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3 years ago

12

For how many weeks would the class need ti sell cookies in order to earn $5000

1 answer:

Effectus [21]3 years ago

8 0

So we know that every 3 weeks they make $500 dollars.

We're trying to find out how many weeks will make $5000 dollars.

Let's start out simple.

3 weeks = $500.
5 x 10 is 50, so if we multiply 500 by 5, you get 5000.
Do the same thing on the other side. (3 weeks) This equals 30.

So it will take 30 weeks to make $5000 dollars.

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I need help....Does anybody know the answer?

Alex_Xolod [135]

Answer:

4/5

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8/15 = 8/15

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11/15 = 11/15

We see that the pattern is that the terms increase by 1/15 as they move up.

Thus, the next term in the sequence is 12/15, which can be simplified to 4/5.

Let me know if this helps!

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3 years ago

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3 years ago

In the radian measure, the angles of a 45-45-90 triangle are:

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45, 45, 90

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4 0

3 years ago

Read 2 more answers

A) Find the value of N if, Kn has 105 edges.

blagie [28]

The number of edges can be calculated from the number of vertices.

<em>There are 14 vertices for 105 edges</em>
<em>There are 200 vertices for 19900 edges</em>

The variable N is used to always represent the number of vertices.

So, we represent the edges as:

$E \to Edges$

<u />

<u>(a) The value of N for 105 edges</u>

The relationship between N and E is:

$E = \frac{N \times (N - 1)}{2}$

Substitute 105 for E

$105 = \frac{N \times (N - 1)}{2}$

Multiply through by 2

$210 = N \times (N - 1)$

$210 = N^2 - N$

Rewrite as:

$N^2 - N - 210 = 0$

Expand

$N^2 +14N - 15N - 210 = 0$

Factorize

$N(N +14) - 15(N + 14) = 0$

Factor out N + 14

$(N - 15) (N + 14) = 0$

Solve for N

$N = 15$ or $N = -14$

The number of vertices (N) cannot be negative. So:

$N = 15$

<u>(b) The value of N for 19900 edges</u>

We have:

$E = \frac{N \times (N - 1)}{2}$

Substitute 19900 for E

$19900 = \frac{N \times (N - 1)}{2}$

Multiply through by 2

$39800 = N \times (N - 1)$

$39800= N^2 - N$

Rewrite as:

$N^2 - N - 39800= 0$

Expand

$N^2 +199N - 200N - 39800= 0$

Factorize

$N(N +199) - 200(N + 199) = 0$

Factor out N + 199

$(N + 199) (N - 200) = 0$

Solve for N

$N = 200$ or $N = -199$

The number of vertices (N) cannot be negative. So:

$N = 200$

<em>Hence, there are 200 vertices for 19900 edges</em>

Read more about vertices and edges at:

brainly.com/question/22118318

7 0

3 years ago