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Doss [256]
3 years ago
14

Solve the following system of equations using matrices.

Mathematics
1 answer:
Alecsey [184]3 years ago
5 0
It is convenient to let a calculator do the math.
.. (x, y) = (0, 3) . . . . . your 3rd selection

_____
I prefer using the RREF function to put the augmented matrix into reduced row-echelon form. If the equations are dependent or inconsistent, using that method tells you so without giving a matrix inversion error.

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Create a trinomial that can be factored and write it in standard form.
Sedbober [7]
Hello : 
<span>a trinomial that can be factored is : (x-1)(x+2)
</span>the <span>standard form is : x²+x-2</span>
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4 years ago
The measure of one angle of a right triangle is 12 degrees more than the measure of the smallest angle. Find the measures of all
Svet_ta [14]

Answer:

90°, 51° and 39°

Step-by-step explanation:

The sum of the 3 angles in a triangle = 180°

Let the smallest angle be x then the other angle is x + 12, thus

x + x + 12 + 90 = 180, that is

2x + 102 = 180 ( subtract 102 from both sides )

2x = 78 ( divide both sides by 2 )

x = 39

The other angle is 39 + 12 = 51

The 3 angles are 90°, 51° and 39°

8 0
3 years ago
Which list shows all the factors of 24?
Artyom0805 [142]

Answer!!

it would be a

Step-by-step explanation:

3 0
2 years ago
A box contains 5 red marbles, 8 green marbles, 4 blue marbles, and 3 yellow marbles. if 2 marbles are randomly selected without
Lilit [14]
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5 0
4 years ago
Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

5 0
3 years ago
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