If the area of parallelogram ABCD = 246mm squared and h= 20.5 mm, what is the length of the base of triangle ABD?

Nate is creating a PSA about building a "green” roof. Which source is most reliable for his research? a blog post written by a c

galina1969 [7]

Answer:

Nate wants to use a .org (or .gov) site about ecological roofs, a book featuring photos of different styles of room because

1. It specifically talks about green roof's

2. Unlike the blog post, the .org/.gov is not attempting to sell a product that may change how the article is written.

7 0

2 years ago

A King in ancient times agreed to reward the inventor of chess with one grain of wheat on the first of the 64 squares of a chess

statuscvo [17]

Let's start by visualising this concept.

Number of grains on square:
1 2 4 8 16 ...

We can see that it starts to form a geometric sequence, with the common ratio being 2.

For the first question, we simply want the fifteenth term, so we just use the nth term geometric form:
$T_n = ar^{n - 1}$
$T_{15} = 2^{14} = 16384$

Thus, there are 16, 384 grains on the fifteenth square.

The second question begs the same process, only this time, it's a summation. Using our sum to n terms of geometric sequence, we get:
$S_n = \frac{a(r^{n} - 1)}{r - 1}$
$S_{15} = \frac{2^{15} - 1}{2 - 1}$
$S_{15} = 2^{15} - 1 = 32767$

Thus, there are 32, 767 total grains on the first 15 squares, and you should be able to work the rest from here.

6 0

2 years ago

Read 2 more answers

Jar A contains four liters of a solution that is 45% acid. Jar B contains five liters of a solution that is 48% acid. Jar C cont

castortr0y [4]

Answer:

k= 80%

Step-by-step explanation:

Jar A contains 4*0.45 L acid, and 4 L of a solution of acid.

Jar B contains 5*0.48 L acid., and 5 L of a solution of acid.

Jar C contains 1*k/100 = k/100 acid, and 1 L of a solution.

50% = 0.5

For jar A.

(2/3)*k/100 L acid is added to jar A.

Now jar A contains 4*0.45 L + (2/3)*k/100 L acid, and it has (4+2/3)L of a solution.

L solute/L solution = 0.5

[4*0.45 L + (2/3)*k/100 L]/(4+2/3)L = 0.5

[1.8 + (2k/300)]/[(12+2)/3] = 0.5

[1.8 + (2k/300)]/[14/3] = 0.5

[1.8 + (2k/300)]= 0.5*(14/3)

(2k/300) = 0.5*(14/3) - 1.8

2k = (0.5*(14/3) - 1.8)*300

k = (0.5*(14/3) - 1.8)*300/2 =80

k= 80%

We also can find k using jar B.

(1/3)k/100 L acid is added to jar B.

Now jar B contains 5*0.48 L+ (1/3)k/100 L acid, and it has (5+1/3) L of a solution.

L solute/L solution = 0.5

[5*0.48 L+ (1/3)k/100 L ]/(5+1/3)L= 0.5

[5*0.48 + (1/3)k/100 ]/(5+1/3)= 0.5

This equation also gives k=80%

Check.

We can check at least for jar A.

Jar A has 4L solution and 4*0.45=1.8 L acid.

2/3 L of the solution from jar C was added, and now we have 4 2/3 L of solution.

(2/3)* 80%= (2/3)*0.8 acid was added from jar C.

Now we have [1.8 +(2/3)*0.8] L acid in jar A.

L solute/L solution = [1.8 +(2/3)*0.8] L /(4 2/3) L = 0.5 or 50% as it is given that jar A has 50% at the end.

7 0

2 years ago

Rajeev has \$175$175dollar sign, 175 that he earned from his summer job. He puts the money in an account that yields 4\%4%4, per

Reptile [31]

It will take 4 yrs for Rajeev to reach 200$

8 0

2 years ago

A scientist mixes x liters of water into a container that has 10 liters of a mixture that is 12% salt and 88% water. The functio

dsp73

Answer:

110 liters of water

Step-by-step explanation:

Let

x ----> liters of water

f(x) ----> the percent of the sat in the new mixture

we have

$f(x)=\frac{1.2}{x+10}$

For $f(x)=1\%=1/100=0.01$

substitute in the equation

$0.01=\frac{1.2}{x+10}$

solve for x

$x+10=\frac{1.2}{0.01}$

$x+10=120\\x=120-10\\x=110\ L$

5 0

2 years ago