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storchak [24]
4 years ago
7

How do you solve this problem?

Mathematics
1 answer:
kykrilka [37]4 years ago
5 0
You can multiply 8 times 9 to get 72 and then move the decimal over to the right to get the answer of 7.2
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What is the measure of ACE in the diagram below?<br><br> A. 46<br> B. 104<br> C. 29<br> D. 58
beks73 [17]

Answer:

C

Step-by-step explanation:

The secant- secant angle ACE is half the difference of the measures of the intercepted arcs, that is

∠ ACE = \frac{1}{2} (AE - BD ) = \frac{1}{2} (104 - 46)° = \frac{1}{2} × 58° = 29° → C

3 0
4 years ago
Read 2 more answers
#2) PLEASE HELP WITH QUESTION. MARKING BRAINLIEST + POINTS :}
djverab [1.8K]
An = -15 + 9(n - 1), since the first term is 15 and for each consecutive term we are adding 9 to the previous term
7 0
4 years ago
A.
Sophie [7]

Answer:

B.

Step-by-step explanation:

I graphed it but you could also use the y intercept to determine which function matches the graph

log(x+1)<u>+3</u>

the +3 at the end of the function makes the y-intercept positive 3

4 0
3 years ago
.A number consists of 3-digit numbers. The ones and the hundreds place digits are both divisible by 3. The tens and ones place d
Westkost [7]

Answer:

946

Step-by-step explanation:

xyz- the searched number where x depicts hundreds, y - tens and z describes ones. Both X and Z are divisible by 3 .Y and Z are divisible by 2

Z is divisible by 3 and divisible by 2, the only digit which is possible is 3*2=6

6-2=4 - tens digit Y

The hundreds place digit is 3*3=9

S0 946- is an answer

8 0
3 years ago
In each situation, write a recurrence relation, including base case(s), for the given function. briefly explain in words why thi
vova2212 [387]
A thumb-wrestling match requires two thumbs, so we can either suppose that we need at least two people (n\ge2), or allow one to thumb-wrestle one's self. In either case, we'd have t(1)=t(2)=1, so let's just say we need a minimum of two players.

If we add one more person to the set of players, then the first two people would need to play 1 additional match each. So t(3)=t(2)+2.

If we add one more person, then the first three people would again each have to play 1 more match with the new person. So t(4)=t(3)+3.

And so on, so that in general, the number of games needed for everyone to play exactly one match with everyone else is given recursively by

\begin{cases}t(2)=1\\t(n+1)=t(n)+n&\text{for }n\ge2\end{cases}
6 0
3 years ago
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