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3 years ago

(-4,r) (-8,3) m=5 Find r please!

Mathematics

1 answer:

viva [34]3 years ago

4 0

Answer:

r = 23

Step-by-step explanation:

m = (y2 - y1)/(x2 - x1)

5 = (3 - r)/(-8 - (-4))

5 = (3 - r)/(-8 + 4)

3 - r = 5 × -4

3 - r = -20

-r = -23

r = 23

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If it takes 1074 digits to number the pages of a book starting with page 1. how many pages are in the book? The book has a total

Charra [1.4K]

Answer:

394 pages

Step-by-step explanation:

1- digit numbers:

1 to 9, number of digits = 9

2- digit numbers:

10 to 99, number of digits = 2(99 - 9) = 180

3- digit numbers:

100 to 999, number of digits = 3(999 - 99) = 2700

This is more than we have, but we know the number of pages is 3-digit.

Let it be x, then we have:

9 + 180 + 3(x - 99) = 1074
3(x - 99) = 1074 - 189
3(x - 99) = 885
x - 99 = 885/3
x - 99 = 295
x = 295 + 99
x = 394

4 0

2 years ago

Helppp please it may be simple for u!

nikitadnepr [17]

3,1,2 replay if i got it wong

5 0

3 years ago

A pudding recipe requires 2/3 of a cup of milk for every 1 3/4 cups of sugar. What is the unit rate of sugar to milk in the pudd

Zinaida [17]

Answer:

2 5/8 to 1.

Step-by-step explanation:

That would be 1 3/4 divided by 2/3.

= 7/4 / 2/3

= 7/4 * 3/2

= 21 / 8

= 2 5/8

Answer 2 5/8 to 1.

8 0

3 years ago

In a right triangle, what is the 3rd angle measure if an angle is 25

melomori [17]

Answer:

Step-by-step explanation:

The sum of the angles in a triangle is 180. In a right triangle, there is one 90 degree angle. If the second angle is 25, and 90+25+the last angle must add up to 180, simple math can find the value of the last angle.

180-90-25=65

5 0

3 years ago

Read 2 more answers

Show that a sequence {sn} coverages to a limit L if and only if the sequence {sn-L} coverages to zero.

Andreyy89

Let ${s_n}_{n\in\Bbb N}$ be a sequence that converges to $L$ . This means for any $\varepsilon>0$ , there is some $N$ such that $|s_n-L|$ for all $n>N$ . From this inequality we see that $|(s_n-L)-0|$ , so it follows that $s_n-L\to0$ .

On the other hand, let ${s_n-L}$ be a sequence that converges to 0. This means $|(s_n-L)-0|$ for all large enough $n$ , and we get the simpler inequality for free, $|s_n-L|$ , so it follows that $s_n\to L$ .

3 0

3 years ago