What is the value of x?

Mathematics

1 answer:

Lelechka [254]3 years ago

5 0

X = 35

use the formula a^2 + b^2 = c^2 in order to find the hypotenuse, which is x.

21^2 + 28^2 = c^2
441 + 784 = c^2
1225 = c^2
35 = c

x = 35

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The volume

Sedaia [141]

$\bf \begin{array}{cccccclllll}
\textit{something}&&\textit{varies directly to}&&\textit{something else}\\ \quad \\
\textit{something}&=&{{ \textit{some value}}}&\cdot &\textit{something else}\\ \quad \\
y&=&{{ k}}&\cdot&x
&& y={{ k }}x
\end{array}\\ \quad \\
$

and also

$\bf \begin{array}{llllll}
\textit{something}&&\textit{varies inversely to}&\textit{something else}\\ \quad \\
\textit{something}&=&\cfrac{{{\textit{some value}}}}{}&\cfrac{}{\textit{something else}}\\ \quad \\
y&=&\cfrac{{{\textit{k}}}}{}&\cfrac{}{x}
&&y=\cfrac{{{ k}}}{x}
\end{array}
$

now, we know that V varies directly to T and inversely to P simultaneously
thus $\bf V=T\cdot \cfrac{k}{P}$

so $\bf V=T\cdot \cfrac{k}{P}\qquad 
\begin{cases}
V=42\\
T=84\\
P=8
\end{cases}\implies 42=\cfrac{84k}{8}\implies 4=k
\\\\\\
V=\cfrac{4T}{P}\qquad now\quad 
\begin{cases}
V=74\\
P=10
\end{cases}\implies 74=\cfrac{4T}{10}\implies 185=T$

7 0

3 years ago

Two competitive neighbours build rectangular pools that cover the same area but are different shapes. Pool A has a width of (x +

GenaCL600 [577]

Answer:

a)Dimensions of pool A are length = 6.667m and width = 3.667 m and dimension of pool B are length = 7.333m and width = 3.333m.

b) Area of pool A is equal to area of pool B equal to 24.44 meters.

Solution:

Let’s first calculate area of pool A .

Given that width of the pool A = (x+3)

Length of the pool A is 3 meter longer than its width.

So length of pool A = (x+3) + 3 =(x + 6)

Area of rectangle = length x width

So area of pool A =(x+6) (x+3) ------(1)

Let’s calculate area of pool B

Given that length of pool B is double of width of pool A.

So length of pool B = 2(x+3) =(2x + 6) m

Width of pool B is 4 meter shorter than its length,