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2 years ago

Which inequality is true?

Mathematics

1 answer:

ICE Princess25 [194]2 years ago

8 0

Answer:

∠6 > ∠3 is TRUE.

Step-by-step explanation:

We know that acute angle is less than 90 degree while obtuse angle is more than 90 degree. By looking at the given angles, we can easily determine which angle is obtuse or acute. then we can decide which angle is bigger to find the answer.

∠10 > ∠7

FALSE;

Because∠10 is acute while ∠7 is obtuse.

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∠6 > ∠3

TRUE;

Because∠3 is acute while ∠6 is obtuse.

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∠9 > ∠11

FALSE;

Because∠9 is acute while ∠11 is obtuse.

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∠8 > ∠11

FALSE;

Because∠8 is acute while ∠11 is obtuse.

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Write 4^3 in standard form

VikaD [51]

Answer:

$64$

60+4

Step-by-step explanation:

Order of Operations

Parenthesis

Exponent

Multiply

Divide

Add

Subtract

First, do exponent.

$4^3=4*4*4=64$

Standard form of 64 is: 60+4

64 is the correct answer.

60+4 is the correct answer.

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3 years ago

What is 9.75 times 3

astra-53 [7]

Step-by-step explanation:

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2 years ago

What is the area of the trapezoid shown below?

anzhelika [568]

Answer:

78 units squared

Step-by-step explanation:

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2 years ago

A line is parallel to the line y = -x + 5 and passes through the point (4. - 1). Which of the following is an equation of the

MariettaO [177]

Answer:

y = -x + 3; could also be y + 1 = -(x - 4) or y = -(x - 4) - 1 or x + y = 3

Step-by-step explanation:

No options given, so I hope this helps.

1. Identify any of the equations that have the same slope as your given equation (-1). If there's only one, stop here.

2. Find the y-intercept by replacing (x,y)=(4,-1) in the equation y = -x + b and solve for b: -1 = -(4) + b ==> 3 = b

3. Replace b in the equation y = -x + b with 3.

5 0

9 months ago

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

3 years ago