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aleksley [76]
3 years ago
11

Round off 10256 to the nearest 1000​

Mathematics
2 answers:
lions [1.4K]3 years ago
7 0

Answer:

Step 1:10256 will fortunately have 17777 nearest

Step-by-step explanation:

Plss answer my Question

SashulF [63]3 years ago
6 0

Answer:

10,000

I  hope im right !!

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What would happen to the slope of the graph of y=2/3×-3 if the line were shifted 6 units up?
yaroslaw [1]

The slope wouldn't change, but the y-intercept would become 3 because you add -3 to 6

5 0
3 years ago
Given: LN⊥KM,KL≅ML Prove: ΔKLN≅ΔMLN
miss Akunina [59]

Answer:

The Proof is given below.

Step-by-step explanation:

Given:

LN⊥KM,

KL≅ML

To Prove:

ΔKLN≅ΔMLN

Proof:

In  Δ KLN and Δ MLN

KL ≅ ML     ....……….{Given i.e Hypotenuse }

LN ≅ LN     …………..{Reflexive Property}

∠ LNK ≅ ∠ LNM  ……….{ LN ⊥ KM i.e Measure of each angle is 90° given}

Δ KLN ≅ Δ MLN     ….{By Hypotenuse Leg Theorem} ....Proved

6 0
3 years ago
* The nth term of sequence A is 3n − 2 The nth term of sequence B is 10 − 2n Sally says there is only one number that is in both
Nina [5.8K]

Answer:

Sally is not right

Step-by-step explanation:

Given the two sequences which have their respective n^{th} terms as following:

Sequence A. 3n - 2

Sequence B. 10 - 2n

As per Sally, there exists only one number which is in both the sequences.

To find:

Whether Sally is correct or not.

Solution:

For Sally to be correct, we need to put the n^{th} terms of the respective sequences as equal and let us verify that.

3n-2=10-2n\\\Rightarrow 3n+2n=10+2\\\Rightarrow 5n=12\\\Rightarrow n = \dfrac{12}{5}

When we talk about n^{th} terms, n here is a whole number not a fractional number.

But as per the statement as stated by Sally n is a fractional number, only then the two sequences can have a number which is in the both sequences.

Therefore, no number can be in both the sequences A and B.

Hence, Sally is not right.

7 0
3 years ago
What is the value of the expression 30 - 9+ 3 - (12 = 4)?
Flauer [41]

Answer:

21

Step-by-step explanation:

30 - 9 +3

30 - 12

18

18 - 12=4

      12/4 = -3

18 - -3

21

8 0
2 years ago
Three vectors A, B, C in three-dimensional space satisfy the following properties
wlad13 [49]

Answer:

\frac{7}{8}\pi

Step-by-step explanation:

observe

||a–b+c|| = ||a+b+c||

(a-b+c)² = (a+b+c)²

(a+b+c)² – (a-b+c)² = 0

((a+b+c)+(a-b+c))((a+b+c)–(a-b+c)) = 0

(2a+2c)(2b) = 0

(a+c)b = 0

a•b + c•b = 0

||a||×||b||×cos(π/8) + ||c||×||b||×cos(θ) = 0

\cos(\theta)=-\frac{||a||\times ||b|| \times \cos(\frac{\pi}{8})}{||c||\times ||b||}=-\cos(\frac{\pi}{8})\\ \theta=\frac{7}{8}\pi

3 0
4 years ago
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