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tigry1 [53]
3 years ago
15

What are the solutions to the following system of equations? Select the correct answer. y=x^2-4x+8 4x+y=12

Mathematics
2 answers:
Murrr4er [49]3 years ago
8 0
Reduce the second equation to y= -4x + 12, then subtract the two equations. (Quadratic on top), to get y=x^2 -4. Then, plug in an x,y coordinate if you have one.
lorasvet [3.4K]3 years ago
7 0
Y=x^2-4x+8 and 4x+y=12 so y=12-4x, insert y in the quadratic to get
12-4x=x^2-4x+8
12=x^2+8
0=x^2-4=(x-2)(x+2)
So the solutions are x=2 and x=-2.
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Nadir saves $1 the first day of a month, $2 the second day, $4 the third day, and so on. He continues to double his savings each
zheka24 [161]

ANSWER

$16,384

EXPLANATION

From the question we have that,

Nadir saves $1 the first day of a month, $2 the second day, $4 the third day, and so on.

This forms a geometric sequence,

1,2,4,...

The first term of this sequence is

a = 1

The common ratio is

r =  \frac{2}{1}  =  \frac{4}{2}  = 2

The general term of a geometric sequence is given by the formula:

f(n) = a {r}^{n - 1}

To find the 15th term, we plug in a=1, r=2 and n=15.

f(15) = 1 {(2)}^{15 - 1}

f(15) =  {2}^{14}

f(15) = 16384

The amount he will save on the 15th day is $16,384

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3 years ago
What is the slope of the line shown below?
melisa1 [442]

Answer:

C. 2

Step-by-step explanation:

Gradient = \frac{change in y}{change in x}

Gradient = \frac{6}{3}

Gradient = 2

7 0
2 years ago
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Find the slope of each graph. Express the answer in simplest form.help ​
Licemer1 [7]

Answer:

I think the slope is, as a point on the graph, (0,4).

Step-by-step explanation:

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HELLOOOO HELP PLEASE
MA_775_DIABLO [31]

Answer:

2*log(x)+log(y)

Step-by-step explanation:

So, there are two logarithmic identities you're going to need to know.

<em>Logarithm of a power</em>:

   log_ba^c=c*log_ba

   So to provide a quick proof and intuition as to why this works, let's consider the following logarithm: log_ba=x\implies b^x=a

   Now if we raise both sides to the power of c, we get the following equation: (b^x)^c=a^c

   Using the exponential identity: (x^a)^c=x^{a*c}

    We get the equation: b^{xc}=a^c

    If we convert this back into logarithmic form we get: log_ba^c=x*c

    Since x was the basic logarithm we started with, we substitute it back in, to get the equation: log_ba^c=c*log_ba

Now the second logarithmic property you need to know is

<em>The Logarithm of a Product</em>:

    log_b{ac}=log_ba+log_bc

    Now for a quick proof, let's just say: x=log_ba\text{ and }y=log_bc

    Now rewriting them both in exponential form, we get the equations:

    b^x=a\\b^y=c

    We can multiply a * c, and since b^x = a, and b^y = c, we can substitute that in for a * c, to get the following equation:

    b^x*b^y=a*c

   Using the exponential identity: x^{a}*x^b=x^{a+b}, we can rewrite the equation as:

 

   b^{x+y}=ac

   taking the logarithm of both sides, we get:

   log_bac=x+y

   Since x and y are just the logarithms we started with, we can substitute them back in to get: log_bac=log_ba+log_bc

Now let's use these identities to rewrite the equation you gave

log(x^2y)

As you can see, this is a log of products, so we can separate it into two logarithms (with the same base)

log(x^2)+log(y)

Now using the logarithm of a power to rewrite the log(x^2) we get:

2*log(x)+log(y)

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I think it might be -76
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