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Sedaia [141]
3 years ago
8

Simplify 7.32a+2.1∙(2.7−18a)

Mathematics
1 answer:
tresset_1 [31]3 years ago
6 0
7.32a + 2.1(2.7 -18a)
7.32a + 5.67 - 37.8a (multiply 2.1 to 2.7 and 18a)
5.67 - 30.48a (combine 7.32a and -37.8a since they are like terms)
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8) An extremely handsome math teacher is on African
viva [34]

Answer:

One less zebra

Step-by-step explanation:

T        z        l

0       200   220

1         160    175       (subtract 40 from 200  and 45 from 220)

2        120    130

3         80      85

4         40  =  40      <===== lion has zebra lunch before the river is reached

Zdistance = 200 - 40s

Ldistance = 220 - 45 s            where  's' is seconds

8 0
2 years ago
Stuffed animals each weighing 1 over 4 pound are on a scale. The scale shows 1 1 over 2 pounds. How many stuffed animals are on
Anna35 [415]

Answer:

6 stuffed animals

Step-by-step explanation:

if 1/4= 1 then 1x6= 1 1/2

4 0
3 years ago
15 Points and ill mark brainliest!
natima [27]
Neat
of turf, 83.33333% of the 1050 were minor
so about 875 minor injuries in turf
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leaving 63 severe field turf
 
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so now we know that grassmionr=938 that that is 1
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I have to go sorry bye hope this helped
7 0
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)$

$\frac{dP}{dt}=bP(\frac{a}{b}-P)$    .................(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
Find the slope of the line through the pair of points.<br> (17,-7) and (13,-4)]
Marysya12 [62]

Answer:

3/4

Step-by-step explanation:

(17,-7) And (13,-4).  First take -4 and subtract it from-7.  When you do that you get 3.  Then Subtract 13-17.  You get 4.  Write 3 and 4 as a fraction.  3/4.  The 3 is the x-value and the 4 is the y value. So 3/4 is your slope

3 0
3 years ago
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