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3 years ago

On average, how many mini marshmellows are in 1 serving? Explain. 9+2x6÷2-8x5-7+124=

Mathematics

1 answer:

ad-work [718]3 years ago

5 0

The answer would be 92 for this problem

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I REALLY NEED HELP MY GRADES GO IN TONIGHT AND I CANT SOLVE THIS I NEED A REAL ANSWER I WILL CHECK BEFORE GIVING POINTS!!!!!!!!!

Helga [31]

Answer:

K. 8

N. 12

A. 29

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7 0

1 year ago

Read 2 more answers

What is 549,423 rounded to the nearest thousand

Lubov Fominskaja [6]

The answer is 549,000

6 0

3 years ago

Solve for p: p/10 = 7.2/36

vovikov84 [41]

P=20 this is because you have to get 2 on each side 7.2/3.6 =2 so 20/10=2 2=2

3 0

2 years ago

A motor club has estimated that a 325 mile car trip will take 6 1/2 hours. What is the club assuming to be the average speed mil

In-s [12.5K]

Answer:

The average speed that the club is assuming is 50 miles per hour

Solution:

We know that Distance = Speed × Time.

Here it is given that distance is 325 miles and the time to cover 325 miles is $6_{2}^{1}$ hours which is equal to $\frac{13}{2}$ hours.

Now to find the average speed mile per hour we use the formula

$\begin{aligned} \text { Speed } &=\frac{\text { Distance }}{\text { time }} \\ &=\frac{325}{\frac{13}{2}} \\ &=\frac{325 \times 2}{13} \end{aligned}$

By cancelling 325 by 13 we get 25

= 25 × 2

= 50 miles per hour

Thus the average speed that the club is assuming is 50 miles per hour.

3 0

3 years ago

39-50 find the limit. 41. <img src="https://tex.z-dn.net/?f=%5Clim%20_%7Bt%20%5Crightarrow%200%7D%20%5Cfrac%7B%5Ctan%206%20t

Katyanochek1 [597]

Write tan in terms of sin and cos.

$\displaystyle \lim_{t\to0}\frac{\tan(6t)}{\sin(2t)} = \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)}$

Recall that

$\displaystyle \lim_{x\to0}\frac{\sin(x)}x = 1$

Rewrite and expand the given limand as the product

$\displaystyle \lim_{t\to0}\frac{\sin(6t)}{\sin(2t)\cos(6t)} = \lim_{t\to0} \frac{\sin(6t)}{6t} \times \frac{2t}{\sin(2t)} \times \frac{6t}{2t\cos(6t)} \\\\ = \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right)$

Then using the known limit above, it follows that

$\displaystyle \left(\lim_{t\to0} \frac{\sin(6t)}{6t}\right) \times \left(\lim_{t\to0}\frac{2t}{\sin(2t)}\right) \times \left(\lim_{t\to0}\frac{3}{\cos(6t)}\right) = 1 \times 1 \times \frac3{\cos(0)} = \boxed{3}$

4 0

1 year ago