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ololo11 [35]
2 years ago
9

Write the slope-intercept form of the line that passes through the point (1, 0) and is parallel to x-y=7.

Mathematics
1 answer:
Mnenie [13.5K]2 years ago
8 0

We can rewrite the equation of the given line to be in slope-intercept form as follows.

x-y=7\\x=y+7\\y=x-7

Thus, the slope of the given line is 1.

Since parallel lines have equivalent slopes, the slope of the line we want to find is also 1.

Substituting into point-slope form,

y-0=1(x-1)\\\\\boxed{y=x-1}

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labwork [276]

Hello!

The figure is made up of a cone and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π . Enter your answer in the box. cm³

A 12 cm cone with a dome on top of it that has an 8 cm diameter

Data: (Cone)

h (height) = 12 cm

r (radius) = 4 cm (The diameter is 8 being twice the radius)

Adopting: \pi \approx 3.14

V (volume) = ?

Solving: (Cone volume)

V = \dfrac{ \pi *r^2*h}{3}

V = \dfrac{ 3.14 *4^2*\diagup\!\!\!\!\!12^4}{\diagup\!\!\!\!3}

V = 3.14*16*4

\boxed{V = 200.96\:cm^3}

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 4 cm

Adopting: \pi \approx 3.14

If: We know that the volume of a sphere is V = 4* \pi * \dfrac{r^3}{3} , but we have a hemisphere, so the formula will be half the volume of the hemisphere V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}

Formula: (Volume of the hemisphere)

V = 2* \pi * \dfrac{r^3}{3}

Solving:

V = 2* \pi * \dfrac{r^3}{3}

V = 2*3.14 * \dfrac{4^3}{3}

V = 2*3.14 * \dfrac{64}{3}

V = \dfrac{401.92}{3}

\boxed{ V_{hemisphere} \approx 133.97\:cm^3}

Now, to find the total volume of the figure, add the values: (cone volume + hemisphere volume)

Volume of the figure = cone volume + hemisphere volume

Volume of the figure = 200.96 cm³ + 133.97 cm³

\boxed{\boxed{\boxed{Volume\:of\:the\:figure = 334.93\:cm^3}}}\end{array}}\qquad\quad\checkmark

_______________________

I Hope this helps, greetings ... Dexteright02! =)

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