All categories

2 years ago

For which value of a are the graphs of –4 = 3x + 6y and ax – 8y = 12 parallel?

Mathematics

2 answers:

zloy xaker [14]2 years ago

6 0

Answer:

The value of a is -4.

Step-by-step explanation:

Since, two lines $a_1 x+b_1 y=c_1$ and $a_2 x+b_2 y=c_2$ are called parallel lines,

If $\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}$

If 3x + 6y = 4 and ax-8y = 12 are parallel lines,

Then, by the above property,

$\frac{3}{a}=\frac{6}{-8}\neq \frac{4}{12}$

$\implies \frac{3}{a}=\frac{6}{-8}$

By cross multiplication,

$-24 = 6a$

$\implies -4=a$ ( Dividing both terms by 6}

Hence, the value of a is -4 for which 4 = 3x + 6y and ax – 8y = 12 are parallel.

Dimas [21]2 years ago

4 0

Answer:

The answer is A) a=-4

Step-by-step explanation:

You might be interested in

Which procedure justifies whether Negative 3 x (5 minus 4) + 3 (x minus 6) is equivalent to Negative 12 x minus 6? The expressio

Nady [450]

Answer:

The expressions are not equivalent because -3(2)(5-4)+3(2-6)=-18 and -12(2)-6=-30

Step-by-step explanation:

Two expressions are said to be equal if after a number is substituted to the expression, they produce the same result (that is they have the same value).

To determine whether -3x(5-4)+3(x-6) is equivalent to -12x-6, we have to substitute the same number to the expression and see if it produces the same result.

Substituting 2 to the first expression gives:

-3×2(5-4) + 3(2 - 6) = -6 - 12 = -18

Substituting 2 to the second expression gives:

-12(2) - 6 = -24 - 6 = -30

Since -3×2(5-4) + 3(2 - 6) = -6 - 12 = -18 and -12(2) - 6 = -24 - 6 = -30, the expressions are not equivalent because they do not produce the same result.

7 0

3 years ago

Read 2 more answers

The cost of a pair of jeans is 4 times the cost of a book

galben [10]

Answer:

what's the cost of the books then

Step-by-step explanation:

3 0

3 years ago

The segment show below form a triangle<br> True or false?

seropon [69]

Answer:

true

Step-by-step explanation:

because it would be

5 0

3 years ago

If D is midpoint of side BC of triangle ABC. P and Q are points lying respectively on side AB and AC such that DP is parallel to

LUCKY_DIMON [66]

See proof below

Step-by-step explanation:

Assume triangle ABC to have vertices at;

A(2,-1), B(2,-7) and C(6,-7)

D is midpoint of BC, thus D is at (4,-7)

The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA

Plot the points on a graph tool and join the points to view the sketch.

To prove area of triangle CPQ is 1/4 area of ABC will be;

Find area ABC and CPQ then compare the areas.

Apply the distance formula to find the length of sides of the triangles then find the areas.

The distance formula is;

$d=\sqrt{x_2-x_1)^2+(y_2-y_1)^2}$

Length of side AB from the sketch is;

$AB=\sqrt{(2-2)^2+(-7--1)^2} =\sqrt{-6^2} =\sqrt{36} =6units$

Length of side BC will be;

$BC=\sqrt{(-7--7)^2+(6-2)^2} =\sqrt{4^2} =\sqrt{16} =4units$

Thus area of triangle ABC will be;

1/2 *base length*height ------because it is a right-triangle

1/2*4*6=12 square units

Find the lengths of all sides of triangle CPQ

Length of side PQ is half that of side BC thus PQ=2 units

Length of side PC is;

$PC=\sqrt{(6-2)^2+(-7--4)^2} =\sqrt{4^2+-3^2} =\sqrt{16+9} =\sqrt{25} =5units$

Length of side QC will be;

$QC=\sqrt{(6-4)^2+(-7--4)^2} =\sqrt{2^2+-3^3} =\sqrt{4+9} =\sqrt{13}$

QC= √13 = 3.6 units

Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;

A=√s(s-a)(s-b)(s-c) where;

A=area of the triangle

s= half the perimeter of the triangle

a=side PQ = 2 units

b=side PC = 5 units

c= side QC = 3.6 units

Finding the perimeter of triangle CPQ will be;

P=sum of all sides

P=2+5+3.6 =10.6 units

s=10.6/2 = 5.3

Area of the triangle CPQ will be;

$A=\sqrt{5.3(5.3-2)(5.3-5)(5.3-3.6)} \\A=\sqrt{5.3(3.3)(0.3)(1.7)} \\A=\sqrt{8.9} =2.98$

A=3.0 (1 decimal place)

Compare the areas;

Area of triangle ABC=12 square units

Area of triangle CPQ = 3 square units

Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4

Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3

Learn More

Area of a triangle ;brainly.com/question/14869984

The Heron's formula : brainly.com/question/10713495

Keywords: midpoint, triangle, sides, parallel, prove , area, equal

#LearnwithBrainly

6 0

2 years ago

What is the equation of the line that passes through (0,-2) and (4,6)

balandron [24]

It the 23,56(375);&:&

5 0

2 years ago