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3 years ago

Which ordered pair is a solution of the given system of equations {x+2y=6 {x-y=3

Mathematics

2 answers:

tino4ka555 [31]3 years ago

8 0

X+2y=6 now subtract the other equation from this...
-x+y=-3 gives you

3y=3, so y =1, making x+2y=6 become:

x+2(1)=6

x+2=6, so x=4

The solution is the point (4,1)

Ainat [17]3 years ago

4 0

The pair (4,1) is the solution.

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The number of Atlantic blue fin tuna in thousands can be modeled by P(x)=230(.881)^x where X represents the number of years sinc

Maurinko [17]

<h3>Answer: 1981</h3>

===================================

Work Shown:

Recall that P is in thousands, so P = 95 means 95,000.

Plug in P(x) = 95. Solve for x. Use logarithms to get this done.

P(x)=230(0.881)^x

95=230(0.881)^x

95/230 = (0.881)^x

0.41304347826087 = (0.881)^x

(0.881)^x = 0.41304347826087

Log( (0.881)^x )= Log( 0.41304347826087 )

x*Log( 0.881 )= Log( 0.41304347826087 )

x= Log( 0.41304347826087 )/Log( 0.881 )

x= 6.97883817154785

x= 7

Approximately 7 years after 1974 is when the population will be around 95,000.

7 years after 1974 = 1974+7 = 1981

8 0

2 years ago

Help me asap! I will give you marks

Law Incorporation [45]

Recall the binomial theorem.

$(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k$

1. The binomial expansion of $\left(1+\frac x3\right)^7$ is

$\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}$

Observe that

$k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x$

$k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2$

When we multiply these by $8-9x$ ,

• $8$ and $\frac73 x^2$ combine to make $\frac{56}3 x^2$

• $-9x$ and $\frac73 x$ combine to make $-\frac{63}3 x^2 = -21x^2$

and the sum of these terms is

$\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}$

2. The binomial expansion is

$\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k$