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larisa [96]
4 years ago
10

Some help would be appreciated

Mathematics
1 answer:
Elis [28]4 years ago
3 0
She was driving an average of 51 mph.
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Verify the identity. tan x plus pi divided by two = -cot x
Misha Larkins [42]
The provided equation is not an identity
4 0
3 years ago
PLZ HELP URGENT GET THE BRAINLIEST AND MORE POINTS
lord [1]

Answer:

  • The 27th term is 22

Step-by-step explanation:

Let the first term is a and common difference is d.

<u>The nth term is:</u>

  • aₙ = a + (n - 1)d

<u>We have:</u>

  • a₆ = 15
  • a₉ = 16

<u>The difference of these terms is:</u>

  • (a + 8d) - (a + 5d) = 16 - 15
  • 3d = 1
  • d = 1/3

<u>Then the first term is:</u>

  • a + 5*1/3 = 15
  • a = 15 - 5/3 = 13 1/3

<u>The nth term equation is:</u>

  • aₙ = 13 1/3 + 1/3(n - 1) = 1/3n + 13

<u>If the nth term is 22, find n:</u>

  • 1/3n + 13 = 22
  • 1/3n = 22 - 13
  • 1/3n = 9
  • n = 9*3
  • n = 27
7 0
3 years ago
What greater -3.509 or -5.20
GalinKa [24]
-3.509 is greater because -3.509 is closer to 0 than -5.2.

Hope this helps!
5 0
4 years ago
Read 2 more answers
At the beginning of year 1, Bode invests $250 at an annual simple interest rate of 3%. He makes no deposits to or withdrawals fr
Dennis_Churaev [7]
The initial investment = $250
<span>annual simple interest rate of 3% = 0.03
</span>
Let the number of years = n
the annual increase = 0.03 * 250
At the beginning of year 1 ⇒ n = 1 ⇒⇒⇒ A(1) = 250 + 0 * 250 * 0.03 = 250

At the beginning of year 2 ⇒ n = 2 ⇒⇒⇒ A(2) = 250 + 1 * 250 * 0.03
At the beginning of year 3 ⇒ n = 3 ⇒⇒⇒ A(2) = 250 + 2 * 250 * 0.03
and so on .......
∴ <span>The formula that can be used to find the account’s balance at the beginning of year n is:
</span>
A(n) = 250 + (n-1)(0.03 • 250)
<span>At the beginning of year 14 ⇒ n = 14 ⇒ substitute with n at A(n)</span>
∴ A(14) = 250 + (14-1)(0.03*250) = 347.5

So, the correct option is <span>D.A(n) = 250 + (n – 1)(0.03 • 250); $347.50 </span>
4 0
3 years ago
Read 2 more answers
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