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3 years ago

Hi, please help me its timed!

Mathematics

1 answer:

Zina [86]3 years ago

8 0

Option 2 and option 4 are linear

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Help me understand this!!!!!!!!!! m

gregori [183]

Answer:

Your question does not make sense

Step-by-step explanation:

Pls give me points for my answer anyways.

We're no strangers to love

You know the rules and so do I

A full commitment's what I'm thinking of

You wouldn't get this from any other guy

I just wanna tell you how I'm feeling

Gotta make you understand

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

We've known each other for so long

Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

And if you ask me how I'm feeling

Don't tell me you're too blind to see

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give, never gonna give

(Give you up)

(Ooh) Never gonna give, never gonna give

(Give you up)

We've known each other for so long

Your heart's been aching but you're too shy to say it

Inside we both know what's been going on

We know the game and we're gonna play it

I just wanna tell you how I'm feeling

Gotta make you understand

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

Never gonna say goodbye

Never gonna tell a lie and hurt you

Never gonna give you up

Never gonna let you down

Never gonna run around and desert you

Never gonna make you cry

7 0

3 years ago

Prove that:

Lorico [155]

A.)

$\csc^2(x) \tan^2 (x)- 1 = \tan^2(x)$

Use the identities $\csc x = 1 / \sin x$ and $\tan x = \sin x / \cos x$ on the left-hand side

$\begin{aligned}
\text{LHS} &= \csc^2(x) \tan^2 (x)- 1 \\
&= \frac{1}{\sin^2 (x)} \cdot \frac{\sin^2 (x)}{\cos^2 (x)} - 1 \\
&= \frac{1}{\cos^2 (x)} - 1
\end{aligned}$

Make 1 have a common denominator to allow for fraction subtraction
Multiply the numerator and denominator of 1 by cos^2 x

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - 1 \cdot \tfrac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{\cos^2 (x)}{\cos^2 (x)} \\
&= \frac{1 - \cos^2 x}{\cos^2 (x)}
\end{aligned}$

Use Pythagorean identity for the numerator.

If $\sin^2 (x) + \cos^2(x) = 1$ then subtracting both sides by $\cos^2 (x)$ yields $\sin^2(x) = 1 - \cos^2(x)$ . We can substitute that into the numerator

$\begin{aligned} \text{LHS} &= \frac{1 - \cos^2 (x)}{\cos^2 (x)} \\
&= \frac{\sin^2 (x)}{\cos^2 (x)} \\
&= \tan^2 (x) && \text{Since } \tan x = \tfrac{\sin x }{\cos x} \\
&= \text{RHS}
\end{aligned}$

======

b.)

$\dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} = 1$

For the left-hand side:
By definition, $\sec(x) = 1/\cos(x)$ and $\tan (x) = 1/\cot (x)$

$\begin{aligned}
\text{LHS} &= \dfrac{\sec(x)}{\cos(x)} - \dfrac{\tan(x)}{\cot(x)} \\
&= \dfrac{ \frac{1}{\cos(x)} }{\cos(x)} - \dfrac{\frac{1}{\cot(x)}}{\cot(x)} \\
&= \frac{1}{\cos^2 (x)} - \frac{1}{\cot^2(x)} 
\end{aligned}$

Since $\cot (x) = \cos (x) / \sin (x)$

$\begin{aligned} \text{LHS} &= \frac{1}{\cos^2 (x)} - \frac{1}{\frac{\cos^2(x)}{\sin^2(x)} } \\ &= \frac{1}{\cos^2 (x)} -\frac{\sin^2(x)}{\cos^2(x)} \\ &= \frac{1 - \sin^2(x)}{\cos^2 (x)} \end{aligned}$

Using Pythagorean identity, $\cos^2(x) = 1 - \sin^2(x)$ so

$\begin{aligned} \text{LHS} &= \frac{\cos^2(x)}{\cos^2 (x)} \\
&= 1 \\
&= \text{RHS}
\end{aligned}$

6 0

3 years ago

What is the distance between the points (1, -4) and (1, 0)

tino4ka555 [31]

For this case we have that the distance between two points is given by:

$d = \sqrt{(y2-y1)^2+(x2-x1)^2}$

Given the points:

$(x1, y1) = (1, -4)(x2, y2) = (1,0)$

We substitute in the formula:

$d = \sqrt{(0 - (- 4)) ^ 2 + (1-1) ^ 2}$

$d = \sqrt{(4 ^ 2 + 0 ^ 2)}$

$d=\sqrt{16}$

$d = 4$

So, the distance is $d = 4$

Answer:

$d = 4$

5 0

3 years ago

ERROR ANALYSIS Describe and correct the error in

valkas [14]

Answer:

The graph should be stretched rather than become narrower.

Step-by-step explanation:

To figure this out, just create some example points.

At x = 0, your y-value will always be 0. But if you were to plug in the value 1, you would get a y-value of 1 in y=x^2, but a value of 0.5 in y=0.5x^2. If you were to plug in a value of 2, you would get a value of 4 in y=x^2, but a value of 2 in y=0.5x^2.

If you continue this pattern for a few more points, then plot them, you will see that adding a coefficient of 0.5 simply stretches the graph

3 0

2 years ago

Y-1= -3(x + 4) Find the equation for y

sineoko [7]

Y=-3x-11. you add the one to the other side making the equation y=-3(x+4)+1
then you distribute making it y=-3x-12+1

8 0

3 years ago

Read 2 more answers

Hi, please help me its timed!​

Hi, please help me its timed!