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stellarik [79]
4 years ago
15

Q ÷ 2 -qp ÷ 6; let p = -6 and q = 10

Mathematics
1 answer:
ozzi4 years ago
5 0

Answer:

15

Step-by-step explanation:

1. Plug in the values.

q=10

p=-6

10÷2-(10)(-6)÷6

2. Use PEMDAS.

10÷2+60÷6

5+10

15

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D. multiplication property

Step-by-step explanation

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Use sigma notation to represent the sum of the first seven terms of the following sequence -4,-6,-8.....
lina2011 [118]

Answer:Answer:

\sum\left {{7} \atop {1}} \right -n(3+n)

Step-by-step explanation:

Given the sequence -4,-6,-8..., in order to get sigma notation to represent the sum of the first seven terms of the sequence, we need to first calculate the sum of the first seven terms of the sequence as shown;

The sum of an arithmetic series is expressed as S_n = \frac{n}{2}[2a+(n-1)d]

n is the number of terms

a is the first term of the sequence

d is the common difference

Given parameters

n = 7, a = -4 and d = -6-(-4) = -8-(-6) = -2

Required

Sum of the first seven terms of the sequence

S_7 = \frac{7}{2}[2(-4)+(7-1)(-2)]\\\\S_7 =  \frac{7}{2}[-8+(6)(-2)]\\\\S_7 =  \frac{7}{2}[-8-12]\\\\\\S_7 = \frac{7}{2} * -20\\\\S_7 = -70

The sum of the nth term of the sequence will be;

S_n = \frac{n}{2}[2(-4)+(n-1)(-2)]\\\\S_n = \frac{n}{2}[-8+(-2n+2)]\\\\S_n = \frac{n}{2}[-6-2n]\\\\S_n =  \frac{-6n}{2} -  \frac{2n^2}{2}\\S_n = -3n-n^2\\\\S_n = -n(3+n)

The sigma notation will be expressed as \sum\left {{7} \atop {1}} \right -n(3+n). <em>The limit ranges from 1 to 7 since we are to  find  the sum of the first seven terms of the series.</em>

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3 years ago
Lara chose a square number. She rounds it to the nearest hundred. Her answer is 200. What are all the possible square number she
gogolik [260]
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4 years ago
Differentiating Exponential functions In Exercise,find the derivative of the function. See Example 2 and 3.
Citrus2011 [14]

Answer:

f'(x)=-5ae^{ax}=a f(x), where a be a constant.

Step-by-step explanation:

Note: The given functions is a constant function because variable term is missing.

Consider the given function is

f(x)=-5e^{ax}

where a be a constant.

We need to find the derivative of the function.

Differentiate with respect to x.

f'(x)=\dfrac{d}{dx}(-5e^{ax})

f'(x)=-5\dfrac{d}{dx}(e^{ax})

f'(x)=-5e^{ax}\dfrac{d}{dx}(ax)            [\because \dfrac{d}{dx}(e^x)=e^x]

f'(x)=-5ae^{ax}

f'(x)=a(-5e^{ax})

f'(x)=a f(x)

Therefore, the derivative of the function is f'(x)=-5ae^{ax}=a f(x).

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