Pls help help Im crying and i need help please im nevous please help me

Mathematics

2 answers:

aliya0001 [1]2 years ago

8 0

Answer:

Look in step by step

Step-by-step explanation:

The way to read coordinates is to know these steps.

(#,#)

the first number is the x number which on the graph is the horizontal line and the second number is the y number which on the graph is vertical.

Theres also negatives which is the same thing but its on the lower side of the graph.

So now for example you take -4,-9.

To make it easy move the dot to the center or the (0,0) cords. Then follow the numbers. When you start on the zero cord you will look at the numbers, it says -4 so drag the dot 4 units to the left where the line says -4. Then drag it down 9 units so its on the cords. Hope this helps!!

Evgesh-ka [11]2 years ago

8 0

Answer:

Ok so for (0,8 1/2), drag the dot from the middle, up to 8 but then move up a little bit more so it is between the 8 and 9. Next, for (-4, -9) drag one of the other dots to the -4 on the horizontal line, but then move it down to the -9 on the vertical line. And finally for (-6,5), you should put the last dot on the -6 on the horizontal line, but move it up five spaces until it is at the 5 on the vertical line.

I hope I was able to help you, if you still do not understand I will try to explain a little bit more.

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Which calculation can be used to find the value of q in the equation q3 = 64?

alexandr402 [8]

To isolate the variable q, divide both sides by 3:
q3 ÷ 3 = 64 ÷ 3
q = 21.33 or 21 1/3

If the q is cubed(I can't really tell), however, you would use the cube root:
³√q³ = ³√64
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8 0

2 years ago

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HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$