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snow_tiger [21]
3 years ago
15

What's the answers for this problem .) 5a + 3(6 + 7a)

Mathematics
2 answers:
Nataly [62]3 years ago
6 0
Let me I recommend you someone
KengaRu [80]3 years ago
5 0
5a+3(6+7a)
5a+18+21a
=26a+18
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Mr. Gardner is making 6 treat bags. He has 185 chocolate covered raisins to share evenly among the teat bags. How many chocolate
Hunter-Best [27]

Answer:

30 Chocolate covered raisins will be in each bag and 5 will be left over.

Step-by-step explanation:

In order to solve this, you must divide the number of Chocolate covered raisins by the number of bags you are making. 185 divided by 6 is 30 with 5 left over.

3 0
2 years ago
a theater is designed with 15 seats in the first row, 19 in the second, 23 in the third, and so on. if this seating pattern cont
marusya05 [52]

Answer:

<u>131 seats</u> are in the 30th row.

Step-by-step explanation:

The theater is designed with the first row there are 15 seats, in second row 19 seats and in the third row there are 23 seats.

Now, to find the number of seats in the 30th row.

So, we get the common difference(d) from the arithmetic sequence first:

19-15=4.

Thus, d=4.

So, the first tem a(1) = 15.

The number of last row (n) = 30.

Now, to get the number of seat in the 30th row we put formula:

a(n) = a(1) + d(n-1)

a(30)=15+4(30-1)

a(30)=15+4\times 29

a(30)=15+116

a(30)=131

Therefore, 131 seats are in the 30th row.

4 0
4 years ago
The one selected is wrong please help meee!
GREYUIT [131]

Answer:

  (a)  ΔARS ≅ ΔAQT

Step-by-step explanation:

The theorem being used to show congruence is ASA. In one of the triangles, the angles are 1 and R, and the side between them is AR. The triangle containing those angles and that side is ΔARS.

In the other triangle, the angles are 3 and Q, and the side between them is AQ. The triangles containing those angles and that side is ΔAQT.

The desired congruence statement in Step 3 is ...

  ΔARS ≅ ΔAQT

5 0
2 years ago
Determine the longest interval in which the given initialvalue problemis certainto have a unique twicedifferentiable solution. D
AlekseyPX

Answer:

t_o = 3, so solution exists on (0,4).  

Step-by-step explanation:

Use Theorem

Divide equation with t(t — 4).

y''+[3/(t-4)]*y'+ [4/t(t-4)]*y=2/t(t-4)

p(t)=3/t-4—> continuous on (-∞, 4) and (4,∞)

q(t) = 4/t(t-4) —> continuous on (-∞,0), (0,4) and (4, ∞)

g(t) = 2/t(t-4)—> continuous on (-∞, 0), (0,4) and (4,∞)

t_o = 3, so solution exists on (0,4).  

6 0
3 years ago
HELP I BEG OF YOU!!!!!!!!!!!
sergeinik [125]
Its 4 because its the gcf not the lcd
8 0
3 years ago
Read 2 more answers
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