PLEASE HELP ASAP!! sin ∠E = A) 20/21 B) 20/29 C) 21/29 D) 21/20

Write the explicit formula that represents the geometric sequence -2, 8, -32, 128

Jet001 [13]

The Geometric Sequence Formula:

$a_n=a_1r^{n-1}$

We have:

$a_1=-2,\ a_2=8,\ a_3=-32,\ a_4=128,\ ...$

$d=\dfrac{a_{n+1}}{a_n}\to d=\dfrac{8}{-2}=-4$

Substitute:

$a_n=-22\cdot(-4)^{n-1}=-22\cdot(-4)^n\cdot(-4)^{-1}=-22\cdot(-4)^n\cdot\left(-\dfrac{1}{4}\right)=5.5(-4)^n$

7 0

3 years ago

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Prove the following limit. lim x → 5 3x − 8 = 7 SOLUTION 1. Preliminary analysis of the problem (guessing a value for δ). Let ε

Temka [501]

$\displaystyle\lim_{x\to5}3x-8=7$

means to say that for any given $\varepsilon>0$ , we can find $\delta$ such that anytime $|x-5|$ (i.e. the whenever $x$ is "close enough" to 5), we can guarantee that $|(3x-8)-7|$ (i.e. the value of $3x-8$ is "close enough" to the limit value).

What we want to end up with is

$|(3x-8)-7|=|3x-15|=3|x-5|$

Dividing both sides by 3 gives

$|x-5|$

which suggests $\delta=\dfrac\varepsilon3$ is a sufficient threshold.

The proof itself is essentially the reverse of this analysis: Let $\varepsilon>0$ be given. Then if

$|x-5|$

and so the limit is 7. QED

8 0

3 years ago

Complete the missing parts of the paragraph proof. Draw a perpendicular from P to AB. Label the intersection C. We are given tha

zvonat [6]

1) We are given that PA = PB, so PA ≅ PB by the definition of the radius.

When you draw a perpendicular to a segment AB, you take the compass, point it at A and draw an arc of size AB, then you do the same pointing the compass on B. Point P will be one of the intersections of those two arcs. Therefore PA and PB correspond to the radii of the arcs, which were taken both equal to AB, therefore they are congruent.

2) We know that angles PCA and PCB are right angles by the definition of perpendicular.

Perpendicularity is the relation between two lines that meet at a right angle. Since we know that PC is perpendicular to AB by construction, ∠PCA and ∠PCB are right angles.

3) PC ≅ PC by the reflexive property congruence.

The reflexive property congruence states that any shape is congruent to itself.

4) So, triangle ACP is congruent to triangle BCP by HL, and AC ≅ BC by CPCTC (corresponding parts of congruent triangles are congruent).

CPCTC states that if two triangles are congruent, then all of the corresponding sides and angles are congruent. Since ΔACP ≡ ΔBCP, then the corresponding sides AC and BC are congruent.

5) Since PC is perpendicular to and bisects AB, P is on the perpendicular bisector of AB by the definition of the perpendicular bisector.

The perpendicular bisector of a segment is a line that cuts the segment into two equal parts (bisector) and that forms with the segment a right angle (perpendicular). Any point on the perpendicular bisector has the same distance from the segment's extremities. PC has exactly the characteristics of a perpendicular bisector of AB.

6 0

3 years ago

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Factor by grouping:

BARSIC [14]

$(x-y)^2=x^2-2xy+y^2\\---------------------------\\\\4-12r+9r^2=2^2-2\cdot2\cdot3r+(3r)^2=(2-3r)^2$

$other\ method:\\\\4-12r+9r^2=4-6r-6r+9r^2=2(2-3r)-3r(2-3r)\\\\=(2-3r)(2-3r)=(2-3r)^2$

3 0

3 years ago

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In circle F with m \angle EFG= 122m∠EFG=122 and EF=16EF=16 units, find the length of arc EG. Round to the nearest hundredth.

kow [346]

Given:

In circle F, $m\angle EFG=122^\circ,\ EF=16$ units.

To find:

The length of arc EG.

Solution:

It is given that, $m\angle EFG=122^\circ,\ EF=16$ units. It means the central angle of arc EG is 122 degrees and the radius of the circle F is 16 units.

Formula for arc length is:

$s=2\pi r\times \dfrac{\theta}{360^\circ }$