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UkoKoshka [18]
3 years ago
11

Need help I don’t know the answer to this.

Mathematics
1 answer:
Bond [772]3 years ago
8 0

Answer:

what grade is this

Step-by-step explanation:

I'm very confused is this algebra?

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a store has a sale on books. The price is 17.55 for one book, 16.70 each for 2 books, 15.85 each for 3 books, and $15 each for 4
svetoff [14.1K]
Do miuness 5 10 15 so now muiness 2o cent for 5 6 is 25 7 is 30 so muiness 3o cents is 14 dollers and 70 cents
7 0
3 years ago
Read 2 more answers
Can someon tell me which one it is
podryga [215]

The correct answer is second one

5 0
3 years ago
Which expression is equivalent to -
dedylja [7]
x^\frac{4}{3} x^ \frac{2}{3}Lets get started :)

Referring to the picture I have uploaded might help you
We have the same base, which in this case is x. 

When multiplication occurs in the same base, we add the powers
x^\frac{4}{3} x^ \frac{2}{3}
   ↓ (I am going to perform the problem separately and then give the final answer as the power to x)
\frac{4}{3} + \frac{2}{3} = \frac{6}{3} = 2
  ↓
x²

(x^2)^ \frac{1}{3}

The power of a power should be performed by multiplying the powers together.
2 x \frac{1}{3} = \frac{2}{3}

x^\frac{2}{3}

Your final answer will be the second option



4 0
3 years ago
If p(x) = x2 – 1 and q(x) = 5(x-1) which expression is equivalent to (p – q)(x)?
lilavasa [31]

Answer:

Option C is correct.

The expression which is equivalent to (p-q)(x)  is, x^2 -1 - 5(x-1)

Step-by-step explanation:

Given: p(x) = x^2 -1 and q(x) = 5(x-1)

To find the expression which is equivalent to (p-q)(x):

here p and q both are function assuming;

we can write the expression :

(p-q)(x) = p(x) - q(x) = x^2 -1 - 5(x-1)

therefore, the expression which is equivalent to (p-q)(x)  is;   x^2 -1 - 5(x-1)

3 0
4 years ago
Read 2 more answers
Determine whether the series is convergent or divergent. 1 2 3 4 1 8 3 16 1 32 3 64 convergent divergent Correct:
Vanyuwa [196]

Answer:

This series diverges.

Step-by-step explanation:

In order for the series to converge, i.e. \lim_{n \to \infty} a_n =A it must hold that for any small \epsilon>0, there must exist n_0\in \mathbb{N} so that starting from that term of the series all of the following terms satisfy that  |a_n-A|n_0 .

It is obvious that this cannot hold in our case because we have three sub-series of this observed series. One of them is a constant series with a_n=1 , the other is constant with a_n=3 , and the third one has terms that are approaching infinity.

Really, we can write this series like this:

a_n=\begin{cases} 1 \ , \ n=4k+1, k\in \mathbb{N}_0\\ 2^{k}\ , \ n=2k, k\in \mathbb{N}_0\\3\ , \ n=4k+3, k\in \mathbb{N}_0\end{cases}

If we  denote the first series as b_n=1, we will have that \lim_{k \to \infty} b_k=1.

The second series is denoted as c_k=2^k and we have that \lim_{k \to \infty} c_k=+\infty.

The third sub-series d_k=3 is a constant series and it holds that \lim_{k \to \infty} d_k=3.

Since those limits of sub-series are different, we can never find such n_0\\ so that every next term of the entire series is close to one number.

To make an example, if we observe the first sub-series if follows that A must be equal to 1. But if we chose \epsilon =1, all those terms associated with the third sub-series will be out of this interval (A-1, A+1)=(0, 2).

Therefore, the observed series diverges.

5 0
4 years ago
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