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3 years ago

Sixty 11th graders were asked to select a favorite pattern out of 6 choices.

Mathematics

1 answer:

Airida [17]3 years ago

4 0

Answer: Graph D

Step-by-step explanation:

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$1.66 x 100 = ? I can't seem to find the correct answer and I must show my work

crimeas [40]

Answer:

166

Step-by-step explanation:

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3 years ago

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Please help if you do not know the answer don’t comment in the answer box just for points

Paul [167]

I don't know all of them but I can tell you that B and E are answers that you should check.

6 0

3 years ago

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What is 12-9×3×4+3 in three different ways

Maurinko [17]

1. 12-9×3×4+3=-93
2.3+4×3×9-12=99
3. 12-9+3×9×3×4=327

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3 years ago

Prove by mathematical induction that 1+2+3+...+n= n(n+1)/2 please can someone help me with this ASAP. Thanks

Iteru [2.4K]

Let

$P(n):\ 1+2+\ldots+n = \dfrac{n(n+1)}{2}$

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

$P(1):\ 1 = \dfrac{1\cdot 2}{2}=1$

So, the base case is ok. Now, we need to assume $P(n)$ and prove $P(n+1)$ .

$P(n+1)$ states that

$P(n+1):\ 1+2+\ldots+n+(n+1) = \dfrac{(n+1)(n+2)}{2}=\dfrac{n^2+3n+2}{2}$

Since we're assuming $P(n)$ , we can substitute the sum of the first n terms with their expression:

$\underbrace{1+2+\ldots+n}_{P(n)}+n+1 = \dfrac{n(n+1)}{2}+n+1=\dfrac{n(n+1)+2n+2}{2}=\dfrac{n^2+3n+2}{2}$

Which terminates the proof, since we showed that

$P(n+1):\ 1+2+\ldots+n+(n+1) =\dfrac{n^2+3n+2}{2}$

as required

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3 years ago

A student solves the following equation for all possible values of x:

bekas [8.4K]

The correct answer is: 
C) his solution for x is correct, but in order for 6 to be an extraneous solution, one denominator has to result in 0 when 6 is substituted for x.

Explanation: 
To solve this rational equation, we cross multiply: 
8*(x-4)=2*(x+2).

Using the distributive property, we have 
8*x-8*4=2*x+2*2;
8x-32=2x+4.

Subtract 2x from each side: 
8x-32-2x=2x+4-2x;
6x-32=4.

Add 32 to each side: 
6x-32+32=4+32;
6x=36.

Divide both sides by 6:
 $\frac{6x}{6}$ = $\frac{36}{6}$ ;
x=6.

Extraneous solutions are solutions that come about in the problem but are not valid solutions to the problem; the only values of x that would give this sort of answer are -2 and 4, since these are the two values that would make one of the denominators 0 (we cannot divide by 0).

6 0

2 years ago

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