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Flauer [41]
3 years ago
9

What is the value of the 23rd term in the sequence 10, 8, 6, 4... ​

Mathematics
1 answer:
zalisa [80]3 years ago
7 0

Answer:

-34

Step-by-step explanation:

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PLS help due by 3pm
dimaraw [331]

Answer:

Step-by-step explanation:

6

6 0
4 years ago
Which unit is acceleration measured in?<br> A. m<br> B. m/h<br> C. m/s<br> D. m/s²
GrogVix [38]
Acceleration is how fast the velosity is changing: it's the rate of change of velocity per second. So that means one needs to divide velocity per time, and the unit will then be meters per second (velocity) per second:
answer <span>D. m/s² </span>
5 0
3 years ago
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Can someone plz help me with this
Ronch [10]
The line is right in the middle of 100 and 110 degrees so the answer should be 105 degrees. This tool is called a protractor. :)
3 0
3 years ago
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V is a vector space. Mark each statement True or False. Justify each answer.
SOVA2 [1]

Answer:

  • False
  • False
  • False
  • False
  • True

Step-by-step explanation:

A) False ; R^2 is not a two dimensional subspace of R^3 because Vector R^2 will have 2 entries while vector R^3 has 3 entries

B) False : The number of variable in the equation is not equal to the Dimension because dim of Null A represents the number of free variables

C) False : This is because you can span R^3 with an infinite set of vectors but that doesn't make it infinite dimensional subspace

D) False : This is only possible when S has exactly n elements/vectors

E) True : Since R^3 is a three dimensional subspace it is spanned with 3 vectors whom are linearly independent

6 0
3 years ago
I will give brainliest to best answer
lorasvet [3.4K]

That's a Congruent Triangle question.

The first thing that we should to analize is it: What figure is EFGH? The question say for us that \overline{EF} \cong \overline{GH} and \overline{EH} \cong \overline{GF}, therefore, EFGH is a parallelogram because its parallel sides are congruents (equals).

Now, we should to proof that \triangle EFH \cong \triangle GHF. The simbol "≅" means congruence, that is, \triangle EFH \cong \triangle GHF means that the triangles EFH and GHF are equals.

Let's proof:

\overline{HF} is a diagonal of the parallelogram EFGH. When we have a diagonal in a parallelogram, the opposite angles are congruents (look at the picture: the blue angles are congruent and the red angles are congruents too). Therefore, E\hat{F}H \cong F\hat{H}G and E\hat{H}F \cong H\hat{F}G.

Both triangle has a comum side: the diagonal \overline{HF}. The diagonal \overline{HF} is between two angles that we know that are congruents, Therefore, by the case ASA (Angle, Side, Angle), we proof that \triangle EFH \cong \triangle GHF.

7 0
3 years ago
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