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3 years ago

Tell whether the sequence is arithmetic if it is identify the common difference -9,-17,-26,-33

Mathematics

2 answers:

aleksklad [387]3 years ago

8 0

No it is not, there is no common difference.

Kazeer [188]3 years ago

4 0

Keywords:

Arithmetic sequence, general term, common difference

For this case we have to identify if the given sequence is arithmetic, for this, we must take into account that by definition the form of the general term of an arithmetic sequence is given by:

$a_ {n} = a_ {1} + (n-1) * d$

Where:

$a_ {n}$ : It is the general term

$a_ {1}$ : Value of the first term

n: Number of terms

d: Difference

We test if the consecutive terms of the given sequence have the same difference.

$-17 - (- 9) = - 8\\-26 - (- 17) = - 9\\-33 - (- 26) = - 7$

It is observed that the difference is not common, therefore, the given sequence is not arithmetic.

Answer:

The given sequence has no common difference, it is not arithmetic.

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Answer:

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Step-by-step explanation:

(a) Given: The given points are $(1,2)$ and $(-3,8)$ .

To find: The parametric equation of line containing points $(1,2)$ and $(-3,8)$ .

We know that the parametric equation of line containing $(x_{1} ,y_{1} )$ and $(x_{2} ,y_{2} )$ is given by $(x(t),y(t))=(x_{1}+(x_{2}-x_{1})t,y_{1}+(y_{2}-y_{1})t)$ where $t$ ∈ $[0,1]$ .

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And, $y(t)=2+(8-2)t$

i.e, $y(t)=2+6t$

Hence, the required parametric equation of the line is $(x(t),y(t))=(1-4t,2+6t)$ .

(b) Given: The radius of circle is 3 and centre is $(4,1)$ .

To find: The parametric equation of circle with radius 3 and centre $(4,1)$ .

We know that parametric equation of circle with radius $r$ and centre $(h,k)$ is given by $F(t)= (x(t),y(t))$ where $x(t)=rcos(t)+h$ and $y(t)=rsin(t)+k$ .

Now, $x(t)=3cos(t)+4$

$y(t)=3sin(t)+1$

So, the parametric equation of circle having radius 3 and centre $(h,k)$ is $F(t)= (3cos(t)+4,3sin(t)+1)$ .

Hence, the required equation of the circle is $F(t)= (3cos(t)+4,3sin(t)+1)$ .

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2 years ago

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Answer:

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2 years ago

Angie and Becky each completed a separate proof to show that the measures of vertical angles AKG and HKB are equal. Who complete

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Vertical angles are angles which are opposite to each other and have equal measures. With respect to the given proof, Angie completed the proof incorrectly.

Two angles are said to be vertical when they are opposite to one another and have equal measures. However, two angles are supplementary if they add up to 180°.

So, comparing Angie's and Becky's proofs, it can be deduced that from Angies proof:

1. Segment GH intersects segment AB at K (1. Given)

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m∠GKB + m∠HKB = 180° (2. Definition of Supplementary Angles)

3. m∠AKG + m∠ GKB = m∠GKB + m∠HKB (3. Substitution Property)

4. m∠AKG = m∠HKB (4. Subtraction Property)

Note:

Step 2 should be Angle Addition Postulate, since:

m∠AKG + m∠GKB = 180°

m∠GKB + m∠HKB = 180°

So that;

m∠AKG + m∠GKB = m∠GKB + m∠HKB

Subtract m∠GKB from both sides to have;

m∠AKG = m∠HKB (which is the proof for step 4)

Therefore, Angie completed the poof incorrectly because of the properties of steps 2 and 4.

A diagram is herewith attached for more clarifications.

Visit: brainly.com/question/24529637

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