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2 years ago

Write the ordered pair for each point

Mathematics

1 answer:

Digiron [165]2 years ago

8 0

to find an ordered pair for each point we need to know what the points are

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Multiply (3)(−4)(2). (5 points) Group of answer choices −48 −24 24 48

vladimir2022 [97]

Answer:

-24

Step-by-step explanation:

3 * -4 =.-12

-12 * 2 = -24

8 0

2 years ago

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HELP I'M SOOOOOO ANGRYYYYY PLEASE HELPPPP determine the values of X for which g(x) is defined

tiny-mole [99]

Answer: Any real number x as long as $x \ne 0$ and $x \ne -\frac{2}{3}$

In other words, anything but 0 or -2/3 is valid.

========================================================

Explanation:

Set the denominator equal to zero and solve for x

2(3x^2 + 2x) = 0

3x^2 + 2x = 0

x(3x + 2) = 0

x = 0 or 3x+2 = 0 .... zero product property

x = 0 or 3x = -2

x = 0 or x = -2/3

If either x = 0 or x = -2/3, then the denominator 2(3x^2 + 2x) will be zero. But recall that we cannot have zero in the denominator. Dividing by zero is not allowed. The expression is undefined when we divide by zero.

Therefore, we must exclude x = 0 and x = -2/3 from the domain. Any other real number is valid as an x input.

7 0

2 years ago

I don't get these what Do I do

max2010maxim [7]

Answer:

No idea

Step-by-step explanation:

3 0

2 years ago

A survey of students at a particular college showed that 33% of them prefer the American Heritage Dictionary (as opposed to Webs

vladimir1956 [14]

Answer:

.066 = 6.6 percent chance

Step-by-step explanation:

1 in 5 chance to get 6, so that is .20 multiply .20 and .33 and you get .066

8 0

2 years ago

Please someone help me to prove this.

dem82 [27]

<h3>Answer :</h3>

We know that,

$\dag\bf\:sin^2A=\dfrac{1-cos2A}{2}$

$\dag\bf\:sin2A=2sinA\:cosA$

___________________________________

Now, Let's solve !

$\leadsto\:\bf\dfrac{sin^2A-sin^2B}{sinA\:cosA-sinB\:cosB}$

$\leadsto\:\sf\dfrac{\frac{1-cos2A}{2}-\frac{1-cos2B}{2}}{\frac{2sinA\:cosA}{2}-\frac{2sinB\:cosB}{2}}$

$\leadsto\:\sf\dfrac{1-cos2A-1+cos2B}{sin2A-sin2B}$

$\leadsto\:\sf\dfrac{2sin\frac{2A+2B}{2}\:sin\frac{2A-2B}{2}}{2sin\frac{2A-2B}{2}\:cos\frac{2A+2B}{2}}$

$\leadsto\:\sf\dfrac{sin(A+B)}{cos(A+B)}$

$\leadsto\:\bf{tan(A+B)}$

5 0

2 years ago

Read 2 more answers