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Tema [17]
2 years ago
5

Samuel and Hayden solved the system of equations –6x – 6y = –6 and 7x + 7y = 7. Samuel says there are infinitely many solutions,

but Hayden says the solution is (1,1). Explain who is correct.
Mathematics
2 answers:
yawa3891 [41]2 years ago
7 0

Answer:

Samuel is correct i.e  there are infinitely many solutions

Step-by-step explanation:

Given that Samuel and Hayden solved the system of equations –6x – 6y = –6 and 7x + 7y = 7. we have to find that whether the system of equations has infinitely many solutions or not.

A system of linear equations has infinite solutions if the graphs are the exact same line i.e the the equations are equivalent.

The first equation:  –6x – 6y = –6 ⇒ x+y=1 ⇒ y=-x+1

∴ the slope of its line is -1 and the y-intercept is 1

The second equation:  7x + 7y = 7 ⇒  x+y=1 ⇒ y=-x+1

∴ the slope of its line is -1 and the y-intercept is 1.

Here, we get the equation which has the same slope and y-intercept as that of the first equation.

In other words, the two equations are represented by the same line. This implies that the lines intersect infinitely many times, or that the system has infinitely many solutions.

Hence, Samuel is correct.

emmainna [20.7K]2 years ago
6 0

Answer:

Step-by-step explanation:

Sample Response: The first step in solving this system is to multiply the first equation by 7 and the second equation by 6. When you add the equations, both variables are eliminated and you are left with 0 = 0, which is a true statement. That means there are infinitely many solutions and Samuel is correct.

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Answer:

17million

Step-by-step explanation:

5+(2m)+m+2+m=54

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This table shows the area in inches of four rectangular scraps of fabric.
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Answer:

<h2>82 in²</h2>

Step-by-step explanation:

The median of the four numbers a, b, c and d. Where a ≤ b ≤ c ≤ d:

\dfrac{b+c}{2}

We have

84in², 72in², 80in², 84in² → 72in², 80in², 84in², 84in²

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To increase an amount by 7% what single multiplier would you use?
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What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootIndex 3 StartRoot 12 x Supers
KengaRu [80]

Answer:

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }

Step-by-step explanation:

Given

\sqrt[3]{16x^7} * \sqrt[3]{12x^9}

Required

Find the products

From laws of indices;

\sqrt[m]{a} * \sqrt[m]{b} = \sqrt[m]{a*b}

So;

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16x^7 * 12x^9}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16* x^7 * 12 * x^9}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^7  * x^9}

From laws of indices

a^m * a^n = a^(m+n); So,

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{7+9}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{16}}

Expand 16 * 12

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4*4*4*3* x^{16}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4^3 *3* x^{16}}

From laws of imdices

a^{\frac{1}{m}} = \sqrt[m]{a}

So;

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^3 *3* x^{16}})^{\frac{1}{3}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^{3*{\frac{1}{3}}} *3^{{\frac{1}{3}}}* x^{16*{\frac{1}{3}}}})

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{\frac{16}{3}}}

Divide 16 by 3 (Write as ,mixed number)

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5\frac{1}{3}}}

Split mixed numbers

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5+\frac{1}{3}}}

Apply law of indices

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5}*{x ^\frac{1}{3}}}

Reorder

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 * x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}

Apply law of indices

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3} *\sqrt[3]{x} }

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3*x} }

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3x} }

\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }

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2 years ago
Anton will be constructing a segment bisector with a compass and straightedge, while Maxim will be constructing an angle bisecto
Genrish500 [490]

The similarities are;

  • Compass and a straight edge required for both construction
  • Both construction includes a line drawn from the intersection of arcs to bisect a segment or an angle
  • The bases for the construction of both bisector are the ends of segment and the angle to be bisected
  • The width of the compass when drawing intersecting arcs, is more than half the width of the segment or angle being bisected

The differences are;

  • Two points of intersection of arcs are used in the segment bisector while only one is requited in an angle bisector
  • The bisecting line crosses the segment in a segment bisector, while it stops at the vertex of the angle being bisected in an angle bisector

The sources of the above equations are as follows;

The steps to construct a segment bisector are;

  • Place the needle of the compass at one of the ends of the line segment to be bisected
  • Widen the compass so as to extend more than half of the length of the segment to be bisected
  • Draw two arcs, one above, and the other below the line
  • Place the compass needle at the other end and with the same compass width draw arcs that intersects with the arcs drawn in the above step
  • Draw a line segment by placing the ruler on the points of intersection of the arcs above and below the line

The steps to construct an angle bisector are;

  • With the compass needle at the vertex, open the pencil end such that arcs can be drawn on the rays (lines) forming the angle
  • Draw an arc on both lines forming the angle
  • Place the compass needle at one of the intersection points and draw an arc in between the lines forming the angle
  • Repeat the above step with the same compass width from the other intersection point with the rays forming the angle
  • Join the point of intersection of the two arcs to the vertex of the angle to bisect the angle

Therefore, we have;

The similarities are;

  • A compass and a straight edge can be used for both construction
  • A straight line is drawn from the point of intersection of arcs to bisect the segment or the angle
  • The arcs are drawn from the ends of the segment or angle to be bisected
  • The width of the compass is more than half the width of the line or angle when drawing the arcs

The differences are;

  • In a segment bisector, the intersection point is above and below the line, while in an angle bisector only one pair of arcs are drawn to intersect above the line
  • The bisecting line passes through the segment being bisected, while the line stops at the vertex in an angle bisector

Learn more about the construction of segment and angle bisectors here;

brainly.com/question/17335869

brainly.com/question/12028523

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