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Answer:

#color(blue)((1/3)x+4 > y >= -x+1#

Explanation:

Given:

We are given the System of Inequalities:

#color(brown)(y < (1/3)x+4 # and

#color(brown)(y >= (-x)+1#

#color(green)(Step.1#

#color(brown)(y < (1/3)x+4 # - Image of the graph created using GeoGebra

enter image source here

#color(green)(Step.2#

#color(brown)(y >= (-x)+1# - Image of the graph created using GeoGebra

enter image source here

#color(green)(Step.3#

#color(brown)(y < (1/3)x+4 # and #color(brown)(y >= (-x)+1# - Image of the combined graphs created using GeoGebra

If you observe closely, you will find the solution in a visual form.

The solution to the system of inequalities is the darker shaded region, which is the overlap of the two individual regions.

enter image source here

#color(green)(Step.4#

If you want to view just the solutions, please refer to the image below:

enter image source here

Note:

If the inequality is < or >, graph of the equation has a dotted line.

If the inequality is ≤ or ≥, graph of the equation has a solid line.

This line divides the xy- plane into two regions: a region that satisfies the inequality, and a region that does not.

Answer:

#7# miles

Explanation:

#s = d/t#

speed = distance / time

#d/t = 6 mi//h#

here, #d# is the total distance of the race

and #t# is the time, in hours, that Beth would have taken, if she ran at #6 mi//h#.

#1h = 60m#
she ran #10# minutes sooner than planned, so the time taken for her to run was #t - 1/6#. (where #t# is in hours)

when she ran for #10# minutes less than planned, her speed was #7 mi//h#.

#d/(t-1/6) = 7 mi//h#

this gives the two equations, #d/t = 6# and #d/(t-1/6) = 7#

you can solve both by isolating distance #d:#

#d/t = 6#

#d = 6t#

#d/(t-1/6) = 7#

#d = 7(t - 1/6)#

#d = 6t, d = 7(t - 7/6)#

this means that #6t = 7(t - 1/6)#.

using this, you can solve for time #t:#

#6t = 7(t - 1/6)#

#6t = 7t - 7/6#

#t - 7/6 = 6t - 6t = 0#

#t = 7/6 h# (or #7h# #10m#)

then #t# can be substituted into the speed equation:

#d/t = 6#

#d = 6t#

#d = 6 * 7/6#

#= 7/1#

#= 7#

the distance of the race is #7# miles.

Let,the coordinate of #B# is #(a,b)#

So,if #AB# is perpendicular to #x=2# then,its equation will be #Y=b# where #b# is a constant as slope for the line #x=2# is #90^@#, hence the perpendicular line will have a slope of #0^@#

Now,midpoint of #AB# will be #((-4+a)/2), ((1+b)/2)#

clearly,this point will lie on #x=2#

So, #(-4+a)/2=2#

or, #a=8#

And this will lie as well on #y=b#

so, #(1+b)/2 =b#

or, #b=1#

So,the coordinate is #(8,1)#

Answer:

#y=21/125#

Explanation:

#"the initial statement is "ypropx/z^2#

#"to convert to an equation multiply by k the constant"#
#"of variation"#

#rArry=kxxx/z^2=(kx)/z^2#

#"to find k use the given condition"#

#y=1/6" when "x=20" and "z=6#

#y=(kx)/z^2rArrk=(yz^2)/x=(1/6xx36)/20=3/10#

#"equation is " color(red)(bar(ul(|color(white)(2/2)color(black)(y=(3x)/(10z^2))color(white)(2/2)|)))#

#"when "x=14" and "z=5" then"#

#y=(3xx14)/(10xx25)=21/125#

Answer:

#50#

Explanation:

To simplify this expression, all we need to do, is use #PEMDAS#.

First thing we need to know is the definition of #PEMDAS#:

#P# - Parenthesis

#E# - Exponents

#MD# - Multiply/Divide (Left to Right)

#AS# - Add/Subtract (Left to Right)

This is the order we must do this, from top to bottom. This process is more commonly known as the Order of Operations .

Now that we know the order of the steps we must take to solve this, this is the solution:

#(3+5)^2-2*7#

#=(8)^2-2*7#

#=64-2*7#

#=64-14#

#=50#

Answer:

#x^3+y^3 = (x+y)(x^2-xy+y^2)#

#color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

Explanation:

Given:

#x^3+y^3#

Note that if #y = -x# then this is zero. So we can deduce that #(x+y)# is a factor and separate it out:

#x^3+y^3=(x+y)(x^2-xy+y^2)#

We can calculate the discriminant for the remaining homogeneous quadratic in #x# and #y# just like we would for a quadratic in a single variable:

#x^2-xy+y^2#

is in standard form:

#ax^2+bxy+cy^2#

with #a=1#, #b=-1# and #c=1#

This has discriminant #Delta# given by the formula:

#Delta = b^2-4ac = color(blue)(1)^2-4(color(blue)(1))(color(blue)(1)) = -3#

Since #Delta < 0#, this quadratic has no linear factors with real coefficients.

We can factor it with complex coefficients by completing the square and using #i^2=-1# as follows:

#x^2-xy+y^2 = (x-1/2y)^2+3/4y^2#

#color(white)(x^2-xy+y^2) = (x-1/2y)^2+(sqrt(3)/2 y)^2#

#color(white)(x^2-xy+y^2) = (x-1/2y)^2-(sqrt(3)/2 y i)^2#

#color(white)(x^2-xy+y^2) = ((x-1/2y)-sqrt(3)/2i y)((x-1/2y)+sqrt(3)/2i y)#

#color(white)(x^2-xy+y^2) = (x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

So:

#x^3+y^3 = (x+y)(x^2-xy+y^2)#

#color(white)(x^3+y^3) = (x+y)(x-(1/2+sqrt(3)/2i)y)(x-(1/2-sqrt(3)/2i)y)#

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