Ask question

Ask question

All categories

2 years ago

13

Karl and Mark are walking through the woods dropping breadcrumbs to help them find their way home. After ten minutes, their brea

dbasket has 120 crumbs left. After thirteen minutes, the basket has 96 crumbs left. What is the rate of change (breadcrumbs per minute)?

1 answer:

givi [52]2 years ago

5 0

Answer:

-8

Step-by-step explanation

$\frac{96-120}{13-10} = -8$

You might be interested in

The value of n for the equation 5n = 5 11 × 5 3

ExtremeBDS [4]

If you add the two terms on the right, you will get 5^14. 14 is your answer.

3 0

3 years ago

A triangular pane of glass has a height of 34 inches and an area of 442 square inches. What is the length of the base of the pan

Katena32 [7]

Answer:

26

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

What will be the multipicative inverse of 2/3?

Eddi Din [679]

Answer:

3/2

Step-by-step explanation:

The multiplicative inverse also known as reciprocal implies it is something that is the opposite.

The given value is 2/3, so the opposite would just mean you flip the numerator and denominator. Thus, the reciprocal of 2/3 is 3/2.

8 0

2 years ago

Read 2 more answers

Plot the line for the equation on the graph. y−4=−1/4(x−5)

icang [17]

You just need two points to graph a line.

When x = 1

y - 4 = -1/4(1 - 5)
y - 4 = -1/4(-4)
y - 4 = 4/4
y - 4 = 1
y = 5

So, we have our first point, (1,5)

When x = 2

y - 4 = -1/4(2 - 5)
y - 4 = -1/4(-3)
y - 4 = 3/4
y = 4 3/4

Now we have the point (2, 4 3/4)

Now just run a line through the two points and you're done!

6 0

2 years ago

Needhdjdjdjdjdjdnhiskdjdjdjdjdjdjjshdjdjdjdjdjdjdjd<br><br>

Lilit [14]

9514 1404 393

Answer:

-(√2)/2

Step-by-step explanation:

The expression evaluated at n=a gives the indeterminate form 0/0, so L'Hopital's rule can be used to find the limit. The second expression comes from differentiating numerator and denominator. Then the form with n=a is no longer indeterminate.

$\displaystyle\lim_{n\to a}{\frac{\sqrt{2n}-\sqrt{3n-a}}{\sqrt{n}-\sqrt{a}}}=\lim_{n\to a}{\frac{\frac{2}{2\sqrt{2n}}-\frac{3}{2\sqrt{3n-a}}}{\frac{1}{2\sqrt{n}}-0}}\\\\=\sqrt{a}\left(\frac{2}{\sqrt{2a}}-\frac{3}{\sqrt{3a-a}}}\right)=\boxed{-\frac{1}{\sqrt{2}}}$

5 0

2 years ago