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3 years ago

This is due tm please help

Mathematics

1 answer:

Mumz [18]3 years ago

8 0

Answer:

25. No slope

Step-by-step explanation:

25. Y2 - Y1 / X2 - X1

= -8 - (-3) / 9 - 9

= -5 / 0

= no slope; straight line

Use this formula (Y2 - Y1 / X2 - X1) for each ques.

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If x>2, then x^2-x-6/x^2-4=

Viefleur [7K]

Consider the expression $\frac{x^{2} -x-6}{ x^{2} -4}$

To factorize the expression in the denominator we use difference of squares: $x^{2} -4=x^{2} - 2^{2} =(x-2)(x+2)$

To factorize $x^{2} -x-6$ we use the following method:

$x^{2} -x-6=(x-a)(x-b)$

where a, b are 2 numbers such that a+b= -1, the coefficient of x,

and a*b= -6, the constant.

such 2 numbers can be easily checked to be -3 and 2

(-3*2=6, -3+2=-1)

So $x^{2} -x-6=(x-a)(x-b)=(x+3)(x-2)$

$
 \frac{x^{2} -x-6}{ x^{2} -4}= \frac{(x+3)(x-2)}{(x-2)(x+2)}= \frac{x+3}{x+2}$

$\frac{x+3}{x+2}= \frac{x+2+1}{x+2}= \frac{x+2}{x+2}+ \frac{1}{x+2}=1+ \frac{1}{x+2}$

for x>2

$\frac{1}{x+2}\ \textless \ \frac{1}{2+2}= \frac{1}{4}$

thus

for x>2,

$1+ \frac{1}{x+2}\ \textless \ 1+ \frac{1}{4}= \frac{5}{4}$

Answer:

for x>2

$\frac{x^{2} -x-6}{ x^{2} -4} = \frac{x+3}{x+2} \ \textless \ \frac{5}{4}$ , (but the expression is never 0)

8 0

4 years ago

Find f(4) when f(x) = 5x^2 - 5x + 7. A) 107 B) 53 C) 67 D) 3

fgiga [73]

I think the answer is 67

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4 years ago

2/3x=4 what is the answer?

GREYUIT [131]

Answer:

x=6

Step-by-step explanation:

7 0

3 years ago

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The volume of a barrel of oil is calculated as 257,166 cm³.

Firdavs [7]

What is the digit at the hundred thousands place? It is 2. Is the digit to the right of it less than 5? No, 5 is 5. So replace the digits to the right of 2 with 0, and change the 2 to 3.
Answer: 300,000

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3 years ago

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Help please ITS OF TRIGONOMETRY PROVE

Flauer [41]

Answer:

The equation is true.

Step-by-step explanation:

In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation. Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

$sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}$

One is given the following equation,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

$sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}$

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

$\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}$

$1+(-1)=0$

$1-1=0$

$0=0$

7 0

3 years ago

This is due tm please help​

This is due tm please help