All categories

3 years ago

Is the open sentence 3z = 2z +5 true or false when z=5?

1 answer:

8 0

If 3z=2z+5 and z=5, then you’d just plug the 5 in.

3(5)=2(5)+5 —> 15=10+5 —> 15=15

This statement is TRUE because 15 does equal 15.

I hope this helps. :)

You might be interested in

Please help I’ll mark you as brainliest if correct

Dominik [7]

Answer:

25%

Step-by-step explanation:

5 0

2 years ago

What are the 3 different forms of a quadratic function and what do the variables in each represent?

Likurg_2 [28]

Answer:

general, factored, and vertex form.

general: y=ax^2+b^2+c

Factored: y=(x+a)(x+b)

Vertex: y=a(x-h)^2+k

Step-by-step explanation:

8 0

3 years ago

How many terms of the arithmetic sequence {1,22,43,64,85,…} will give a sum of 2332? Show all steps including the formulas used

MA_775_DIABLO [31]

There's a slight problem with your question, but we'll get to that...

Consecutive terms of the sequence are separated by a fixed difference of 21 (22 = 1 + 21, 43 = 22 + 21, 64 = 43 + 21, and so on), so the n-th term of the sequence, a (n), is given recursively by

• a (1) = 1

• a (n) = a (n - 1) + 21 … … … for n > 1

We can find the explicit rule for the sequence by iterative substitution:

a (2) = a (1) + 21

a (3) = a (2) + 21 = (a (1) + 21) + 21 = a (1) + 2×21

a (4) = a (3) + 21 = (a (1) + 2×21) + 21 = a (1) + 3×21

and so on, with the general pattern

a (n) = a (1) + 21 (n - 1) = 21n - 20

Now, we're told that the sum of some number N of terms in this sequence is 2332. In other words, the N-th partial sum of the sequence is

a (1) + a (2) + a (3) + … + a (N - 1) + a (N) = 2332

or more compactly,

$\displaystyle\sum_{n=1}^N a(n) = 2332$

It's important to note that N must be some positive integer.

Replace a (n) by the explicit rule:

$\displaystyle\sum_{n=1}^N (21n-20) = 2332$

Expand the sum on the left as

$\displaystyle 21 \sum_{n=1}^N n-20\sum_{n=1}^N1 = 2332$

and recall the formulas,

$\displaystyle\sum_{k=1}^n1=\underbrace{1+1+\cdots+1}_{n\text{ times}}=n$

$\displaystyle\sum_{k=1}^nk=1+2+3+\cdots+n=\frac{n(n+1)}2$

So the sum of the first N terms of a (n) is such that

21 × N (N + 1)/2 - 20N = 2332

Solve for N :

21 (N ² + N) - 40N = 4664

21 N ² - 19 N - 4664 = 0

Now for the problem I mentioned at the start: this polynomial has no rational roots, and instead

N = (19 ± √392,137)/42 ≈ -14.45 or 15.36

so there is no positive integer N for which the first N terms of the sum add up to 2332.

4 0

2 years ago

Name 3 numbers between 0.33 and 0.34

Tamiku [17]

0.331

0.332

0.333

There you go.

3 0

3 years ago

Read 2 more answers

17. Evaluate each pair of expressions. a. (-3)-8 and -3-8 b. (-3)-9 and -3-9

Semmy [17]

A = -11
B= -12

This is because of how calculations work in negatives. Because they are both negative they work similarly to addition

5 0

3 years ago