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3 years ago

Which of the following orderd pairs is a solution of x + 1/2 y = 1

Mathematics

1 answer:

lorasvet [3.4K]3 years ago

3 0

Answer:

(-2,6)

Step-by-step explanation:

For an ordered pair (x,y) to be a solution to the equation given, we have to plug in x-coordinate in "x" and y-coordinate in "y" of the following equation and see if it is "1".

Whichever makes it true, is the correct answer.

Checking:

$x+\frac{1}{2}y\\=(-2)+\frac{1}{2}(6)\\=-2+3\\=1$

This is right.

We don't need to check the other two, the first one is the right option.

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A conditional statement is logically equivalent to a biconditional statement. True False pls help i have a test and i was absent

andreyandreev [35.5K]

Answer:

false.

Step-by-step explanation:

A conditional statement is something like:

If P, then Q.

This means that if a given proposition P is true, then another proposition Q is also true.

An example of this is:

P = its raining

Q = there are clouds in the sky.

So the conditional statement is

If its raining, then there are clouds in the sky.

A biconditional statement is:

P if and only if Q.

This means that P is only true if Q is true, and Q is only true if P is true.

So, using the previous propositions we get:

Its raining if and only if there are clouds in the sky.

This statement is false, because is possible to have clouds in the sky and not rain.

(this statement implies that if there are clouds in the sky, there should be rain)

Then we could see that for the same propositions, the conditional statement is true and the biconditional statement is false.

Then these statements are not logically equivalent.

The statement is false.

3 0

3 years ago

1+secA/sec A = sin^2 A / 1-cos A

Fofino [41]

Answer: see proof below

<u>Step-by-step explanation:</u>

$\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^2 A}{1-\cos A}$

Use the following Identities:

sec Ф = 1/cos Ф

cos² Ф + sin² Ф = 1

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \dfrac{1+\sec A}{\sec A}$

$\text{Identity:}\qquad \qquad \dfrac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\cos A+1}{\cos A}}{\frac{1}{\cos A}}\\\\\\.\qquad \qquad \qquad =\dfrac{1+\cos A}{1}$

$\text{Multiply:}\qquad \qquad \dfrac{1+\cos A}{1}\cdot \bigg(\dfrac{1-\cos A}{1-\cos A}\bigg)\\\\\\.\qquad \qquad \qquad =\dfrac{1-\cos^2 A}{1-\cos A}$