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2 years ago

HELP IM BEING TIMED 20+ points!!

Mathematics

1 answer:

Pani-rosa [81]2 years ago

4 0

Answer:

1: +

2: +

3: -

4: +

5: -

6: -

Step-by-step explanation:

hope this helps!

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Where do the graphs of the linear equation y=2x+5 and y=-2x-3 intersect?

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the coordinates of the vertices of a triangle are (3, 7), (3, 1), and (-5, 1). determine if the following statements are true or

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Step-by-step explanation:

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1 year ago

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Which of the following relations has an inverse that is not a function?

sertanlavr [38]

Answer:

Option B is correct.

The relation which has an inverse that is not a function is :(1, 2), (1, 3), (4, 3)

Step-by-step explanation:

Inverse of a function is the set of ordered pairs obtained by interchanging the first and second elements of each pair in the original function.

Let f(x) ={(1, 2), (1, 3), (4, 3)}

To find the inverse:

Inverse of f(x)=g(x)= {(2, 1), (3, 1), (3, 4)}

Function states that a relation in which each input value has only one output value.

Since, g(x) is not a function because the input 3 has multiple output values,.

Therefore, the only relation which has an inverse that is not a function is :(1, 2), (1, 3), (4, 3)

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2 years ago

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1 year ago

Verify the following identity:

Paraphin [41]

We will simplify the left hand side of the equation to look like the right

$\displaystyle \frac{1-sin(t)}{cos^{2}(t)} + \frac{1}{1 - sin(t)}$

taking the LCM

$\displaystyle \frac{[1-sin(t)]^{2} + cos^{2}(t)}{[cos^{2}(t)][1 - sin(t)]}$

expanding the binomial in the numerator

$\displaystyle \frac{[1 + sin^{2}(t) - 2sin(t) + cos^{2}(t)]}{[cos^{2}(t)][1 - sin(t)]}$

Since sin²x + cos²x = 1

$\displaystyle \frac{[2 - 2sin(t)]}{[cos^{2}(t)][1 - sin(t)]}$

factoring out the 2 from the numerator

$\displaystyle \frac{2[1 - sin(t)]}{[cos^{2}(t)][1 - sin(t)]}$

1-sin(t) will cancel out

$\displaystyle \frac{2}{cos^{2}(t)}$

since cos²(t) = 1/(sec²(t))

$2 sec^{2}(t)$

which is equal to the right hand side of our given equation.

So we verified the identity!

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2 years ago

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