All categories

2 years ago

What is the value of |3.25| ?

Mathematics

1 answer:

Gre4nikov [31]2 years ago

3 0

Answer:

3.25

Step-by-step explanation:

The answer is 3.25 because the absolute value of a number is basically that same number but it ALWAYS comes out positive.

Hope this helps :)

You might be interested in

Show the distributive property of multiplicat<br> ion to solve 4×18

AVprozaik [17]

Answer: 4(10 +8) = 72

Step-by-step explanation:

You want to distribute the 18 and make it smaller so you change it to 10 + 8. Then you do 4(10) + 4(8) = 40 + 32 = 72

8 0

3 years ago

Read 2 more answers

24(9) = 2x i need help with this

Degger [83]

Answer:

24×9 = 2x

216 = 2x

216÷2 = x

108 = x

x=108

4 0

3 years ago

Pam signed up for a new gym. She had to pay a $45 sign-up fee and will pay dues of $30 each month for 6 months. Which expression

erastovalidia [21]

Answer:

45 + (30*6)

Step-by-step explanation:

3 0

2 years ago

For any real number c, √² =<br> A. ²<br> B. cl<br> C. 1<br> D. C

aliina [53]

Answer:

answer D ( if i interpreted your question correctly )

Step-by-step explanation:

Sqrt (c^2) = √(c^2) = C

8 0

2 years ago

Prove the following by induction. In each case, n is apositive integer.<br> 2^n ≤ 2^n+1 - 2^n-1 -1.

frutty [35]

<h2>Answer with explanation:</h2>

We are asked to prove by the method of mathematical induction that:

$2^n\leq 2^{n+1}-2^{n-1}-1$

where n is a positive integer.

Let us take n=1

then we have:

$2^1\leq 2^{1+1}-2^{1-1}-1\\\\i.e.\\\\2\leq 2^2-2^{0}-1\\\\i.e.\\2\leq 4-1-1\\\\i.e.\\\\2\leq 4-2\\\\i.e.\\\\2\leq 2$

Hence, the result is true for n=1.

Let us assume that the result is true for n=k

i.e.

$2^k\leq 2^{k+1}-2^{k-1}-1$

Now, we have to prove the result for n=k+1

i.e.

<u>To prove:</u> $2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Let us take n=k+1

Hence, we have:

$2^{k+1}=2^k\cdot 2\\\\i.e.\\\\2^{k+1}\leq 2\cdot (2^{k+1}-2^{k-1}-1)$

( Since, the result was true for n=k )

Hence, we have:

$2^{k+1}\leq 2^{k+1}\cdot 2-2^{k-1}\cdot 2-2\cdot 1\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{k-1+1}-2\\\\i.e.\\\\2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-2$

Also, we know that:

$-2$

(

Since, for n=k+1 being a positive integer we have:

$2^{(k+1)+1}-2^{(k+1)-1}>0$ )

Hence, we have finally,

$2^{k+1}\leq 2^{(k+1)+1}-2^{(k+1)-1}-1$

Hence, the result holds true for n=k+1

Hence, we may infer that the result is true for all n belonging to positive integer.

i.e.

$2^n\leq 2^{n+1}-2^{n-1}-1$ where n is a positive integer.

6 0

3 years ago