HELP PLEASE I DON´T UNDERSTAND RATIOS!

Mathematics

2 answers:

jeyben [28]2 years ago

7 0

Answer:

Jake made 7 bead strings

Step-by-step explanation:

The ratio is 2:6 next would be 4:12 then 6:18 and so on. You just add what you started with to the corresponding side. 8:24, 10:30, 12:36, 14:42. The right side is 3x the left side.

kipiarov [429]2 years ago

5 0

Answer:

Step-by-step explanation:

You don't really need to use ratios for this question. You need to understand that for each necklace he uses 2 red and 6 whites.

Let the number of necklaces he can make = x

The total number of beads = 56

2x + 6x = 56

8x = 56 Divide by 8

8x/8 = 56/8

x = 7

So the number of necklaces possible is 7.

If you insist on doing this with ratios then you could do it as

2/8 = x/56 What this means is that 2 reds to the total = x / 56. The 56 is the total number of beads used. The x represents the number of reds.

2/8 = x/56 Cross multiply

2 * 56 = 8x Divide both sides by 8

2 * 56/8 = x

x = 14.

That means of the 56 beads used, 14 of them are red.

6/8 = y / 56 Cross multiply

6*56 = 8y

336 = 8y Divide by 8

336/8 = 8y/8

y = 42

This means that 42 beads were white

The total is 56. But here's the catch. The highest prime common to both is 7 and that's how many necklaces can be made.

You might be interested in

Suppose that triangle TUV is an isosceles with the base VU.

vekshin1

Set them equal to each other and solve
3x + 39 = 4x +34
-x=-5
X=5
Plug it in
3(5) + 39 = 54
4(5) +34 = 54
For the final angle ,
54 + 54 = 108
180-108 =72
The angle measures are 54,54, and 72

3 0

3 years ago

What did sin theta/theta whisper worksheet answer key worksheet?

djyliett [7]

To evaluate
$\lim_{\theta \to 0} \frac{\sin\theta}{\theta}$

First, we input 0, for theta in the function to obtain:
$\frac{\sin0}{0} = \frac{0}{0}$

This is an indeterminate form.

So, we apply L'Hopital's rule by differentiating the numerator and the denominator as follows:

$\lim_{\theta \to 0} \frac{\sin\theta}{\theta}=\lim_{\theta \to 0} \frac{ \frac{d}{d\theta} (\sin\theta)}{\frac{d}{d\theta}\theta} \\ \\ =\lim_{\theta \to 0} \frac{\cos\theta}{1}=\cos0=1$