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SVETLANKA909090 [29]
3 years ago
15

Write an explicit formula for an, the nth term of the sequence 11,2,-7

Mathematics
1 answer:
abruzzese [7]3 years ago
8 0

Answer:

a_{n} = 20 - 9n

Step-by-step explanation:

There is a common difference d between consecutive terms, that is

d = 2 - 11 = - 7 - 2 = - 9

This indicates the sequence is arithmetic with nth term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 11 and d = - 9 , then

a_{n} = 11 - 9(n - 1) = 11 - 9n + 9 = 20 - 9n

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Sonya’s average so far is 84 for 3 math tests. What must she get on the fourth test to have an average of 88
ioda

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she will need to get a 100 on the fourth test

Step-by-step explanation:

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2 years ago
Find the measure of angle 1
kolbaska11 [484]

Answer:

44°

Step-by-step explanation:

8x - 4 = 6x + 8 (This is because they are corresponding angles)

8x - 6x = 8 + 4

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3 years ago
5^(-x)+7=2x+4 This was on plato
Setler79 [48]

Answer:

Below

I hope its not too complicated

x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

Step-by-step explanation:

5^{\left(-x\right)}+7=2x+4\\\\\mathrm{Prepare}\:5^{\left(-x\right)}+7=2x+4\:\mathrm{for\:Lambert\:form}:\quad 1=\left(2x-3\right)e^{\ln \left(5\right)x}\\\\\mathrm{Rewrite\:the\:equation\:with\:}\\\left(x-\frac{3}{2}\right)\ln \left(5\right)=u\mathrm{\:and\:}x=\frac{u}{\ln \left(5\right)}+\frac{3}{2}\\\\1=\left(2\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)-3\right)e^{\ln \left(5\right)\left(\frac{u}{\ln \left(5\right)}+\frac{3}{2}\right)}

Simplify\\\\\mathrm{Rewrite}\:1=\frac{2e^{u+\frac{3}{2}\ln \left(5\right)}u}{\ln \left(5\right)}\:\\\\\mathrm{in\:Lambert\:form}:\quad \frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}

\mathrm{Solve\:}\:\frac{e^{\frac{2u+3\ln \left(5\right)}{2}}u}{e^{\frac{3\ln \left(5\right)}{2}}}=\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}:\quad u=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)\\\\\mathrm{Substitute\:back}\:u=\left(x-\frac{3}{2}\right)\ln \left(5\right),\:\mathrm{solve\:for}\:x

\mathrm{Solve\:}\:\left(x-\frac{3}{2}\right)\ln \left(5\right)=\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right):\\\quad x=\frac{\text{W}_0\left(\frac{\ln \left(5\right)}{2e^{\frac{3\ln \left(5\right)}{2}}}\right)}{\ln \left(5\right)}+\frac{3}{2}

3 0
3 years ago
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