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ra1l [238]
2 years ago
5

Witcch point lies on the line of the point-slpoe equation y-2=5(x-6)

Mathematics
2 answers:
Marina86 [1]2 years ago
7 0

Answer:

See below.

Step-by-step explanation:

<u>Simplify</u>

y - 2 = 5(x - 6)

y - 2 = 5x - 30

y = 5x - 28

Points which lie on the line

(6, 2)

(5, -3)

(7, 7)

juin [17]2 years ago
6 0

Answer:

(6, 2 )

Step-by-step explanation:

the equation of a line in point- slope form is

y - b = m(x - a)

where m is the slope and (a, b ) a point on the line

y - 2 = 5(x - 6) ← is in point- slope form

with (a, b ) = (6, 2 )

then (6, 2 ) is a point on the line

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svetoff [14.1K]
The answer is 14.137 cubic inches.

For this case, the radius of whole coconut is 2.5 inches (which is 5 inches / 2).
You have to subtract the thickness of the coconut from this radius for you to get the volume of the space for coconut milk.

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3 years ago
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Nady [450]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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Amanda's parents had a birthday dinner for her at Sushi Yama restaurant. The bill came to $74 before the tip. They left an 18% t
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Answer:

<h2>$87.32</h2>

Step-by-step explanation:

Step one:

given data

we are told that the initial bill is $74

thereafter they gave a tip worth 18% of the bill

let us compute what 18% of $74 will amount to

Step two:

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Hence the total cost is

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3 0
3 years ago
Find the sum of 10 and -10.
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A ball is thrown from an initial height of 1 meter with an initial upward velocity of 15m/s. The ball's height h (in meters) aft
lakkis [162]

\bf \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{h}=1+15t-5t^2}\implies \stackrel{\textit{ball's height}~\hfill }{\stackrel{\downarrow }{6}=1+15t-5t^2}\implies 0=-5+15t-5t^2 \\\\\\ ~~~~~~~~~~~~\textit{using the quadratic formula} \\\\ \stackrel{\stackrel{a}{\downarrow }}{5}t^2\stackrel{\stackrel{b}{\downarrow }}{-15}t\stackrel{\stackrel{c}{\downarrow }}{+5}=0 \qquad \qquad t= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}

\bf t=\cfrac{-(-15)\pm\sqrt{(-15)^2-4(5)(5)}}{2(5)}\implies t=\cfrac{15\pm\sqrt{225-100}}{10} \\\\\\ t=\cfrac{15\pm\sqrt{125}}{10}\implies t=\cfrac{15\pm\sqrt{5^2 \cdot 5}}{10}\implies t=\cfrac{\stackrel{3}{~~\begin{matrix} 15 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\pm ~~\begin{matrix} 5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~\sqrt{5}}{\underset{2}{~~\begin{matrix} 10 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}}

\bf t=\cfrac{3\pm \sqrt{5}}{2}\implies t= \begin{cases} \frac{3+ \sqrt{5}}{2} \approx 2.618\\\\ \frac{3- \sqrt{5}}{2}\approx 0.382 \end{cases}

8 0
2 years ago
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