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2 years ago

What is the least common denominator of the fractions 2/3, 1/4, 1/8 and ?

Mathematics

1 answer:

KiRa [710]2 years ago

3 0

Answer:

Step-by-step explanation:

That's the first number that all your denominators have in common.

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A triangle has one side that measures 12 inches and another side that measures 33 inches. Which are possible side lengths of the

In-s [12.5K]

Answer:

Greater than 21

Less than 45

Step-by-step explanation:

You add and subtract the two numbers you are given.

33 - 12 = 21

The third side has to be bigger than 21.

33 + 12 = 45

The third side has to be less than 45.

It can be written in one math sentence:

21 < x < 45

5 0

1 year ago

By the Interior Angles Theorem, if Angle A is 25° and Angle B is greater than 51° but less than 57°, what are the possible measu

BARSIC [14]

98 < C < 104

180 - 25 - < C < 180 - 25 - 51
98 < C < 104

Measures the interior angles of a triangle sum to 180

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2 years ago

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What is 91(3456+9878kh)-3(4564-9087)

goldfiish [28.3K]

The answer is 328065+ 898898hk hope I helped

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2 years ago

WILL MAKE BRAINLIEST!

Stolb23 [73]

1) <span>x + 24=95

2) 1 / 5x = 10
5x = 10/1
5x = 10
x = 10/5
x = 2

3) b/5 + 15 = 45
b = 5(45-15)
b = 150

</span>

8 0

2 years ago

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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

2 years ago

Read 2 more answers