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Answer:

The first number is 10 and the second number is 7

Explanation:

Let's define the two numbers we are looking for as #x# and #y#.

The first number added to three times the second number is 31 can be written as:

#x + 3y = 31#

Three time the first less twice the second is 16 can be written as:

#3x - 2y = 16#

Solve the first equation for #x#L

#x + 3y - 3y = 31 - 3y#

#x = 31 - 3y#

Substitute #31 - 3y# for #x# in the second equation and solve for #y#:

#3(31 - 3y) - 2y = 16#

#93 - 9y - 2y = 16#

#93 - 11y = 16#

#93 - 11y + 11y - 16 = 16 + 11y - 16#

#93 - 16 = 11y#

#77 = 11y#

#77/11 = (11y)/11#

#y = 7#

Now substitute #7# for #y# in the solution to the first equation to calculate #x#:

#x = 31 - 3*7#

#x = 31 - 21#

#x = 10#

Answer:

The answer is #x in ] 3,9/2 ]#

Explanation:

Let's rearrange the equation

#6/(x-3)>=4#

#6/(x-3)-4>=0#

#(6-4(x-3))/(x-3)>=0#

#(18-4x)/(x-3)>=0#

#(2(9-2x))/(x-3)>=0#

Let, #f(x)=(2(9-2x))/(x-3)#

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##3##color(white)(aaaa)##9/2##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-3##color(white)(aaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##9-2x##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##-##color(white)(aaa)##+##color(white)(aaaa)##-#

So, #f(x)>=0# when #x in ] 3,9/2 ]#

graph{(18-4x)/(x-3) [-32.47, 32.47, -16.23, 16.26]}

Answer:

# 2x^2-3x-9 = (2x+3)(x-3) #

Explanation:

The rule to factorise any quadratic is to find two numbers such that

#"product" = x^2 " coefficient "xx" constant coefficient"#
#"sum" \ \ \ \ \ \ = x " coefficient"#

So for #2x^2-3x-9# we seek two numbers such that

#"product" = (2)*(-9) = -18#
#"sum" \ \ \ \ \ \ = -3#

So we look at the factors of #-18#. As the product is negative one of the factors must also be negative and the other positive, We compute their sum we get

# {: ("factor1", "factor2", "sum"), (18,-1,17), (9,-2,7), (6,-3,3), (-18,1,-17),(-9,2,-7), (-6,3,-3) :} #

So the factors we seek are #color(blue)(-6)# and #color(green)(3)#

Therefore we can factorise the quadratic as follows:

# \ \ \ \ \ 2x^2-3x-9 = 2x^2 color(blue)(-6)x + color(green)(3)x -9 #
# :. 2x^2-3x-9 = 2x(x-3) + 3(x -3) #
# :. 2x^2-3x-9 = (2x+3)(x-3) #

This approach works for all quadratics (assuming it does factorise) , The middle step in the last section can usually be skipped with practice.

Answer:

Undefined/indeterminate

Explanation:

How to classify #0/0# depends both on context and intent.


First, as this question is in prealgebra, let's look at it without any reference to higher level mathematics. In this context, we should treat it as undefined.

When we perform the operation #x/y=x-:y#, it is exactly equivalent to multiplying #x# by the multiplicative inverse of #y^(-1)# of #y#, where #y^(-1)# is the value satisfying #yxxy^(-1) = 1#.

As there is no value which can be multiplied by #0# to produce #1#, #0# has no multiplicative inverse, and so multiplying by the multiplicative inverse of #0# is not possible. It is neither a finite value nor infinite, but rather is undefined, because there is no way to perform the operation.


As the question also mentions division by #0# resulting in infinity, let's also address how that can occur and what is meant when it does. To do so will require concepts from calculus, however as this is the prealgebra section, this answer will avoid using calculus terminology or assume prior calculus knowledge.

In calculus, we consider what happens when values get very large or very close to specific values. Rather than dividing by #0#, then, we can look at what happens when a dividend gets very close to #0#.

  • #1/1 = 1#
  • #1/(1/2) = 2#
  • #1/(1/10) = 10#
  • #1/(1/10000) = 10000#
    #...#

Notice that as the dividend gets closer to #0#, then quotient gets larger and larger. This pattern continues, and we can make the quotient larger than any given number by making the dividend close enough to #0#, i.e. the quotient becomes infinitely large as the dividend approaches #0#. Sometimes a calculus shorthand for this is writing #1/0 = oo#, even though we are not dividing by #0# nor is #oo# actually a number.

It gets trickier when we have #0# as a divisor, though.

  • #0/1 = 0#
  • #0/(1/2) = 0#
  • #0/(1/10) = 0#
    #...#

It seems if we just plug in #0# as the divisor, then the quotient is #0# every time, and so we should have #0/0=0#. But remember that we aren't actually dividing, but are looking at what happens very close to certain numbers. Consider another pattern:

  • #1/1 = 1#
  • #(1/2)/(1/2) = 1#
  • #(1/10)/(1/10) = 1#
    #...#

Both the divisor and dividend are getting close to #0#, but this time the quotient is #1# each time. Other patterns could lead to the quotient being any given finite number, or even growing infinitely large. In the case when we are considering the divisor and the dividend getting close to #0#, there isn't a general rule for how the quotient will behave, and so we say that #0/0# is an indeterminate form.


In general, different branches of math and physics treat #0/0# differently, depending on how they arrive there and the desired outcome.

Answer:

#sqrt(225x^18y^22) = abs(15x^9y^11)#

Explanation:

Note that:

#225x^18y^22 = 15^2*(x^9)^2*(y^11)^2 = (15x^9y^11)^2#

So #15x^9y^11# is a square root of #225x^18y^22#.

We also find:

#(-15x^9y^11)^2 = 225x^18y^22#

So #-15x^9y^11# is also a square root of #225x^18y^22#.

Note that if #a >= 0# then #sqrt(a)# denotes the non-negative square root.

So if, for example, #x < 0# and #y > 0# then #15x^9y^11 < 0# would be the wrong square root.

We can express that we want the non-negative square root by taking the modulus:

#sqrt(225x^18y^22) = abs(15x^9y^11)#

Answer:

Vertex#->(x,y)=(0,0)#

The intercepts are only at 1 point, the origin

The axis of symmetry is the x-axis ie #y=0#

Explanation:

This is a quadratic in #y# instead of #x#

So instead of form type #y=ax^2+bx+c ....EquationType(1)#
#" we have: " x=ay^2+by+c...EquationType(2)#

Type 1#->nn" or "uu#
Type 2#-> sub" or "sup#

Instead of the horse shoe type shape being up or down it is left or right.

As the coefficient of #y^2# is negative the general shape is #sup#

Consider equationtype(1): The #b# from #bx# moves the graph left or right in that #x_("vertex")=(-1/2)xxb#

Consider equationtype(2); The #b# from #by# moves the graph up or down in that #y_("vertex")=(-1/2)xxb#

However, in this graph #b=0# so the axis of symmetry coincides with the x-axis.

y-intercept is at #x=0#

#=>0=-1/32 y^2" "=>" "y=0# which is only 1 value (single point) so the graph does not #ul("cross")# the y-axis but the axis is tangential to the vertex.

Tony B

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