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Yanka [14]
3 years ago
9

Solve the equation - 7s-9 = - 37 for s

Mathematics
2 answers:
Ksju [112]3 years ago
7 0

Answer: s = 4

Step-by-step explanation:

frosja888 [35]3 years ago
4 0

Answer:

s = 4

Step-by-step explanation:

- 7s - 9 = - 37

- 7s = 9 - 37

- 7s = - 28

7s = 28

s = 28/7

s = 4

Thus, The value of s is 4

<u>-TheUnknown</u><u>S</u><u>cientist</u>

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Law Incorporation [45]

Recall the binomial theorem.

(a+b)^n = \displaystyle \sum_{k=0}^n \binom nk a^{n-k} b^k

1. The binomial expansion of \left(1+\frac x3\right)^7 is

\left(1 + \dfrac x3\right)^7 = \displaystyle\sum_{k=0}^7 \binom 7k 1^{7-k} \left(\frac x3\right)^k = \sum_{k=0}^7 \binom 7k \frac{x^k}{3^k}

Observe that

k = 1 \implies \dbinom 71 \left(\dfrac x3\right)^1 = \dfrac73 x

k = 2 \implies \dbinom 72 \left(\dfrac x3\right)^2 = \dfrac73 x^2

When we multiply these by 8-9x,

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• -9x and \frac73 x combine to make -\frac{63}3 x^2 = -21x^2

and the sum of these terms is

\dfrac{56}3 x^2 - 21x^2 = \boxed{-\dfrac73 x^2}

2. The binomial expansion is

\left(2a - \dfrac b2\right)^8 = \displaystyle \sum_{k=0}^8 \binom 8k (2a)^{8-k} \left(-\frac b2\right)^k = \sum_{k=0}^8 \binom 8k 2^{8-2k} a^{8-k} b^k

We get the a^6b^2 term when k=2 :

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1 year ago
F (x) = - eˣ Baseline (x - 4)<br> What​ is(are) the critical​ point(s) of​ f?
Ber [7]

Answer

given,

   f(x) = \dfrac{-e^x}{x - 4}

to find the critical point of the given expression

fist differentiating the function

f'(x) = -\dfrac{(x-4)e^x+ e^x}{(x - 4)^2}

f'(x) = \dfrac{-(x-4)e^x- e^x}{(x - 4)^2}

f'(x) = \dfrac{-e^x(x-3)}{(x - 4)^2}

now equating differential equation to zero

\dfrac{e^x(-x+3)}{(x - 4)^2}=0

e^x(-x+3)=0

now,

-x + 3 = 0            and eˣ ≠ 0

x = 3          

the critical number will be equal to x = 3

y = \dfrac{-e^3}{3 - 4}

y =e^3

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