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sineoko [7]
2 years ago
5

Can somebody help with this question please ?

Mathematics
1 answer:
Tom [10]2 years ago
6 0
It is DF and DE. Hope I helped.

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What are three numbers that are greater than 543,000 but less than 544,000
Bumek [7]
543,356 and 543,456 so it is pretty simple right
8 0
2 years ago
Help me on this please !
inna [77]

Answer:

24.08

Step-by-step explanation:

We use the Pythagorean theorem and we get (16)^2 + (18)^2 = (x)^2

256 + 324 = x^2

x^2 = 580

x is about 24.08.

7 0
2 years ago
The algebraic expression of The sum of four times a number and -2 and six times a number and 3
riadik2000 [5.3K]

Hey!

The sum of four times a number and -2

Let us take the number as ' x '

( 4x - 2 )

Six times a number and 3

( 6x + 3 )

The sum = ( 4x - 2 ) + ( 6x + 3 )

Hope helps!

6 0
3 years ago
The following are scores for students in Ms. Kennedy's math class.What is the range of the scores?
Eddi Din [679]

Answer:

The answer will be D. 181

Step-by-step explanation:

603 - 422 = 181

<em>Please mark me as the brainliest.</em>

3 0
2 years ago
Find the vector that has the same direction as 3, 2, −6 but has length 2.
Nata [24]

Answer:

The vector is \vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right).

Step-by-step explanation:

We can determine the equivalent vector (\vec r), dimensionless, by means of the following formula:

\vec r = \frac{\vec u}{\|\vec u\|} \cdot \|\vec r\| (1)

Where:

\vec u - Original vector, dimensionless.

\|\vec u\| - Norm of the original vector, dimensionless.

\|\vec r\| - Norm of the new vector, dimensionless.

The norm of the original vector is determined by the following definition:

\|\vec u\| = \sqrt{\vec u\,\bullet \,\vec u} (2)

If we know that \vec u = (3, 2, -6), then the norm of the original vector is:

\|\vec u\| = \sqrt{(3)\cdot (3)+(2)\cdot (2)+(-6)\cdot (-6)}

\|\vec u\| = 7

If we know that \|\vec r\| = 2, then the new vector is:

\vec r = \frac{2}{7}\cdot (3,2,-6)

\vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right)

The vector is \vec r = \left(\frac{6}{7},\frac{4}{7},-\frac{12}{7}\right).

7 0
3 years ago
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