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2 years ago

Find the 120th term, given the first term and the common difference. a =52 and d=12 1

1 answer:

6 0

The first term is 52, the second term is 52+(1*12), the third term is 52+(2*12), the fourth term is 52+(3*12)... the 120th term is 52+((120-1)*12)
Therefore the 120 term is: 1480

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ad-work [718]

Answer:

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Step-by-step explanation:

6 0

2 years ago

PLS HELP ONLINE SCHOOL!! I DONT GET IT

lapo4ka [179]

Answer:

1:1

Step-by-step explanation:

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8 0

3 years ago

Help.? geometry basics homework 2: segment addition postulate

Neporo4naja [7]

The value of x is 3 given that ST = 8x + 11 and TU = 12x-1

The point that bisects a line divides the line into two equal parts

If T is the midpoint of SU, the following are true:

ST = TU
ST + TU = SU

Given the following

ST = 8x + 11

TU = 12x-1

Since ST = TU

8x+11 = 12x-1

8x - 12x = -1 - 11

-4x = -12

x = 12/4

x = 3

Hence the value of x is 3 given that ST = 8x + 11 and TU = 12x-1

Learn more here: brainly.com/question/17204733

5 0

2 years ago

Read 2 more answers

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Lana71 [14]

Answer:

(a) -2-(-9)=-2+9

(b) -2-9=-11

6 0

2 years ago

For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

Nastasia [14]

Answer:

$k=0$ and $k=3$ .

Step-by-step explanation:

To this relation be a function, the horizontal coordinates can't be equal. So, let's find the value of $k$ that makes those coordinates equal.

$k^{2}=3k\\ k^{2}-3k=0\\k(k-3)=0\\$

Using the null factor property, we have

$k=0\\k-3=0 \implies k=3$

Therefore, the given relation is not a function when $k=0$ and $k=3$ .

7 0

2 years ago