1/7 de catorce es dos, ahora Miriam tiene doce dólares restantes. 5/8 de 12 es 7.5, entonces todos juntos, Miriam gastó nueve dólares.

¡Espero que esto ayude! ¡Si necesitas más ayuda házmelo saber! :-)

6 0

3 years ago

Suppose m is in the line given by the equation 6x-3y=7, and suppose n is the line perpendicular to m and passing Nd through the

stealth61 [152]

Answer:

The x co-ordinate o intersection of line k and n is $\frac{8}{11}$

Step-by-step explanation:

Given as :

The equation of line m is 6 x - 3 y = 7

So, in the standard form , line equation is

y = a x + c , where a is the slope

So, 6 x - 3 y = 7 can be written as

3 y = 6 x - 7

or, y = 2 x - $\frac{7}{3}$ ........1

So, slope of this line = a = 2

Now, The line n is perpendicular to line m and passing through line ( 6 , 2 )

So, Slope of line n = b

For , perpendicular lines , products of slope = - 1

Or, a × b = -1

∴ b = - $\frac{1}{a}$

I.e b = - $\frac{1}{2}$

So,equation of line n with slope b and passing through line ( 6 , 2 ) is

y - $y_1$ = b ( x - $x_1$ )

or, y - 2 = - $\frac{1}{2}$ ( x - 6 )

or, 2 × ( y - 2 ) = - 1 ( x - 6 )

or, 2 y - 4 = - x + 6

or, x + 2 y -10 = 0 ........2

Again, equation of line k with slope 5 and y intercept = 1

For y intercept , x coordinate = 0

y = c x + c

or, 1 = c× ( 0 ) + c

Or, c = 1

Or, equation of line k is

y = 5 x + 1 ..........3

Now intersection of line k and n is

put the value of y from eq 3 into eq 2

I.e x + 2 × ( 5 x + 1 )-10 = 0

Or, x + 10 x + 2 - 10 = 0

or, 11 x - 8 = 0

or 11 x = 8

∴ x = $\frac{8}{11}$

Hence The x co-ordinate o intersection of line k and n is $\frac{8}{11}$ Answer

4 0

3 years ago