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3 years ago

I need help ASAP . I have an important test.

Mathematics

2 answers:

Taya2010 [7]3 years ago

7 0

Answer:

**Double Check because the labeling for the number line is not visible in your screenshot**

Step-by-step explanation:

$3x\leq -6$

$3x/3\leq -6/3$

$x\leq -2$

BigorU [14]3 years ago

4 0

Answer:

Step-by-step explanation:

Divide 3 by both sides to get x ≤ -2

(side note: the picture was difficult to see but -2 is behind the 0 so I could tell it was D)

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Write 5 multiples of 3 means write 5 numbers counting by 3 PLZ HELP

Alex73 [517]

Answer:

ummm....like this?

Step-by-step explanation:

3*2=6

3*3=9

3*4=12

3*5=15

3*6=18

3,6,9,12,15,18,21,24

i don't know if this is what you mean by here you go

6 0

3 years ago

Read 2 more answers

A cylinder diameter of 6cm and length of 20cm. Using n3 what is volume of cylinder cm3

grandymaker [24]

Use the basic formula v=nr^2h to get the answers
180n or 565.2
Hope this helps! :D

5 0

4 years ago

A volleyball team played 28 matches and won 21 of them. What percent of their matches did they win ?

Mazyrski [523]

They won 75% of the matches.

21/28= 0.75

5 0

3 years ago

Nikki and her four friends had lunch at their favorite restaurant. The total bill was $29.00, and they wanted to leave a 15% tip

Butoxors [25]

Answer:

$4.35

Step-by-step explanation:

Knowns:

Total bill: $29.00

Tip of 15% ($29.00)

Work:

$29.00 (0.15) = $4.35 tip

Hope this helps!

7 0

3 years ago

If sinx = p and cosx = 4, work out the following forms :<br><br><br>

Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16}$

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

$\boxed{\sin(-x)=-\sin(x)}$

$\boxed{\cos(-x)=\cos(x)}$

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

$\sin(x)=p$

$\cos(x)=4$

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]$

Using the sum and difference identities:

$\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}$

$\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}$

$\sin(\pi -x) = \sin(x)$

$\sin(2\pi +x)=\sin(x)$

$\cos(2\pi-x)=\cos(x)$

Therefore,

$[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]$

$\implies [p+4] \cdot [p \cdot 4]=4p^2+16p$

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16}$

8 0

3 years ago