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attashe74 [19]
3 years ago
15

Which of the following expression cannot be factored out using difference of two square

Mathematics
1 answer:
Ratling [72]3 years ago
5 0

Answer:

d

Step-by-step explanation:

Factoring using difference of squares

a² - b² = (a - b)(a + b)

a

x² - 1

= x² - 1²

= (x - 1)(x + 1)

b

x² - y²

= (x - y)(x + y)

c

x² - 4

= x² - 2²

= (x - 2)(x + 2)

d

4 - 16y ← 16y is not a perfect square and so cannot be factored

If it was 4 - 16y² then it could be factored using difference of squares

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Which elements make up the stars​
leonid [27]

Answer:

Most stars have small amounts of heavy elements like carbon, nitrogen, oxygen and iron. But the shine from the stars comes from burning hydrogen into helium in their cores.

‍♀️ if this helps

7 0
3 years ago
I need help with the attached​
Aleksandr-060686 [28]

Answer: super sorry! i have no idea but im sure someone else knows (:

Step-by-step explanation:

sorry

8 0
2 years ago
HELP PLEASE!!! I need help with 94 if you could show the steps that would be very helpful!
aksik [14]
A combination is an unordered arrangement of r distinct objects in a set of n objects. To find the number of permutations, we use the following equation:

n!/((n-r)!r!)

In this case, there could be 0, 1, 2, 3, 4, or all 5 cards discarded. There is only one possible combination each for 0 or 5 cards being discarded (either none of them or all of them). We will be the above equation to find the number of combination s for 1, 2, 3, and 4 discarded cards.

5!/((5-1)!1!) = 5!/(4!*1!) = (5*4*3*2*1)/(4*3*2*1*1) = 5

5!/((5-2)!2!) = 5!/(3!2!) = (5*4*3*2*1)/(3*2*1*2*1) = 10

5!/((5-3)!3!) = 5!/(2!3!) = (5*4*3*2*1)/(2*1*3*2*1) = 10

5!/((5-4)!4!) = 5!/(1!4!) = (5*4*3*2*1)/(1*4*3*2*1) = 5

Notice that discarding 1 or discarding 4 have the same number of combinations, as do discarding 2 or 3. This is being they are inverses of each other. That is, if we discard 2 cards there will be 3 left, or if we discard 3 there will be 2 left.

Now we add together the combinations

1 + 5 + 10 + 10 + 5 + 1 = 32 choices combinations to discard.

The answer is 32.

-------------------------------

Note: There is also an equation for permutations which is:

n!/(n-r)!

Notice it is very similar to combinations. The only difference is that a permutation is an ORDERED arrangement while a combination is UNORDERED.

We used combinations rather than permutations because the order of the cards does not matter in this case. For example, we could discard the ace of spades followed by the jack of diamonds, or we could discard the jack or diamonds followed by the ace of spades. These two instances are the same combination of cards but a different permutation. We do not care about the order.

I hope this helps! If you have any questions, let me know :)








7 0
3 years ago
Please helppppppp ❤️
liberstina [14]

Answer:

Inverse of f exists.

Step-by-step explanation:

From the graph attached,

If we do the horizontal line test for the function graphed,

We find the function as one to one function.

In other words for every input value (x-value) there is a different output value.

Since, for one-to-one functions, inverse of the functions exist.

Therefore, the answer will be,

The inverse of 'f' exists.

7 0
3 years ago
Kent multiplies both sides of the equation below by an expression.
Korolek [52]
The steps below are presented in order to arrive to the value of k of the given equation. 
First, multiply both sides of the equation by the variable k since the left-hand side of the equation has it in the denominator. This will be,  
            (k + 12/ k)(k) = 8(k)

Then, we simplify,
             k + 12 = 8k
We then, subtract 8k to both sides of the equation,
           k - 8k + 12 = 8k - 8k

Simplifying,
                  -7k + 12 = 0

Then, subtract 12 from both sides of the equation and divide both sides by -7. This will us the final answer of,                  
                     k = 12/7
8 0
4 years ago
Read 2 more answers
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