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Marina86 [1]
3 years ago
5

Lainey bought a set of 20 markers for $6 dollar sign, 6. What is the cost of 1 marker?

Mathematics
2 answers:
marin [14]3 years ago
5 0
Hello!

To find the price of one marker you divide the price paid by the amount of markers bought

6 / 20 = 0.30

A marker cost $0.30

The answer is $0.30

Hope this helps!
iVinArrow [24]3 years ago
3 0

Answer:

$0.30

Step-by-step explanation:

Lainey bought a set of 20 markers.

Since the price of  the set of 20 markers = $6.00

To calculate the price of 1 marker, we will divide $6 by 20.

Therefore, the price of 1 marker = \frac{6}{20}

                                                     = $0.30

The cost of 1 marker would be $0.30.

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3x + 7 equal 13x equal -2X equal to 2X equals 5
lianna [129]

3x + 7 = 13

     -7      -7

     3x = 6

     /3     /3

       x=2

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3 years ago
What is the answer to 9÷2/3
Karo-lina-s [1.5K]

Answer:

27/2

Step-by-step explanation:

9 : 2/3 =

9 * 3/2 =

(3*9=27)

27/2 Is your answer



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Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
juin [17]

Answer:

The Taylor series of f(x) around the point a, can be written as:

f(x) = f(a) + \frac{df}{dx}(a)*(x -a) + (1/2!)\frac{d^2f}{dx^2}(a)*(x - a)^2 + .....

Here we have:

f(x) = 4*cos(x)

a = 7*pi

then, let's calculate each part:

f(a) = 4*cos(7*pi) = -4

df/dx = -4*sin(x)

(df/dx)(a) = -4*sin(7*pi) = 0

(d^2f)/(dx^2) = -4*cos(x)

(d^2f)/(dx^2)(a) = -4*cos(7*pi) = 4

Here we already can see two things:

the odd derivatives will have a sin(x) function that is zero when evaluated in x = 7*pi, and we also can see that the sign will alternate between consecutive terms.

so we only will work with the even powers of the series:

f(x) = -4 + (1/2!)*4*(x - 7*pi)^2 - (1/4!)*4*(x - 7*pi)^4 + ....

So we can write it as:

f(x) = ∑fₙ

Such that the n-th term can written as:

fn = (-1)^{2n + 1}*4*(x - 7*pi)^{2n}

6 0
3 years ago
Which real numbers are zeros of the function?
monitta
<h3><u>The roots are -1/2, 0, 2, and 3.</u></h3>

Let's trying factoring this polynomial.

We can factor an x out of each term to start.

x(2x^3 - 9x^2 + 7x + 6)

We now know one of the roots is going to be zero.

Using the rational roots theorem, and the remainder theorem, we can try to find some more roots that way.

Factors of 6: 1, 2, 3, 6.

Factors of 2: 1

Possible rational roots: +/-1/2, +/-1, +/-2, +/-3/2, +/-3, +/-6

Using the remainder theorem, we can plug these values into the polynomial, and if we get a remainder of zero, we know it's a root.

-1/2(2(-1/2)^3 - 9(-1/2)^2 + 7(-1/2) + 6) = 0

Our first root is -1/2.

We can successfully factor a -1/2 out of this polynomial.

After diving the polynomial by -1/2, we're left with: 2x^2 - 10x + 12.

We can now try using the AC method to get our last two roots.

First, we can factor this polynomial to simplify it.

Divide all terms by 2.

2(x^2 - 5x + 6)

Now let's try using the AC method.

The digits -3 and -2 satisfy the criteria.

2(x - 3)(x - 2)

We now have all of our roots: 0, -1/2, 2, and 3.


6 0
3 years ago
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