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Harlamova29_29 [7]
3 years ago
13

...anybody know this one

Mathematics
1 answer:
Debora [2.8K]3 years ago
4 0
The answer is 1/(x^2 y)

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Please help!
Lubov Fominskaja [6]

Answer:

The answer is d (-21/2 , 2)

Step-by-step explanation:

That's where the lines meet up

8 0
3 years ago
For thousands of years, gold has been considered one of the Earth's most precious metals. One hundred percent pure gold is 24-ka
Troyanec [42]
The answer is on bitly im about to put the link
7 0
3 years ago
Hello, I am struggling a bit with this question please help
adelina 88 [10]

Answer:

-7.4

-2/3

⅔-0.1

0

4/8

2.7

5 1/9

29/4

Step-by-step explanation:

I hope this helped and if it did I would appreciate it if you marked me Brainliest. Thank you and have a nice day!

4 0
3 years ago
Read 2 more answers
Help on number 10 pls
Reika [66]

9514 1404 393

Answer:

  (D) 9/5

Step-by-step explanation:

One way to work this problem is to find % per cup for cereal and cups per % for milk, then multiply the product of those by 2.

The way we'll use here is to find the ratio of cups of milk to cups of cereal for the same % RDA.

We're given an amount of milk that has 25% of the RDA, but we want to find an amount that has 15% RDA. That amount of milk will be ...

  (15%)/(25%)(1 cup milk) = 3/5 cup milk

This tells us that 3/5 cup milk has the same %RDA as 2/3 cup cereal. That ratio is ...

  milk/cereal = (3/5)/(2/3) = (3/5)(3/2) = 9/10 cup of milk per cup of cereal

Then for 2 cups of cereal, we need ...

   (9/10)(2 cups cereal) = 9/5 cups milk . . . . matches choice D

6 0
3 years ago
Find the area of the region between the curve y= 2ln(x) and the horizontal axis for 1<= x <= 4
marissa [1.9K]

Answer:

5.09 units

Step-by-step explanation:

Given equation

y=2\ln x=f(x) in the interval 1\le x\le 4

So we integrate y in the given interaval

\int f(x)=2\int\limits^4_1 {\ln x}dx

Let us integrate \ln x first.

let

u=\ln x, dv=dx

du=\dfrac{1}{x}, v=x

\int\ln x dx=udv

Using integration by parts we get

uv-\int vdu

=x\ln x-\int x\dfrac{1}{x}dx

=x\ln x-dx

=x\ln x-x+C

So here

\int f(x)=2\int\limits^4_1 {\ln x}dx\\ =2(x\ln x-x)_1^4\\ =2[(4\ln 4-4)-(1\ln 1-1)]\\ =2[4\ln 4-4+1]\\ =5.09\ units

The area of the the region between the curve and horizontal axis is 5.09 units.

5 0
3 years ago
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