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3 years ago

12

The first five terms of a sequence are 8, 27, 64, 125, and 216. Which of the following describes how to generate the next term i

n the sequence?

2 answers:

oksano4ka [1.4K]3 years ago

8 0

2^3, 3^3,4^3, 5^3, 6^3. The next term in the sequence would be 7^3 which is 343

maksim [4K]3 years ago

7 0

Answer:

the answer is A, or add 1 to the term number, then cube the result

Step-by-step explanation:

got it right on e2020

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vlada-n [284]

Answer:

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Step-by-step explanation:

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3 years ago

What is the value of the function at x=−3?<br><br> A) y = 0<br> B) y=−3<br> C) y = 4<br> D) y = 3

Alexus [3.1K]

Answer:

y=4

Step-by-step explanation:

From the graph we can see that-

when, x=-3 , y=4

That is , the graph passes through (-3 , 4)

So,

for x=-3 ,

we get y=4

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3 years ago

The triangle and rectangle

valentinak56 [21]

2x-4=3x-11 so add 4 to 11 and that’s -7 then subtract 2x from 3x so that’s 1x so 1x divides by -7 equals -7 so x=-7

7 0

3 years ago

Conjunction and disconjuction

maks197457 [2]

Step-by-step explanation:

When two statements are connected with an 'and,' you have a conjunction.

For conjunctions, both statements must be true for the compound statement to be true.

When the connector between two statements is "or," you have a disjunction. In this case, only one statement in the compound statement needs to be true for the entire compound statement to be true.

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2 years ago

Triangle $ABC$ has a right angle at $B$. Legs $\overline{AB}$ and $\overline{CB}$ are extended past point $B$ to points $D$ and

Lisa [10]

Answer:

Given :

ABC is a right triangle in which ∠ABC = 90°,

Also, Legs AB and CB are extended past point B to points D and E,

Such that,

$\angle EAC = \angle ACD = 90^{\circ}$

To prove :

$EB\times BD=AB\times BC$

Proof :

In triangles AEC and EBA,

∠EAC= ∠ABE ( right angles )

∠CEA = ∠AEB ( common angles )

By AA similarity postulate,

$\triangle AEC \sim \triangle EBA$ ,

Similarly,

$\triangle AEC \sim \triangle ABC$

$\implies\triangle EBA\sim \triangle ABC-----(1)$

Now, In triangles ADC and CBD,

∠ACD = ∠CBD ( right angles )

∠ADC= ∠BDC ( common angles )

By AA similarity postulate,

$\triangle ADC \sim \triangle CBD$ ,

Similarly,

$\triangle ADC \sim \triangle ABC$

$\implies \triangle CBD\sim \triangle ABC-----(2)$

From equations (1) and (2),

$\triangle EBA\sim \triangle CBD$

The corresponding sides of similar triangles are in same proportion,

$\frac{EB}{BC}=\frac{AB}{BD}$

$\implies EB\times BD=AB\times BC$

Hence, proved....

5 0

3 years ago