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dexar [7]
4 years ago
6

Find the 15th term of the arithmetic sequence.

Mathematics
2 answers:
krek1111 [17]4 years ago
7 0
 The answer is
 c. 15a + 1
Aneli [31]4 years ago
5 0
It's all about matching patterns.

The coefficient of "a" in the given sequence is the term number (1, 2, 3, ...) and the constant remains constant at 1. Thus the 15th term can be expected to be ...
  c. 15a + 1
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There are 790 cal in 610 ounce bottles of apple juice how many calories are there in one 10 ounce bottle of apple juice
Akimi4 [234]
I would say 18 because 790-610= 180 and 180÷10= 18
7 0
3 years ago
Help me, please !!!!!
vova2212 [387]

Answer: 100

Hi!

ok so the Pythagorean theorem is

c=\sqrt{x} a^2+b^2=\sqrt{x} 8^2+6^2=10

so the other side is 10.

the square 8 is 8*8

and square 6 is 6*6

so the other square is 10*10=100.

so the area of the blank square is 100.

Step-by-step explanation:

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Thanks!

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4 0
3 years ago
at elm elementary school for every 5 students who walk to school there are 7 students who take the bus.if 420 students attend el
Nesterboy [21]
175 students walk to school. 7+5=12 and 420 divided by 12 is 35. I checked how I did it was correct because of 5*35+7*35=420. Then I found the product of 5*35 which is 175.
4 0
3 years ago
Allen has a 75 yard spool of yarn. He needs to cut the spool of yarn into 6 1/4
Gwar [14]

Answer:

B. 12 pieces

Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
In ΔIJK, the measure of ∠K=90°, KJ = 65, IK = 72, and JI = 97. What is the value of the cosine of ∠J to the nearest hundredth?
kow [346]

Answer:

\cos(J) = 0.67

Step-by-step explanation:

Given

\angle K = 90^o

KJ = 65

IK = 72

JI = 97

Required

\cos(J)

The question is illustrated with the attached image.

From the image, we have:

\cos(J) = \frac{KJ}{JI}

This gives:

\cos(J) = \frac{65}{97}

\cos(J) = 0.67010309278

\cos(J) = 0.67 --- approximated

7 0
3 years ago
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