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natali 33 [55]
3 years ago
5

OLEASE DO THIS ITS EASY PKEAS EI NEED TO FINISH IT NOW PKEASE

Mathematics
1 answer:
KATRIN_1 [288]3 years ago
8 0
For x y chart 19.36.2.27.13
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Need help with 6,7,8,9,10,11,12,15
Ainat [17]

Answer:

6. 1.01 x -10^5

7. 6 x -10^14

8. 550,000

9. 60,700,000

10. 0.000204

11. 0.0004

12. 7,000,000,000,000 > 3,500,000,000

     7,000,000,000,000 (or 7 x 10^12) is greater by 2,000 times

15. 10^3 and 10^4

7 0
3 years ago
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What the rule for the transformation is<br>​
AleksandrR [38]
Answer: (x + 3, y - 4)

Explanation: The shape in the middle goes to the right three and down four to match the shape at the bottom
8 0
3 years ago
Multiply or Divide<br><br> (i need this for school tomorrow)
lions [1.4K]

Answer:

Ok so here are the simple rules of doing it (very easy) cause I’m not doing it all so . when multiplying a power with The same base keep the base but add the exponents. Dividing, keep the base (if their the same if not then its already simplified same with multiplication) but SUBTRACT the exponents. Also keep the parenthesis if it’s a negative number base.

I’ll do a few.

11) a^10.           11b) 5^4

12) (-2)^2.        

13) 10^2.          13b) s^6

14) -4s^5(t^6) <- [Im not sure of this one)

15) x^3(y^3)

7 0
3 years ago
7)PLEASE HELP WITH QUESTION. MARKING BRAINLIEST + POINTS.
yanalaym [24]
No triangle is formed

7 0
3 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
3 years ago
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