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3 years ago

Can I get some help here?

Mathematics

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

AB, CB, AC

Step-by-step explanation:

All the angles in a triangle added up equal 180 degrees

Start by finding the measure of angle A

A + 86 + 27 = 180

A + 113= 180

A = 67 degrees

Now order them

86 is greater than 67 and 27 so B is the greatest

27 is less than 86 and 67 so C is the smallest

67 is less than 86 but greater than 27 so A is in the middle

Now for the sides..

AB is opposite angle C so it's the shortest

CB is opposite angle A so it's in the middle

AC is opposite angle B so it's the longest

I hope this helps!!!

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.. pls help me....!!!!! Determine whether the point (-2,4) satisfy y<3x+5,y=3x+5 or y >3x+5..

nasty-shy [4]

Answer:

$y > 3\cdot x + 5$

Step-by-step explanation:

Let $A(x,y) = (-2,4)$ and $y = 3\cdot x + 5$ . Now we evaluate the given function at $x = -2$ :

$y = 3\cdot x + 5$ (1)

$y = 3\cdot (-2)+5$

$y = -6 +5$

$y = -1$

Which means that $y$ is less than the y-component of A. Therefore, the right answer is $y > 3\cdot x + 5$ .

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3 years ago

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3 years ago

a teacher and 10 students are to be seated along a bench in the bleachers at a basketball game. In how many ways can this be don

Veronika [31]

Wow !

OK. The line-up on the bench has two "zones" ...

-- One zone, consisting of exactly two people, the teacher and the difficult student.
   Their identities don't change, and their arrangement doesn't change.

-- The other zone, consisting of the other 9 students.
   They can line up in any possible way.

How many ways can you line up 9 students ?

The first one can be any one of 9.   For each of these . . .
The second one can be any one of the remaining 8. For each of these . . .
The third one can be any one of the remaining 7. For each of these . . .
The fourth one can be any one of the remaining 6. For each of these . . .
The fifth one can be any one of the remaining 5. For each of these . . .
The sixth one can be any one of the remaining 4. For each of these . . .
The seventh one can be any one of the remaining 3. For each of these . . .
The eighth one can be either of the remaining 2. For each of these . . .
The ninth one must be the only one remaining student.

   The total number of possible line-ups is

   (9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1) = 9! = 362,880 .

But wait ! We're not done yet !

For each possible line-up, the teacher and the difficult student can sit

-- On the left end,
-- Between the 1st and 2nd students in the lineup,
-- Between the 2nd and 3rd students in the lineup,
-- Between the 3rd and 4th students in the lineup,
-- Between the 4th and 5th students in the lineup,
-- Between the 5th and 6th students in the lineup,
-- Between the 6th and 7th students in the lineup,
-- Between the 7th and 8th students in the lineup,
-- Between the 8th and 9th students in the lineup,
-- On the right end.

That's 10 different places to put the teacher and the difficult student,
in EACH possible line-up of the other 9 .

So the total total number of ways to do this is

   (362,880) x (10) = 3,628,800 ways.

If they sit a different way at every game, the class can see a bunch of games
without duplicating their seating arrangement !

4 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 area of the rhombus.<br> A=<br> square units

mylen [45]

Answer: A=70 square units

Step-by-step explanation:

The area of a triangle can be found using the formula $A=\frac{1}{2} bh$

Let 5=h and 7=b

Plug in the values and solve

$A=\frac{1}{2} (7)(5)$

$A=17.5$

However, since four of these triangles make up this rhombus, we can multiply the area of one of these triangles by four to find the area of the whole rhombus.

$A=17.5*4$

$A=70$

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3 years ago