}{dt^2} = \frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675}$

Now as $t \to [\infty]$

$\frac{d^2 P}{dt^2} \to 0$

So

$\frac{(1 - \frac{P}{135} )}{5} - \frac{P}{675} = 0$

=> $\frac{(1 - \frac{P}{135} )}{5} = \frac{P}{675}$

=> $P = 67.5$

5 0

3 years ago