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hjlf
3 years ago
9

Solve for x? A 19.8 C 32.7 B 12.8 D 7

Mathematics
1 answer:
Art [367]3 years ago
6 0

Answer:

Step-by-step explanation:

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Use Cramers Rule to solve the system <br> 2x - 3y = 11<br> -6x + 8y = 34
Nastasia [14]

Step-by-step explanation:

given :

2x - 3y = 11

-6x + 8y = 34

find : the solutions of the system by using Cramers Rule.

solutions:

in the matrix 2x2 form =>

[ 2 -3] [x] [11]

=

[-6 8] [ y] [34]

D =

| 2 -3 |

|-6 8 |

= 8×2 - (-3) (-6)

= 16-18 = -2

Dx = | 11 -3 |

| 34 8 |

= 11×8 - (-3) (34)

= 88 + 102

= 190

Dy = | 2 11 |

|-6 34 |

= 2×34 - (-6) (11)

= 68 + 66

= 134

x = Dx/D = 190/-2 = -95

y = Dy/D = 134/-2 = -67

the solutions = {-95, -67}

8 0
3 years ago
Read 2 more answers
Write each sum using summation notation.<br> 1+ 2 + 3 + 4 + ⋯ + 1000
AVprozaik [17]

Answer:

\sum_{k=1}^{1000} k

Step-by-step explanation:

You have to use the summation notation formula:

\sum_{k}^{n} f(k) = f(1) + f(2) +...+f(n)

where k is the starting number, n is the ending number and f(k) is the function or the expression to be added.

In this case, you have the sum of the integer numbers from 1 to 1000. Therefore, k=1 and n=1000.

Now, you have to obtain the function f(k) which is the representation of the expression needed to obtain the correct result of the sum.

f(k=1) = 1

f(k=2)=2

f(k=3)=3 ...

You can notice that the value of k corresponds to te value of f(k) therefore f(k) = k

Replacing the values of k, n and f(k) in the formula:

\sum_{k=1}^{1000} k = 1+2+3+4+...+1000

7 0
4 years ago
Can someone plz help me ASAP??? my hw in hard and I don't get it
Fynjy0 [20]

9514 1404 393

Answer:

  see attachments

Step-by-step explanation:

These are "one-step" equations.

To solve the ones involving addition or subtraction, identify the constant on the same side with the variable. Add its opposite to both sides.

To solve the ones involving multiplication or division, identify the coefficient of the variable. Multiply both sides by its reciprocal (or, equivalently, divide by that coefficient).

Here are more details for the problems on the first page.

  • w +8 = -3  ⇒  w = -3 -8 = -11
  • z +(-7) = -20  ⇒  z = -20 +7 = -13
  • 1/5x = 11  ⇒  x = 5·11 = 55
  • 1u = -88  ⇒  u = -88/11 = -8
  • 1/9y = -2  ⇒  y = 9(-2) = -18
  • 12 +n = 7  ⇒  n = 7 -12 = -5
  • q +(-4) = 8  ⇒  q = 8 +4 = 12

_____

The second page is solved the same way. It is shown in the second attachment.

7 0
3 years ago
there are 78 tourists going on a sightseeing trip.the tour company's buses hold 12 people each how many buses will they need?
vovangra [49]

Answer:

6.5

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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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