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murzikaleks [220]

3 years ago

11

if a farmer has one square mile of land if he divides his land into square fields that are 1/4 mile long and 1/4 mile wide how m

any fields does he have

1 answer:

dem82 [27]3 years ago

6 0

Four wide and four deep
4 ×4=16
sixteen

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frosja888 [35]

I don’t know if this is right but the way I did it I got 7.5

3 0

2 years ago

Read 2 more answers

Find the median of the following set of numbers:<br> 3,12,4,6,8,5,4

natulia [17]

Answer:

el 5

Step-by-step explanation:

La mediana es el número intermedio de un grupo de números; es decir, la mitad de los números

Primero se acomodan de menor a mayor y luego ubicaras el número que queda en medio

6 0

2 years ago

Simplify each expression. Assume that all variables are positive.

kozerog [31]

Q1. The answer is $\frac{8x^{3}y^{6} }{27}$

$( \frac{16 x^{5} y^{10}}{81x y^{2} } )^{ \frac{3}{4} }= ( \frac{16}{81}* \frac{ x^{5} }{x}* \frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} } \\ \\ 
 \frac{ x^{a} }{ x^{b} }= x^{a-b} \\ \\ 
( \frac{16}{81}* \frac{ x^{5} }{x}*\frac{ y^{10} }{y^{2}} )^{ \frac{3}{4} }}=( \frac{16}{81 }* x^{5-1}* y^{10-2})^{ \frac{3}{4} }=( \frac{16}{81 }* x^{4}* y^{8})^{ \frac{3}{4} }= \\ \\ = (\frac{16}{18} )^{ \frac{3}{4} }*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=$
$\frac{(16)^{ \frac{3}{4} }}{(18)^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} }=\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } \\ \\ 
 (x^{a} )^{b} = x^{a*b} \\ \\ 
\frac{( 2^{4} )^{ \frac{3}{4} }}{( 3^{4} )^{ \frac{3}{4} }}*(x^{4})^{ \frac{3}{4} }*(y^{8})^{ \frac{3}{4} } = \frac{ 2^{4* \frac{3}{4} } }{ 3^{4* \frac{3}{4} } } * x^{4* \frac{3}{4} } * y^{8*\frac{3}{4}} = \frac{ 2^{3} }{ 3^{3} } * x^{3} *y^{6} =$
$= \frac{8x^{3}y^{6} }{27}$

Q2. The answer is 1/16

$(-64) ^ \frac{-2}{3} =(-1* 2^{6} ) ^ \frac{-2}{3}=(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} \\\\x^{-a} = \frac{1}{ x^{a} } \\\\(-1)^ \frac{-2}{3} *(2^{6} ) ^ \frac{-2}{3} = \frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{(2^{6})^ \frac{2}{3}} \\ \\ (x^{a} )^{b}=x^{a*b} \\\\x^{ \frac{a}{b} = \sqrt[b]{ x^{a} } } \\ \\ 
$
$\frac{1}{(-1)^ \frac{2}{3}} *\frac{1}{2^{6*\frac{2}{3}}} = \frac{1}{ \sqrt[3]{(-1)^{2} } } * \frac{1}{ 2^{4} } = \frac{1}{ \sqrt[3]{1} } * \frac{1}{16} = \frac{1}{1} * \frac{1}{16}= \frac{1}{16}$

Q3. The answer is $a^{ \frac{7}{6} }$

$a^{ \frac{2}{3} } * a^{ \frac{1}{2} } \\ \\ 
 x^{a}* x^{b} =x^{a+b} \\ \\ 
a^{ \frac{2}{3} } * a^{ \frac{1}{2} }= a^{ \frac{2}{3} + \frac{1}{2} } =a^{ \frac{2*2}{3*2} + \frac{1*3}{2*3} }=a^{ \frac{4}{6} + \frac{3}{6} }=a^{ \frac{4+3}{6} }=a^{ \frac{7}{6} }$

7 0

2 years ago

Determine the function which corresponds to the given graph. (3 points)

damaskus [11]

Answer:

The center of the circle is c=50 and radius of the circle is $r=\sqrt{3}$

Step-by-step explanation:

Given circle equation is

$x^2-4x+y^2+14y=-50\hfill(1)$

Equation (1) can be written as $x^2-4x+y^2+14y+50=0\hfill(2)$

we know that the equation of the circle is of the form

$x^2+y^2+2gx+2fy+c=0\hfill(3)$

with centre (-g,-f) and radius= $\sqrt{g^2+f^2-c}$

when, g,f and c are constants

Now comparing the (2) and (3) equations we get 2g=-4

$g=\frac{-4}{2}$

$g=-2$

$2fy=14$

$f=\frac{14}{2}$

$f=7$

and $c=50$

Now to find the centre and radius of the given circle equation, substituting the values of g,f,c in the formulae of centre and radius

centre=(-g,-f)

=(-(-2),-7)

centre=(2,7)

Radius= $\sqrt{g^2+f^2-c}$

= $\sqrt{(-2)^2+(7)^2-50}$

= $\sqrt{4+49-50}$

= $53-50$

Radius= $\srqt{3}$

The center of the circle is c=50 and the radius of the circle equation $r=\sqrt{3}$

6 0

2 years ago

Determine the equation of a line that is parallel to Y +7 equals 2X goes through the point parentheses -3, one parentheses write

jekas [21]

The Answer Is:

y-7=2x

6 0

2 years ago