All categories

3 years ago

Writing during a thurdestorm yesterday. 600 millimeters of rain fell in 30 minutes ?

Mathematics

1 answer:

defon3 years ago

5 0

600/30= 20
20 millimeters×30 minutes= 600 millimeters
Unit rate= 20 millimeters per minute

You might be interested in

If this particular arithmetic sequence, a7=34 and a15=74. What is the value of a21

Ilya [14]

This is an arithmetic sequence, meaning, to get the next term, you simply add some value to the current one, namely the "common difference" "d".

now, we know the 17th term is 34, let's go to the 15term then

34 + d
34 + d + d
34 + d + d + d
34 + d + d + d + d
34 + d + d + d + d + d
34 + d + d + d + d + d + d
34 + d + d + d + d + d + d + d
34 + d + d + d + d + d + d + d + d

now, notice, we got to the 15th term, which is 34 + d + d + d + d + d + d + d + d, or namely 34 + 8d, it just so happen we know is 74, therefore,

$\bf 34+8d=74\implies 8d=40\implies d=\cfrac{40}{8}\implies \boxed{d=5}$

ok, now that we know what the common difference is, let's find the first term in the sequence using the 7th term of 34,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=7\\
d=5\\
a_7=34
\end{cases}
\\\\\\
a_7=a_1+(7-1)5\implies 34=a_1+(7-1)5
\\\\\\
34=a_1+30\implies \boxed{4=a_1}$

and now let's use those 2 found folks, to get the 21st term,

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
a_1=4\\
d=5\\
n=21
\end{cases}
\\\\\\
a_{21}=4+(21-1)5\implies a_{21}=4+100\implies a_{21}=104$

3 0

3 years ago

According to the text, approximately what percentage of college students drop out before obtaining a degree

BabaBlast [244]

Answer:

40% but there is no article so I looked it up

Step-by-step explanation:

4 0

2 years ago

If the width of a pool is 60 feet and the perimeter is 200 feet, how long is the pool?

velikii [3]

Answer:

140

Step-by-step explanation:

200-60 = 140

8 0

3 years ago

Read 2 more answers

The perimeter of a rectangle is 30.8 km and it’s diagonal length is 11 km. Find it’s length and width

blsea [12.9K]

Answer:

Length of the rectangle is 15.0325 km and width is 0.3765 km.

Explanation:

Given:

Perimeter of a rectangle = 30.8 km

Length of diagonal of rectangle = 11 km

To find:

The length and width of rectangle=?

Solution:

Lets assume length of the rectangle = x km

And assume width of the rectangle = y km

Lets first create equation using given perimeter

perimeter of rectangle = 2 ( length + width )

=> 30.8 km = 2 ( x + y )

=> $x + y = \frac{30.8}{2}$

=> y = 15.4 – x ------(1)

As diagonal and two sides of rectangle forms right angle triangle whose hypoteneus is diagonal ,

=> $length^2 + width^2 = diagonal^2$

=> $x^2 + y^2 = 11^2$

=> $x^2 + y^2 = 121$

On substituting value of y from (1) in above equation we get

=> $x^2 + (15.4-x)^2 = 121$

=> $x^2 + (15.4)^2 + x^2 – 2 x 15.4 \times x = 121$

=> $2x^2-30.8x + 237.16 -121 = 0$

=> $2x^2-30.8x + 116.16 = 0$

Solving above quadratic equation using quadratic formula

General form of quadratic equation is

$ax^2 +bx +c = 0$

And quadratic formula for getting roots of quadratic equation is

$x= \frac{ -b\pm\sqrt{(b^2-4ac)}}{2a}$

As equation is $2x^2-30.8x + 116.16 = 0$ , in our case

a = 2 , b = -30.8 and c = 116.16

Calculating roots of the equation we get

$x=\frac{ -(-30.8)\pm\sqrt{(-30.8)^2-4(2)( 11)} } {(2\times2)}$

$x=\frac{30.8\pm\sqrt{(948.64-88)}}{4}$

$x=\frac{30.8\pm\sqrt{860.64}}{4}$

$x=\frac{(30.8\pm29.33)}4$

$x=\frac{(30.8+29.33)}{4}$

$x=\frac{(30.8-29.33)}{4}$

=> x = 15.0325 or x = 0.3675

As generally length is longer one ,

So x = 1.0325

From equation (1) y = 15.4 – x = 0.3765

Hence length of the rectangle is 15.0325 km and width is 0.3765 km.

6 0

3 years ago

Enter the values of h and I so that y=x2+6x+10 is in vertex form<br> Y=(x+__)^2+___

DiKsa [7]

$Use\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\y=x^2+6x+10=x^2+2(x)(3)+10=\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2+10\\\\=(x+3)^2-9+10=(x+3)^2+1\\\\Answer:\ \boxed{y=(x+3)^2+1}$

3 0

3 years ago