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3 years ago

5

12a-4(5a-1) = 2(3a +6)-4a

1 answer:

Ipatiy [6.2K]3 years ago

7 0

Answer:

The answet to the following question is a = -5/4.

You distribute and then combine similar terms.

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gulaghasi [49]

Answer:

4

Step-by-step explanation:

AO = OC

3x - 4 = 8

x = 4

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3 years ago

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tangare [24]

Answer: 177.5

Step-by-step explanation:so to find to area u need to multiply 16 times 16,which gives u a area of 256. Now to find to Area of circle u need to do 5*5 becuse 5 is the radius,and that is 25. 25*3.14=78.5. So we need to subtract 256-78.5=177.5.

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3 years ago

Evaluate 12x - 1 when x = 4

Slav-nsk [51]

12(4)-1

48-1

47 is the answer

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3 years ago

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20 POINTS AND BRAINLIEST<br> QUESTION ABOUT GRAPHS, PLEASE ANSWER

Rudik [331]

Answer:

The answer is straight

Step-by-step explanation:

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3 years ago

F(x) = x^2+x-2/x^2-3x-4

Andrej [43]

i. Domain and Range

The given function is

$f(x)=\frac{x^2+x-2}{x^2-3x-4}$

The domain of this function is,

$x^2-3x-4\ne 0$

$(x-4)(x+1)\ne 0$

$x\ne4,xne -1$

The range refers to the y-values for which x is defined. x is defined for all values of y.

The range is all real numbers. See graph

ii. x-and-y-intercept

For x- intercept intercept we put $f(x)=0$

This implies that;

$\frac{x^2+x-2}{x^2-3x-4}=0$

This will give us

$x^2+x-2=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow x^2+2x--x-2=0$

$\Rightarrow x(x+2)-1(x+2)=0$

$\Rightarrow (x+2)(x-1)=0$

$\Rightarrow (x+2)=0,(x-1)=0$

$\Rightarrow x=-2,x=1$

The x-intercepts are $(-2,0),(1,0)$

For y-intercept, we put

$x=0$ to obtain;

$f(0)=\frac{0^2+0-2}{0^2-3(0)-4}$

$f(0)=\frac{1}{2}$

The y-intercept is

$(0,\frac{1}{2})$

iii. Horizontal asyptote

Since degree of the numerator and the denominator are the same, there is a horizontal asymptote

To find the horizontal asymptote.

We divide the leading coefficient of the numerator by the leading coefficient of the denominator.

The horizontal asymptote is $y=\frac{1}{1}=1$

iv. Vertical asymptote

To find the vertical asymptote, we equate the denominator to zero to get;

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

This implies that;

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

Split the middle term

$x(x+1)-4(x+1)=0$

Factor

$(x+1)(x-4)=0$

Factor further

$(x+1)=0,(x-4)=0$

$x=-1,x=4$

The vertical asymptotes are $x=-1,x=4$

8 0

3 years ago

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