Find the common difference of the sequence shown. 1/2, 1/4, 0, ... -1/8 -1/4 -1/2

1 answer:

8 0

Common difference in Arithmetic sequences is defined as the fixed amount <span>referring to the fact that the difference between two successive terms yields the constant value that was added. In order to know the common difference of the given sequence above, we just have to deduct 1/2 from 1/4 and the answer is -1/4. Therefore, the correct answer would be the third option. Hope this answer helps.</span>

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PLEASE HELP?!

Anon25 [30]

The middle point of BC is (0,2), the midpoint of CD is (1,0)
to prove two lines are parallel, prove their slopes are the same.
slope of BD: m=rise/run=4/-2=-2 (the run is negative because to get to B from D on the grid, you have to move from right to left, then upward. If the horizontal move is from left to right, the run is positive)
slope of EG: m=rise/run=2/-1=-2
EG and BD have the same slope, so they are parallel.

3 0

3 years ago

After proving two triangles are congruent, then one can prove parts of a triangle are congruent using the reasoning CPCTC.

KatRina [158]

Answer:

The answer is True because CPCTC stands for Corresponding Parts of Congruent Triangles are Congruent.

7 0

3 years ago

Find all solutions of the equation: 2cos^2x-cosx=1

Art [367]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166243

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Solve the trigonometric equation:

$\mathsf{2\,cos^2\,x-cos\,x=1}\\\\ \mathsf{2\,cos^2\,x-cos\,x-1=0}$

Make a substitution:

$\mathsf{cos\,x=t\qquad (-1\le t\le 1)}$

and the equation becomes

$\mathsf{2t^2-t-1=0}$

Rewrite conveniently – t as + t – 2t, and then factor the left-hand side by grouping:

$\mathsf{2t^2+t-2t-1=0}\\\\ \mathsf{t\cdot (2t+1)-1\cdot (2t+1)=0}$

Factor out 2t + 1:

$\mathsf{(2t+1)\cdot (t-1)=0}\\\\ \begin{array}{rcl} \mathsf{2t+1=0}&~\textsf{ or }~&\mathsf{t-1=0}\\\\ \mathsf{2t=1}&~\textsf{ or }~&\mathsf{t=1}\\\\ \mathsf{t=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{t=1} \end{array}$

Substitute back for t = cos x:

$\begin{array}{rcl}\mathsf{cos\,x=\dfrac{\,1\,}{2}}&~\textsf{ or }~&\mathsf{cos\,x=1}\\\\ \mathsf{cos\,x=cos\,60^\circ}&~\textsf{ or }~&\mathsf{cos\,x=cos\,0} \end{array}$

Therefore,

$\begin{array}{rcl} \mathsf{x=\pm\,60^\circ+k\cdot 360^\circ}&~\textsf{ or }~&\mathsf{cos\,x=0+k\cdot 360^\circ} \end{array}$

where k is an integer.

Solution set:

$\mathsf{S=\left\{x\in\mathbb{R}:~~x=-\,60^\circ+k\cdot 360^\circ~~or~~x=60^\circ+k\cdot 360^\circ~~or~~x=k\cdot 360^\circ,~~k\in\mathbb{Z}\right\}}$

I hope this helps. =)

3 0

3 years ago

Which monomial is a perfect cube? <br> 16x6<br> 27x8 <br> 32x12 <br> 64x6

miss Akunina [59]

The fastest way is to just use a scientific calculator.

To do it without a calculator, use prime factorization.

I'll only do the second one which is the answer.

27 x 8

= 3^3 x 2^3 = 6^3

Which is a perfect cube.

5 0

3 years ago

Read 2 more answers

WILL GIVE BRAINLIEST!!

dexar [7]

Peter reflecting trapezoid ABCD across the y-axis would not change the degree measurement of angle A

The degree measurement of angle A is 115 degrees

<h3>How to determine the degree measurement of angle A?</h3>

From the question, we have:

A = 115 degrees

B = 65 degrees

The transformation is a reflection across the y-axis

Reflection is a rigid transformation; and it does not change the angle measure or side lengths.

After the transformation; we have:

A = 115 degrees

B = 65 degrees

Hence, the degree measurement of angle A is 115 degrees