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pantera1 [17]
3 years ago
10

Need help with this answer plz

Mathematics
1 answer:
Sergeu [11.5K]3 years ago
6 0

Answer:

12

Step-by-step explanation:

(5+2)^2 -40 +3 =

PEMDAS

Parentheses first

(7)^2 -40 +3 =

Exponents

49 - 40 +3

Since there are no multiply or divide, we add and subtract from left to right

9+3

12

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BRAINLEST Find the sum of the first 6 terms of the infinite series: 1 - 2 + 4 - 8+...
arsen [322]

Answer:

-21

Step-by-step explanation:

1-2+4-8+16-32

=-21

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3 years ago
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Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?
9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

f(x)=\frac{(x-1)(x^2-4)}{x^2-a}

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that x=\sqrt{a} will annulate the denominator, i.e

(\sqrt{a})^2-a = a-a = 0. But, if a = 1 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}

so, the value x=\sqrt{a} = \sqrt{1} = 1 won't annulate the denominator.

Now, for a = 4 we have:

f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1

so, the value x=\sqrt{a} = \sqrt{4} = 2 won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

4 0
3 years ago
X - 18 = -2. Solve this equation. Check your solution
Alika [10]

Answer:

X=16

Step-by-step explanation:

X-18=-2

+18  +18

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X=16

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MArishka [77]

Step-by-step explanation:

f(x) = 5x - 8

g(x) = 9 - 6x²

= (fog)(-3)

= f( g(3) )

= 5x - 8

= 5(9 - 6x²) - 8

= 5(9 - 6(-3)² ) - 8

= 5(9 - 54) - 8

= 5(-45) - 8

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= -233

(fog)(-3) = -233

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