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3 years ago

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100 points. solve the equation and check the solution

1 answer:

jeka943 years ago

3 0

$a)\\a+2.4=7.8\\a=7.8-2.4\\a=5.4\\\\b)\\b-\dfrac{1}{4} =3\dfrac{1}{2} \\b-0.25=3.5\\b=3.5+0.25\\b=3.75=3\dfrac{3}{4} \\\\c)\\\dfrac{1}{3} s=\dfrac{1}{5} \\s=\dfrac{1}{5}\div\dfrac{1}{3}\\s=\dfrac{1}{5}\times3\\s=\dfrac{3}{5} \\\\d)\\3x=27\\x=27\div3\\x=9\\\\e)\\\dfrac{t}{4} =16|\times4\\t=64\\\\f)\\\dfrac{w}{7} =0.3|\times7\\w=2.1$

Check:

$a)\\5.4+2.4=7.8\\correct\\\\b)\\3\dfrac{3}{4}-\dfrac{1}{4} -3\dfrac{2}{4}=3\dfrac{1}{2}\\correct\\\\c)\\\dfrac{1}{3} \times \dfrac{3}5=\dfrac1{5}\\correct\\\\d)\\3\times9=27\\correct\\\\e)\\\dfrac{64}{4} =16\\corect\\\\f)\\\dfrac{2.1}{7} =0.3\\correct$

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fenix001 [56]

Answer:

$y = -\frac{3}{2}x+6$

Step-by-step explanation:

Given:

The given equation of the line $y = \frac{2}{3} x + 5$ that passes through the point (4, 0).

Part A.

The equation of the line.

$y=mx+c$ -----------(1)

Where:

m = Slope of the line

c = y-intercept

The given equation of the line.

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Comparing the given equation with equation 1.

The slope of the line is $m=\frac{2}{3}$ and y-intercept $c = 5$

We know that the slope of the perpendicular line is $(-\frac{1}{m})$ .

So the slope of the perpendicular line is $(m=-\frac{1}{\frac{2}{3}}=-\frac{3}{2})$ .

Using point slope formula we write the equation of the perpendicular line that passes through the point (4, 0).

$y-y_{1} = m(x-x_{1}$

Now we substitute the slope of the perpendicular line $m=-\frac{3}{2}$ and $y_{1} = 0, x_{1}=4$ from point (4, 0) in above equation.

$y-0 = -\frac{3}{2}(x-4)$

$y = -\frac{3}{2}x-(-\frac{3}{2}\times 4)$

$y = -\frac{3}{2}x-(-3\times 2)$

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Therefore the perpendicular line equation is $y = -\frac{3}{2}x+6$ .

Part B.

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3. The equation of the line $y = -\frac{3}{2}x+6$ is perpendicular to the equation of line $y = \frac{2}{3} x + 5$ .

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