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Wewaii [24]
4 years ago
10

Find the 11th term of the sequence 11, 19, 27, 35

Mathematics
1 answer:
iragen [17]4 years ago
7 0

Answer:

  91

Step-by-step explanation:

Each term is 8 more than the last, so you can write the terms of the sequence until you get there:

  11, 19, 27, 35, 43, 51, 59, 67, 75, 83, 91

__

Or, you can use the formula for the general term of an arithmetic sequence. You have first term a1=11 and common difference d=8.

  an = a1 +d(n -1)

  a11 = 11 + 8(11 -1) = 11 + 80

  a11 = 91

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20 POINTS AND BRAINLIEST
emmainna [20.7K]
To solve this equation we must first plug in 5 for x.

\frac{1}{3}<span> (5 - 4x)
               </span>↓ 
\frac{1}{3} (5 - 4 · 5)

The next step would be to multiply 4 by 5 to get rid of the parenthesis within our main parenthesis.

\frac{1}{3} (5 - 4 · 5)

\frac{1}{3} (5 - 20)

The next step would be to subtract both of the numbers within the parenthesis.

\frac{1}{3} (-15)

Now we must multiply.

- \frac{1}{3} · 15 =  - \frac{1×15}{3}

- \frac{15}{3}

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8 0
3 years ago
22/5 times 2 times 1/3
LUCKY_DIMON [66]

Answer:

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Step-by-step explanation:

4 0
3 years ago
Romb ma obwód równy 32 cm. Jakie jest pole tego rombu, jeżeli jego wysokość ma 3 cm?
Goshia [24]

Answer:

24 cm²

Step-by-step explanation:

Z zadanego powyżej pytania uzyskano następujące dane:

Obwód (obwód) = 32 cm

Wysokość (h) = 3 cm

Obszar rombu (A) =?

Następnie określimy długość boku rombu. Można to uzyskać w następujący sposób:

Obwód, czyli obwód (P) = 32 cm

Strona (y) =?

P = 4 s

32 = 4 s

Podziel obie strony przez 4

s = 32/4

s = 8 cm

Stąd długość boku rombu wynosi 8 cm

Na koniec określimy obszar rombu w następujący sposób:

Długość boków = 8 cm

Wysokość (h) = 3 cm

Obszar rombu (A) =?

A = sh

A = 8 × 3

A = 24 cm²

Dlatego obszar rombu jest

24 cm².

8 0
3 years ago
Show work please thanks
Tju [1.3M]
Be more specific on the first one
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3 years ago
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2.
you had nothing for 1. and I am not sure what you mean by 4.
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5 0
3 years ago
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