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3 years ago

31. Larry weighed twice as much as his sister Mary. Together they weighed 210 pounds. How much did each

Mathematics

2 answers:

Pavel [41]3 years ago

6 0

Answer:

Lary weighs 140 Maur weighs 70

Step-by-step explanation:

210÷105 150÷2×3=70

Norma-Jean [14]3 years ago

5 0

Answer:

210 = 3x

/3 /3

70 = x

Mary is 70

Larry is 2x

so the weight for Larry is 140

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Solve using the quadratic formula. h2 + 8h + 8 = 0 Write your answers as integers, proper or improper fractions in simplest form

Marina86 [1]

A=1 b=8 c=8
h = [-8 +-sq root(64 - 4*1*8)] / 2
h = [-8 +-sq root(32)] / 2
h1 = -4 + 2.8284271247 
h1 = -1.1715728753 
h2 = -4 - 2.8284271247
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3 years ago

Give a geometric description of the following system of equations.a. 2x−4y=12 −3x+6y=−15.b. 2x−4y=12 −5x+3y=10.a. 2x−4y=12 −3x+6

Reptile [31]

Answer:

a. No solution, parallel lines.

b. One solution.

Step-by-step explanation:

Given the system of equations:

a. $2x-4y=12$

$-3x+6y=-15$

b. $2x-4y=12$

$-5x+3y=10$

To give a geometric description of the given system of equations.

The geometric description of a system of equations in 2 variables mean the system of equations will represent the number of lines equal to the number of equations in the system given.

i.e.

Number of planes = Number of variables

Number of lines = Number of equations in the system.

Here, we are given 2 variables and 2 equation in each system.

So, they can be represented in the xy-coordinates plane.

And the number of solutions to the system depends on the following condition.

Let the system of equations be:

$A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0$

1. One solution:

There will be one solution to the system of equations, If we have:

$\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}$

2. Infinitely Many Solutions: (Identical lines in the system)

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}= \dfrac{C_1}{C_2}$

3. No Solution:(Parallel lines)

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}$

Now, let us discuss the system of equations one by one:

a. $2x-4y=12$ OR $2x-4y-12=0$

$-3x+6y=-15$ OR $-3x+6y+15=0$

$A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -3, B_2 = 6, C_2= 15$

Here, the ratio:

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2} = -\dfrac{2}{3}\\\dfrac{C_1}{C_2} = -\dfrac{4}{5}$

$\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}$

Therefore, no solution i.e. parallel lines.

b. $2x-4y=12$ OR $2x-4y-12=0$

$-5x+3y=10$ OR $-5x+3y-10=0$

$A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -5, B_2 = 3, C_2 = -10$

$\dfrac{A_1}{A_2}= -\dfrac{2}{5}\\\dfrac{B_1}{B_2} = -\dfrac{4}{3}\\\dfrac{C_1}{C_2} = -\dfrac{6}{5}$

$\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}$

So, one solution.

Kindly refer to the images attached for the graphical representation of the given system of equations.

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Oxana [17]

Answer:

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Step-by-step explanation:

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