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4 years ago

Solving linear systems with graphing

Mathematics

1 answer:

pantera1 [17]4 years ago

6 0

Answer:

All you have to do is graph both the linear functions on a calculator or on paper and the point where both lines intersect is the solution. So that means if it intersects at a point, your solution would be a coordinate.

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Help me please i dont understand

kipiarov [429]

Answer:

C=75.36

A=452.16

Step-by-step explanation:

C=2x3.14xr

$2$ × $\pi$ × $12=75.36$

A=3.14× $r^{2}$

3.14x12x12=452.16

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3 years ago

A proportion is an equation showing the equivalence of two ratios or rates true or false

Lynna [10]

The answer is true- A proportion is an equation showing equivalent of two ratios or rates

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3 years ago

Domain Range The above is ______

IRISSAK [1]

Answer:

Function.

Domain: {-3, 5, 3, -5}

Range: {-6, 2, 1}

Step-by-step explanation:

The domain of the relation shown here is {-3, 5, 3, -5}. Note how each of these elements is linked to ONLY ONE value in the range {-6, 2, 1}. Because of that, we conclude that the table shown represents a function.

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3 years ago

How do you know if you have to divide or multiply in a ratio problem?

romanna [79]

Answer: you looking for ways to calculate ratio problems quickly and accurately? Learn the best methods here, with useful diagrams – and find out what to avoid too. … Ratios can be scaled up or down by multiplying both parts of the ratio … the ratio by the largest number that they can both be divided by

Step-by-step explanation:

<h2> you looking for ways to calculate ratio problems quickly and accurately? Learn the best methods here, with useful diagrams – and find out what to avoid too. ... Ratios can be scaled up or down by multiplying both parts of the ratio ... the ratio by the largest number that they can both be divided by $did this help$ </h2>

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3 years ago

Help i dont understand how to complete this

Luda [366]

Answer: Josh = 19.8 hours, Danny = 10.8 hours

Step-by-step explanation:

$Josh: \dfrac{1}{x+9}\\\\\\Danny: \dfrac{1}{x}\\\\\\Together: \dfrac{1}{7}\\\\\\Josh\quad + \quad Danny\quad =\quad Together\\\dfrac{1}{x+9}\quad +\qquad \dfrac{1}{x}\qquad = \qquad \dfrac{1}{7}\\\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (x+9)(x)(7)}}\\\\7(x) + 7(x+9)=x(x+9)\\7x + 7x + 63 = x^2+9x\\14x+63=x^2+9x\\0=x^2-5x-63\\\\\\\underline{\text{Use the quadratic formula to solve for x: }}x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(1)(-63)}}{2(1)}\\\\\\.\quad =\dfrac{5\pm\sqrt{25+252}}{2}\\\\\\.\quad =\dfrac{5\pm\sqrt{277}}{2}\\\\\\.\quad =\dfrac{5\pm 16.6}{2}\\\\\\x =\dfrac{5+16.6}{2}\qquad x=\dfrac{5-16.6}{2}\\\\\\x=\dfrac{21.6}{2}\qquad \qquad x=\dfrac{-11.6}{2}\\\\\\x=10.8 \qquad \qquad x=-5.8$

Since time cannot be negative, x=-5.8 is an extraneous solution (not valid) so x = 10.8

Josh: x + 9 --> 10.8 + 9 = 19.8

Danny: x --> 10.8

****************************************************************************************

2a) k = 9.45 & 0.55

2b) x = 3/2

2c) x = 2/3 (x = -1 is an extraneous solution so is not valid)

2d) No Solution (x = 1 is an extraneous solution so is not valid)

Here is the work for 2a. Follow this format for b, c, & d

$\dfrac{3}{k^2-8x+12}=\dfrac{k}{k-2}-\dfrac{4}{k-6}\\\\\text{Since the denominator cannot equal zero, then } k \neq2\ and\ k\neq6\\\text{If either of the solutions are 2 or 6, then that solution needs to be crossed out}\\\\\underline{\text{Clear the denominator by multiplying by the LCM: (k-2)(k-6)}}\\\\3 = k(k-6)-4(k-2)\\3=k^2-6k-4k+8\\0=k^2-10k+5\\\\\\\underline{\text{Use the quadratic formula to solve for k}}$

$x=\dfrac{-(-10)\pm \sqrt{(-10)^2-4(1)(5)}}{2(1)}\\\\\\.\quad =\dfrac{10\pm\sqrt{100-20}}{2}\\\\\\.\quad =\dfrac{10\pm\sqrt{80}}{2}\\\\\\.\quad =\dfrac{10\pm8.9}{2}\\\\\\x=\dfrac{10+8.9}{2}\qquad x=\dfrac{10-8.9}{2}\\\\\\x=\dfrac{18.9}{2}\qquad x=\dfrac{1.1}{2}\\\\\\x=9.45\qquad x=0.55\\\\\\\text{Both answers are valid!}$

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3 years ago