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4 years ago

12: f(a) = a + 3. Find f(7) Help please

Mathematics

1 answer:

Karo-lina-s [1.5K]4 years ago

8 0

It is 3 because you cancel out the a so it leaves you with 3

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Find how many solutions there are to the given equation that satisfy the given condition. X1 + x2 + x3 = 22, each x; is a positi

DiKsa [7]

Answer:

Step-by-step explanation:

Given that there are three variables satisfying the equation

$X1 + x2 + x3 = 22$

Here each x is given to be a positive integer

i.e. solution set for each of the variable can be any integer from 1 to 20 at most.(because if two other integers are 1 each third has to be 20)

Hence solution set can be of the form

$(x1,x2,x3) =(1,1,22) (1,2,21) (1,3,20).....$

$=(2,1,19) (2,2,18),...\\=(3,1,18) (3,2,17),....\\...\\...\\=(20,1,1)$

If x1 =1, there are 20 solution sets

If x1 =2,there are 19

...

If x1 =20 there is 1 set

Hence total solutions can be $= 20+19+...+1\\=\frac{20(21)}{2} =210$

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3 years ago

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9. simplify the radical expression pleasseee and thank you :)

Vinil7 [7]

Answer:

sqrt 10 / 9

Step-by-step explanation:

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3 years ago

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(x^2−x−6x^2−8x+16)/(x2−x−12)

erma4kov [3.2K]

-3 and 4 cannot be the values of x in the given expression

The expression is given as:

$\frac{(x^2-x-6x^2-8x+16)}{(x^2-x-12)}$

Start by equating the denominator to 0

$x^2 - x - 12 = 0$

Expand the above equation

$x^2 - 4x + 3x - 12 = 0$

Factorize the equation

$x(x - 4) + 3(x - 4) = 0$

Factor out x - 4

$(x + 3) (x - 4) = 0$

Split

$(x + 3)= 0\ or\ (x - 4) = 0$

Remove brackets

$x + 3= 0\ or\ x - 4 = 0$

Solve for x

$x =- 3\ or\ x = 4$

Hence, -3 and 4 cannot be the value of x in the given expression